Spaceflight Mechanics - Quiz(MCQ)

A)

Helium

B)

Nitrogen

C)

Hydrogen

D)

Carbon dioxide

Correct Answer : Helium

Explanation : Helium is the most common gas in the atmosphere of our solar system. It comprises of a significant portion in the atmosphere of Sun, Mercury, Moon, Jupiter, Saturn, Uranus and Neptune. The second common gas in the solar system is Hydrogen gas, followed by Nitrogen and Carbon dioxide gas.

A)

Oxygen

B)

Helium

C)

Nitrogen

D)

Hydrogen

Correct Answer : Hydrogen

Explanation : Stellar atmosphere refers to the atmosphere of the stars. It is mainly comprised of **Hydrogen**, which is about **70%** of the total atmosphere by mass. The second major component of stellar atmosphere is Helium, which is about **25%** of the total atmosphere by mass.

A)

It is a liquid present on some planets

B)

It is a hard shell of gases around planets

C)

It is a temporary phenomenon due to solar radiations

D)

It is a thin blanket of gases surrounding a planet or an object

Correct Answer : It is a thin blanket of gases surrounding a planet or an object

Explanation : The term ‘Atmosphere’ can be defined as a thin blanket of gases surrounding a planet or an object in outer space. As it is made up of gases, it cannot be a hard shell. It is not caused due to solar radiations.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : False

Explanation : Atmosphere is not limited only to planets. Stars also have an outer atmosphere, generally referred to as Stellar Atmosphere. Celestial objects like comets also possess their own atmosphere.

A)

Weather effects on thermodynamic properties only

B)

Horizontal variations of thermodynamic properties only

C)

Vertical variations of thermodynamic properties only

D)

Horizontal & Vertical variations of thermodynamic properties

Correct Answer : Vertical variations of thermodynamic properties only

Explanation : Horizontal effects, caused by weather & planetary rotation, have negligible variation in thermodynamic properties as compared to the Vertical variations. The thermodynamic properties of the atmosphere vary greatly with the altitude from the surface of the planet. More the change in the altitude, more is the variation in properties. So, only vertical variations are considered in all planetary atmospheric models.

A)

It is a local phenomenon which occurs only at low altitudes

B)

It is regarded to be in a thermal equilibrium, with negligible external influences

C)

It has very little impact on mean thermodynamic properties of the atmospheres

D)

It is under constant perturbation due to winds and two-phase non-equilibrium of water vapour & carbon dioxide vapour on Mars and Earth respectively

Correct Answer : It is under constant perturbation due to winds and two-phase non-equilibrium of water vapour & carbon dioxide vapour on Mars and Earth respectively

Explanation : Weather is a local phenomenon which occurs only at low altitudes of the atmosphere. At low altitudes, the atmosphere is assumed to be in thermal equilibrium & is not influenced by external factors like electromagnetism or chemical reactions. But it is constantly disturbed by horizontal winds and two-phase non-equilibrium due to water vapour on Earth & carbon dioxide vapour on Mars.

A)

Oxygen

B)

Nitrogen

C)

Sulphur oxide

D)

Carbon dioxide

Correct Answer : Nitrogen

Explanation : Carbon dioxide comprises of 96% of the atmosphere on Venus. It also contains small traces of nitrogen & sulphur oxides. So, the most abundant gas in the atmosphere of Venus is Carbon dioxide gas. This makes the atmosphere more denser & hotter as compared to Earth.

A)

Core of the planet

B)

Planet's Gravity

C)

Planetary magnetic ?eld

D)

Chemical composition of atmosphere

Correct Answer : Core of the planet

Explanation : Out of the options, Core of the planet is incorrect. The planetary atmosphere varies with difference in gravity, chemical composition, planetary magnetic field, solar radiation, planetary rotation etc. The core of the planet does not play a major role as a variable in atmosphere.

A)

Mercury

B)

Earth

C)

Jupiter

D)

Saturn

Correct Answer : Mercury

Explanation : The atmosphere of Mercury has 42% of oxygen, which makes it the highest percentage in the solar system. Earth’s atmosphere contains 21% of oxygen whereas Jupiter & Saturn contains negligible amounts of oxygen in their atmosphere.

A)

21% and 78%

B)

25% and 70%

C)

70% and 25%

D)

78% and 21%

Correct Answer : 78% and 21%

Explaination : Earth’s atmosphere contains about 78% Nitrogen and 21% Oxygen. It also has other gases such as 0.97% Argon and 0.03% Carbon dioxide/Water

A)

Exosphere

B)

Mesosphere

C)

Troposphere

D)

Stratosphere

Correct Answer : Troposphere

Explanation : Troposphere is where most of the weather phenomena occur. This layer is situated between the mean sea level and 18 km above mean sea level if the tropopause is included.

A)

202.01 K

B)

210.11 K

C)

216.65 K

D)

250.71 K

Correct Answer : 216.65 K

Explaination : The standard atmosphere temperature will remain the same between geopotential altitudes of 12 km and 13 km since both heights are a part of the tropopause which is an isothermal layer of our atmosphere. Therefore, the answer is 216.65 K, the same as at 12 km.

A)

98.46 km

B)

100 km

C)

102.3 km

D)

107.2 km

Correct Answer : 98.46 km

Explaination : Given,

Geometric altitude (h_{g}) = 100 km

Radius of earth (R_{E}) = 6378 km

Geopotential altitude (h) = (R_{E} / (R_{E} + h_{g})). h_{g}

= (6378/(6378+100))*(100)

= 98.46 km

A)

90% of gravity at MSL

B)

93.2% of gravity at MSL

C)

95.5% of gravity at MSL

D)

99% of gravity at MSL

Correct Answer : 95.5% of gravity at MSL

Explaination : Given,

Gravity at MSL (g_{0}) = 9.81 m/s^{2}

Geopotential altitude (h) = 300 km

Radius of earth (R_{E}) = 6378 km

Gravity at 300 km above MSL (g) = g_{0}.(R_{E}/(R_{E} + h))^{2}

= 9.81*(6378/(6378+300))^{2}

= 9.718 m/s^{2}

Percentage = (9.718/9.81) x 100% = 95.5% of MSL gravity

A)

202.12 K

B)

215.68 K

C)

228.65 K

D)

250.17 K

Correct Answer : 250.17 K

Explaination : Given,

Temperature gradient (a) = 0.001 K/m = 1 K/km

At point 1, h_{1} = 20 km, T_{1} = 216.65 K

At point 2, h_{2} = 32 km, T_{2} =?

T_{2} = a.(h_{2} – h_{1}) + T_{1}

= 1*(32-20) + 216.65

= 228.65 K

A)

0.146 kg/m^{3}

B)

0.252 kg/m^{3}

C)

0.365 kg/m^{3}

D)

1.365 kg/m^{3}

Correct Answer : 0.365 kg/m^{3}

Explaination : Given,

At MSL, Temperature (T_{1}) = 288.16 K

Pressure (p_{1}) = 101325 Pa

Density (ρ_{1}) = 1.225 kg/m^{3}

At 11 km, Temperature (T) = 216.66 K

Pressure (p) = 22700 Pa

Density (ρ) =?

From the equation of state,

p/p_{1} = (ρT)/(ρ_{1}/T_{1})

ρ = (pρ_{1}T_{1})/(p_{1}T)

= (22700*1.225*288.16)/(101325*216.66)

= 0.365 kg/m^{3}

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : False

Explanation : False because the stratosphere contains the ozone and cabin pressurization in the stratosphere could lead to ozone poisoning. Also, because the density at that altitude is very low.

A)

Rocket Engine

B)

Turbojet Engine

C)

Turbofan engine

D)

Internal Combustion Engine

Correct Answer : Rocket Engine

Explanation : All engines except the rocket engine require air to function. Since, there is no air/atmosphere within the mesosphere only a rocket engine will work. Rocket engine has its own propellant and oxidizer.

A)

Exopause

B)

Tropopause

C)

Mesopause

D)

Stratopause

Correct Answer : Mesopause

Explanation : Mesopause is the coldest isothermal layer. It has a temperature of about 190 K or -83°C. Tropopause and Stratopause are relatively warmer. Whereas, exopause doesn’t exist.

A)

1.03 x 10^{-4} per meter

B)

1.12 x 10^{-3} per meter

C)

1.54 x 10^{-4} per meter

D)

1.96 x 10^{-3} per meter

Correct Answer : 1.54 x 10^{-4} per meter

Explaination : Given,

Acceleration due to gravity (g) = 9.81 m/s^{2}

Specific gas constant (R) = 287 J/kg.K

Temperature (T) = 222.6 K

Combining hydrostatic equation and equation of state we get,

Change in pressure with altitude, (dp/p)/dh = g/(RT)

= 9.81/(287*222.6)

= 1.54 x 10^{-4} per meter

A)

125 ^{0}R

B)

225 ^{0}R

C)

250 ^{0}R

D)

390 ^{0}R

Correct Answer : 390 ^{0}R

Explaination : Given,

Pressure (p) = 629.66 lb/ft^{2}

Density (ρ) = 0.9408 x 10^{-3} slug/ft^{3}

Specific gas constant (R) = 1716 lb.ft/slug.^{0}R

Using equation of state,

Temperature, T = p/(ρR)

= 629.66/(0.9408*10^{-3}*1716)

= 390 ^{0}R

A)

Karman line

B)

Van Allen belt

C)

Beginning of exosphere

D)

Beginning of mesosphere

Correct Answer : Karman line

Explanation : The beginning of space is officially denoted at an altitude of 100 km which is also known as the Kármán line. After this line the aerodynamic effects are minimal. Cold gas thrusters are used to correct orientation after this line as aerodynamic control surfaces are useless.

A)

231 K

B)

250 K

C)

255 K

D)

262 K

Correct Answer : 262 K

Explanation : Toussaint’s formula is given by T = 15 – 0.0065h

Where, T is the temperature in Celsius and h is the geopotential altitude in metres.

Therefore, T = 15 – 0.0065*(4000)

= – 11°C

= -11 + 273 K

= 262 K

A)

Center of the mass of the Sun

B)

Center of mass of the Earth

C)

At the fixed point on the Earth’s surface

D)

At the center of gravity of an aircraft/spacecraft

Correct Answer : Center of mass of the Earth

Explanation : The origin of ECI system is located at the mass center of the earth but does not rotate with the Earth. Thus, this is fixed with the relative stars and is hence inertial.

A)

21.5°

B)

22°

C)

23.4°

D)

25°

Correct Answer : 23.4°

Explanation : The angle formed between the Earth’s ecliptic plane/orbit plane and the equatorial plane is 23.4°. This is because of the fact that Earth does not rotate perpendicularly in the orbital plane and is therefore inclined.

A)

Angle between ascending node and perigee

B)

Angle between the orbital and equatorial plane

C)

Angle between the perigee and satellite in orbital plane

D)

Angle between vernal equinox and ascending node on equatorial plane

Correct Answer : Angle between vernal equinox and ascending node on equatorial plane

Explanation : Right ascension is the angle formed between the vernal equinox vector to the ascending node on the equatorial plane. Angle between ascending node and perigee is known as Argument of perigee, angle between the orbital and equatorial plane is called Inclination and finally the perigee and satellite in orbital plane’s angle is known as True anomaly.

A)

Vernal equinox

B)

Normal to the equatorial plane

C)

Points 90 degrees to the west of the x-axis in equatorial plane

D)

Points 90 degrees to the east of the X-axis in equatorial plane

Correct Answer : Points 90 degrees to the east of the X-axis in equatorial plane

Explanation : The Y-axis of the ECI system points to 90 degrees to the east of the X-axis in the equatorial plane. Y satisfies the cross product of Z and X thus completing the right-handed coordinate system.

A)

Vernal equinox

B)

Perihelion

C)

Local horizon

D)

Along the intersection of Earth’s equatorial plane and the plane of ecliptic

Correct Answer : Vernal equinox

Explanation : The X-axis of the ECI system points towards the vernal equinox. Vernal equinox marks the first day of the spring in the northern hemisphere (21 March). It is the line joining the Earth and the sun when the sun passes through earth’s equatorial plane.

A)

ECI system

B)

Horizontal system

C)

Ecliptic system

D)

Equatorial system

Correct Answer : Horizontal system

Explanation : **Azimuth Elevation** system is also known as the Horizontal system because the horizontal plane is taken as reference in it. The equatorial system is another type of coordinate system, which uses the celestial equatorial plane as reference. Ecliptic system is another type of coordinate system, which uses the ecliptic plane as reference. Ecliptic plane is the path traversed by the sun around the earth, keeping earth fixed at the centre of the celestial sphere.

A)

Vernal equinox

B)

Celestial North Pole

C)

Celestial Horizon

D)

Celestial South Pole

Correct Answer : Celestial Horizon

Explanation : The reference for Azimuth Elevation system is taken as the Celestial Horizon. Celestial Horizon is defined as the great circle halfway of zenith & nadir, which are the points just above the observer on the celestial sphere & its opposite point on the celestial sphere.

A)

It is fixed & with respect to the observer

B)

It is based on the direction of rotation of the earth

C)

Horizon is divided into 4 quadrants: North, East, South and West

D)

Earth & the observer are imagined as a point at the centre of the celestial sphere

Correct Answer : It is based on the direction of rotation of the earth

Explanation : In the celestial sphere, the Earth is imagined to be fixed and rotating. But the Celestial Sphere rotates along with the Earth. So the Azimuth Elevation system is based on the direction of Earth’s gravity. In this system, the celestial horizon is divided into 4 quadrants: North, East, South and West. It is fixed & with respect to the observer. Earth & the observer are imagined as a point at the centre of the celestial sphere.

A)

Vertical circle from North through Nadir to South point is called the Upper Meridian

B)

LCM is used to measure local time in the Azimuth Elevation system

C)

Vertical circle from North through Zenith to South point is called the Upper Meridian

D)

Local Celestial Meridian (LCM) is the vertical circle passing through North point of the horizon, Zenith & the South point of the horizon

Correct Answer : Vertical circle from North through Nadir to South point is called the Upper Meridian

Explanation : **Local Celestial Meridian (LCM)** is the vertical circle passing through **North point of the horizon, Zenith & the South point of the horizon**. It is used to measure the local time. The vertical circle from **North** through Zenith to South point is called the **Upper Meridian** and the vertical circle from North through Nadir to South point is called the **Lower Meridian.**

A)

East point of the horizon, Zenith & the South point of the horizon

B)

North point of the horizon, Zenith & the West point of the horizon

C)

North point of the horizon, Zenith & the South point of the horizon

D)

East point of the horizon, Zenith & the West point of the horizon

Correct Answer : East point of the horizon, Zenith & the West point of the horizon

Explanation : The Prime Vertical passes through the East point of the horizon, Zenith & the West point of the horizon. The vertical circle needs two opposite points on the horizon to connect; otherwise, it will become triangular in shape instead of a half-circle. So the vertical circle cannot pass through East South & North West pair of points. The vertical circle passing through North point of the horizon, Zenith & the South point of the horizon is known as Local Celestial Meridian (LCM).

A)

Celestial North Pole

B)

The celestial South Pole

C)

The centre of the Celestial sphere

D)

The point of sky overhead the observer, on the celestial sphere

Correct Answer : The point of sky overhead the observer, on the celestial sphere

Explanation : Zenith is defined as the point of sky overhead the observer, on the celestial sphere. It can be also described as the intersection point of the celestial sphere & the normal drawn to the celestial horizon. As the celestial horizon depends on the position of the observer on the Earth, it can coincide with the Celestial North Pole & Celestial South Pole, but that is not always the case. Zenith can never be the centre of the celestial sphere because it does not lie on the sphere boundary.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : False

Explanation : Nadir is the point opposite to the Zenith, on the celestial sphere. Zenith is defined as the point of sky overhead the observer, on the celestial sphere. As the celestial horizon depends on the position of the observer on the Earth, it can coincide with the Celestial North Pole & Celestial South Pole, but that is not always the case.

A)

Latitude

B)

Horizon

C)

Vertical circles

D)

Longitude

Correct Answer : Vertical circles

Explanation : Vertical circles are the circles on the celestial sphere that passes through Zenith or Nadir from the celestial horizon. They are half circles made on the celestial sphere. Latitudes & Longitudes are the features pertaining to the planet Earth. Horizon is the great circle halfway of zenith & nadir.

A)

Altitude, Azimuth

B)

Azimuth, Declination

C)

Altitude, Declination

D)

Right ascension, Declination

Correct Answer : Altitude, Azimuth

Explanation : To identify the location of any celestial object, the Azimuth Elevation system uses two parameters, Altitude & Azimuth. In the Equatorial Coordinate system, the two parameters used are Right ascension & Declination.

A)

0° = North point

B)

90° = East point

C)

180° = West point

D)

270° = West point

Correct Answer : 180° = West point

Explanation : The North point of the horizon is considered 0° Azimuth. From that point, we measure distance eastwards. So that 90° Azimuth is the East point of the horizon whereas 180° & 270° represents the South point & the West point respectively.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : The North point of the horizon is considered 0° Azimuth. From that point, we measure distance eastwards. The value of Azimuth increases as we go towards east from 0° point. So that 90° Azimuth is the East point of the horizon whereas 180° & 270° represents the South point & the West point respectively.

A)

It is measured in hours

B)

It is analogous to latitude

C)

It is the angular distance measured from the East point of horizon

D)

It is the angular distance measured from the North point of horizon

Correct Answer : It is the angular distance measured from the North point of horizon

Explanation : Azimuth is the second parameter of the Azimuth Elevation system & is defined as the angular distance measured from the North point of horizon in the horizontal direction. It is analogous to the longitude system of the Earth & is measured in degrees.

A)

0 hour = Celestial Horizon

B)

0° = Celestial Horizon

C)

-90° = Zenith

D)

+90° = Nadir

Correct Answer : 0° = Celestial Horizon

Explanation : 0° Altitude represents the Celestial Horizon as the vertical distance from the horizon is zero. The angular distance is measured in degrees and not in hours. Above the horizon is considered as positive sign & below the horizon is considered as negative sign. So the Zenith can be represented as +90° whereas the Nadir is represented as -90° altitude.

A)

It is measured in hours

B)

It is analogous to longitude

C)

Above the horizon is considered as negative sign

D)

It is the angular distance of a point from the horizon

Correct Answer : It is the angular distance of a point from the horizon

Explanation : Altitude is the angular distance of a point from the horizon in vertical direction. So it is analogous to latitude system of the Earth & it is measured in degrees. Above the horizon is considered as positive sign whereas below the horizon is considered as negative sign.

A)

Altitude = -20° & Azimuth = 90°

B)

Altitude = +20° & Azimuth = 90°

C)

Altitude = 90° & Azimuth = +20°

D)

Altitude = 90° & Azimuth = -20°

Correct Answer : Altitude = +20° & Azimuth = 90°

Explanation : Azimuth is defined as the angular distance measured from the North point of horizon in the horizontal direction. As the star lies on the vertical circle passing through East point, it represents the Azimuth value. East point of horizon has 90° Azimuth. So the Azimuth value of the star is 90°. Altitude is the angular distance of a point from the horizon in vertical direction. As the star is located at 20° celestial latitude, the Altitude is considered as +20°.

A)

Heliocentric system

B)

ECI system

C)

Selenocentric system

D)

Azimuth Elevation system

Correct Answer : Heliocentric system

Explanation : The Ecliptic Coordinate system is also known as the heliocentric system as it takes the Sun as the reference. The ECI and the Azimuth Elevation system take the Celestial Equator and the Celestial Horizon as the reference, respectively. The Selenocentric Coordinate system is based on Moon as the reference.

A)

Center of the Sun

B)

Center of the Earth

C)

Celestial South Pole

D)

Celestial North Pole

Correct Answer : Center of the Sun

Explanation : The Ecliptic Coordinate system takes the center of the Sun as the origin. For the ECI system, the Earth’s center is considered as the origin. Celestial North Pole and Celestial South Pole are not taken as the point of origin in the coordinate systems.

A)

Celestial Horizon

B)

Celestial Equator

C)

Ecliptic Plane

D)

Earth’s equatorial plane

Correct Answer : Ecliptic Plane

Explanation : The Ecliptic plane is the fundamental reference plane of the Ecliptic Coordinate system. The Ecliptic plane is the plane representing the Sun’s path of travel around the Earth in the Celestial sphere. In the Celestial sphere, the Earth is located at the center and is fixed. The Sun revolves around the Earth in a pre-defined path, which is known as the Ecliptic plane.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : False

Explanation : The Ecliptic plane is not parallel to the Celestial Equator. It is inclined to the Celestial Equator because the same side of the Earth doesn’t always face the Sun as the Earth rotates on a tilted axis. To account for that, the Ecliptic plane is inclined to the Celestial Equator.

A)

It lies between ±90°

B)

It is taken as positive towards the Celestial North Pole

C)

It is the angular distance of the object from the ecliptic plane

D)

It is taken as positive towards the Celestial South Pole

Correct Answer : It is taken as positive towards the Celestial South Pole

Explanation : The Ecliptic latitude is the angular distance of the object from the ecliptic plane. It is taken as positive towards the Celestial North Pole. The ecliptic plane is referred to as 0° ecliptic latitude. The value increases to up to 90° on both up and down side.

A)

It ranges from 0° to 360°

B)

It is constant for a fixed object

C)

It is measured eastward from Vernal equinox

D)

It is the angular distance of the object along the ecliptic plane

Correct Answer : It is constant for a fixed object

Explanation : The Ecliptic longitude is defined as the angular distance of the object along the ecliptic plane. It is measured eastward from Vernal equinox and it ranges from 0° to 360°. Because of precession, the value for a fixed object changes by 50.3 arc-seconds per year.

A)

Azimuth

B)

Distance

C)

Ecliptic latitude

D)

Ecliptic longitude

Correct Answer : Azimuth

Explanation : The Ecliptic system uses 3 parameters to locate a celestial object in the celestial sphere. The three parameters are Ecliptic longitude, Ecliptic latitude and Distance. Azimuth is a parameter used in the Azimuth Elevation system.

A)

In direction of the Celestial North Pole

B)

In direction of the Celestial South Pole

C)

In the direction of the Vernal Equinox

D)

In direction of the position of the moon

Correct Answer : In the direction of the Vernal Equinox

Explanation : The Ecliptic Coordinate system defines its X axis in the direction of the Vernal Equinox. Vernal Equinox is the point of intersection of the Celestial Equator and the Ecliptic plane. When Sun is at Vernal Equinox, all Earth locations experience identical durations of daylight and darkness. Position of moon is not used to define axis in the Ecliptic system.

A)

In direction of the Celestial South Pole

B)

In direction of the Celestial North Pole

C)

In the direction of the Winter Solstice

D)

In direction of the position of the moon

Correct Answer : In the direction of the Winter Solstice

Explanation : The Ecliptic Coordinate system defines its Y axis in the direction of the Winter Solstice. Winter Solstice is the first day of winter. It is the shortest day of the year for the Northern Hemisphere when the Sun is lowest in the sky. This point is used as a reference for the Ecliptic system.

A)

In the direction of the Winter Solstice

B)

In direction of the Celestial South Pole

C)

In direction of the position of the moon

D)

In direction of the Celestial North Pole

Correct Answer : In direction of the Celestial North Pole

Explanation : The Ecliptic Coordinate system defines its Z axis in the direction of the Celestial North Pole. It is chosen as the reference because it is constant in the Celestial sphere.

A)

Length

B)

Direction

C)

Magnitude

D)

Angular rotation

Correct Answer : Angular rotation

Explanation : The coordinate transformation changes only the basis of vector, the rest i.e. length and direction after coordinate transformation remains the same.

A)

(x, y)

B)

(-x, y)

C)

(x, -y)

D)

(-x, -y)

Correct Answer : (-x, y)

Explanation : When a point has to be reflected across the y-axis, the y value stays the same and the x value becomes opposite of what it is.

A)

Orthogonal matrix

B)

Skew matrix

C)

Identical matrix

D)

Rotational matrix

Correct Answer : Orthogonal matrix

A)

0

B)

1

C)

2

D)

3

Correct Answer : 3

Explanation : The coordinate transformation from XYZ frame to X’Y’Z’ frame is carried out by a single rotation about one axis. The other two axes remain in the same plane. But to transform from the ECI coordinate system to the perifocal system, there have to be successive 3 rotations. The first rotation is about Z axis, second is through X’ axis and the final rotation is bout Z’’ axis.

A)

TV Broadcasting

B)

Fixed Wireless

C)

Weather Forecasting

D)

Global Positioning System

Correct Answer : Fixed Wireless

Explanation : Fixed Wireless is a type of communication that occurs between two fixed devices or between two fixed locations. An example of fixed wireless is communication between a computer and a Wi-Fi router. All other options are well implemented through satellites.

A)

5.85 km/s

B)

6.01 km/s

C)

7.91 km/s

D)

11.1 km/s

Correct Answer : 5.85 km/s

Explaination : Given,

Distance between Earth and satellite (r) = 5250 + 6378

= 11,628 km

Standard gravitational parameter of Earth (μ) = 398,600 km^{3}/s^{2}

Since, gravitational attraction between Earth and satellite = centrifugal force of the satellite, we can derive,

Tangential velocity of satellite (V) = (μ/r)^{1/2}

= (398,600/11,628)^{1/2}

= 5.85 km/s

60 .

The standard parameter of Earth is 398,600 km^{3}/s^{2}, distance between Earth and Mars is 225 x 10^{6} km, force of gravitational attraction between Earth and Mars is 6.39 x 10^{23} N. What is the standard gravitational parameter of Mars? Given, universal gravitational constant is 6.67408 × 10^{-11} m^{3} kg^{-1} s^{-2}.

A)

12,890 km^{3}/s^{2}

B)

34,567 km^{3}/s^{2}

C)

42,647 km^{3}/s^{2}

D)

78,901 km^{3}/s^{2}

Correct Answer : 42,647 km^{3}/s^{2}

Explaination : Given,

Force of attraction between Earth and Mars (F) = 6.39 x 10^{23} N

Distance between Earth and Mars (r) = 225 x 10^{6} km

Standard Gravitational Parameter of Earth (μ_{E}) = 398,600 km^{3}/s^{2}

Universal gravitational constant (G) = 6.67408 × 10^{-11} m^{3}/(kg.s^{2})

From Newton’s law of gravitation,

Mass of Mars (m) = (Fr^{2})/μ_{E}

= (6.39*10^{23}*(225*10^{6})^{2})/(398,600*10^{9})

= 6.39 x 10^{23} kg

Standard Gravitational Parameter of Earth (μ_{M}) = G.m

= 6.67408*10^{-11}*6.39*10^{23}

= 42,647 km^{3}/s^{2}

A)

1.771 x 10^{26} N

B)

1.982 x 10^{17} N

C)

2.042 x 10^{12} N

D)

6.192 x 10^{17} N

Correct Answer : 1.771 x 10^{26} N

Explaination : Given,

Standard gravitational parameter of earth (μ) = 398,600 km^{3}/s^{2}

Mass of moon (m) = 7.348 x 10^{22} kg

Distance between earth and moon (r) = 384,400 km

Force of attraction (F) = (μ.m) / (r)^{2}

= (398,600 x 7.348 x 10^{22}) / (384,400)^{2}

= 1.982 x 10^{17} N

A)

9.68 m/s^{2}

B)

9.78 m/s^{2}

C)

9.99 m/s^{2}

D)

10 m/s^{2}

Correct Answer : 9.78 m/s^{2}

Explaination : Given,

Acceleration due to gravity at MSL (g_{0}) = 9.81 m/s^{2}

Radius of Earth (R_{E}) = 6378 km

Height above the surface (h) = 10 km

Acceleration due to gravity at 10 km (g) = g_{0} / (1 + h / R_{E})^{2}

= 9.81/(1+10/6378)^{2}

= 9.78 m/s^{2}

A)

3.25 m/s^{2}

B)

4.2 m/s^{2}

C)

5.85 m/s^{2}

D)

7.1 m/s^{2}

Correct Answer : 5.85 m/s^{2}

Explaination : Given,

Thrust (T) = 31 x 10^{6} N

Mass of rocket (m) = 1.98 x 10^{6} kg

Resultant Force (F) = T-mg

= 31*10^{6}-(1.98*10^{6}*9.81)

= 11.57 x 10^{6} N

From Newton’s second law,

Resultant instantaneous acceleration (a) = F/m

=(11.57*10^{6})/(1.98*10^{6})

= 5.85 m/s^{2}

A)

Euler problem

B)

Kepler problem

C)

Two-body problem

D)

N-body problem

Correct Answer : N-body problem

Explanation : When we have o predict and analyze the motion of multiple celestial objects in the sky as a result of their gravitational forces acting on each other, we use N-body problem to solve the dynamics.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : False

Explanation : In N-body problems, where there’s mutual forces between several celestial bodies, the net external force and torque are both zero due to the Newton’s third law.

A)

N

B)

2N

C)

3N

D)

6N

Correct Answer : 6N

Explanation : In case of N-body problem, there are 3N second order differential equations which results in 6N motion variables.

A)

N

B)

2N

C)

3N

D)

6N

Correct Answer : 3N

Explanation : The N-body problem has 3N degrees of freedom which makes use of 3N second order differential equations for finding the position of the particle.

A)

Orbits are circular

B)

Mass of one of the bodies is negligible

C)

Their positions don’t change over time t

D)

The gravitational force exerted on each other cancels out

Correct Answer : Mass of one of the bodies is negligible

Explanation : In case of restricted three-body problem, it is a simplified version of three-body problem where it is assumed that one of the masses is almost negligible compared to the other two. This m1 and m2 move in Keplerian orbit which is unaffected by m3.

A)

10^{-2} g’s

B)

10^{-3} g’s

C)

10^{-4} g’s

D)

10^{-5} g’s

Correct Answer : 10^{-3} g’s

Explaination : There are certain variations introduced due to the earth’ oblateness effect. This force is of the order 10^{-3} g’s. Newton’s law is only applicable for spherical objects.

A)

One-body problem

B)

Two-body problem

C)

Three-body problem

D)

Five-body problem

Correct Answer : One-body problem

Explanation : The simplest case of the N-body problem is a case which has only one body. In the entire space, there lies only one body with mass m. It experiences no force, acceleration and moves with a constant velocity.

A)

Result converge

B)

It is time consuming

C)

Lack of initial condition data

D)

Closed form solution does not exist

Correct Answer : Closed form solution does not exist

Explanation : N-body problems are usually chaotic for initial conditions and require numerical method to compute the result. Analytical results cannot be obtained wince there no closed form solution existing. So far, two body problems and restricted 3-body problem can be solved.

A)

6.23 km/s

B)

7.08 km/s

C)

7.97 km/s

D)

9.01 km/s

Correct Answer : 7.08 km/s

Explaination : Polar coordinates are represented as (r, θ)

Therefore, r = 1200 km

θ(t) = 0.0059t rad

v_{T} = r(dθ(t)/dt)

= 1200*0.0059

= 7.08 km/s

A)

2.01 km/s

B)

2.19 km/s

C)

3.47 km/s

D)

3.99 km/s

Correct Answer : 3.47 km/s

Explaination : Standard gravitational parameter (μ) = 398,600 km^{3}/s^{2}

Specific angular momentum (h) = 57,112 km^{2}/s

Eccentricity for a parabolic orbit (e) = 1

True anomaly (θ) = 0.52 rad

Radial velocity can then be given by:

v_{r} = (μ/h)(esinθ)

= (398,600/57,112)(1*sin(0.52 rad))

= 3.47 km/s

A)

4.99 km/s

B)

5.23 km/s

C)

6.01 km/s

D)

6.98 km/s

Correct Answer : 6.98 km/s

Explaination : Given,

Standard gravitational parameter (μ) = 398,600 km^{3}/s^{2}

Specific angular momentum (h) = 57,112 km^{2}/s

Eccentricity for a circular orbit (e) = 0

Tangential velocity can then be given by:

v_{T} = (μ/h)(1 + ecosθ)

= (398,600/57,112)(1 + 0)

= 6.98 km/s

A)

vis-viva

B)

simplex visa

C)

majores orbita

D)

minores orbita

Correct Answer : vis-viva

Explanation : Vis-viva equation is the other name for energy equation for orbits. It is Latin for “living force”. The equation describes the conservation of energy of each orbit. Kinetic energy and Gravitational energy compensate for each other to conserve the overall energy.

A)

-10.9 km^{3}/s^{2}

B)

-30.3 km^{3}/s^{2}

C)

12.9 J/kg

D)

40.2 km^{3}/s^{2}

Correct Answer : -30.3 km^{3}/s^{2}

Explaination : Given,

Standard gravitational parameter (μ) = 398,600 km^{3}/s^{2}

Radius of earth (R_{E}) = 6378 km

Semi-major axis (a) = (525 + 6250 + 6378)/2

= 6576.5 km

Specific energy (ε) = -μ/(2a)

= -398,600/(2*6576.5)

= -30.3 km^{3}/s^{2}

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : True. Kinematic properties of the moving body relative to the stationary body can be calculated by assuming the bodies as point masses concentrated at their respective center of masses. Despite the bodies being rigid bodies, knowledge of point mass dynamics is all that is required. Rigid body dynamics is applied to calculate more complex data which is beyond the purpose of a two-body problem.

A)

22.62 x 10^{6} km^{2}

B)

29.89 x 10^{6} km^{2}

C)

117.9 x 10^{4} km^{2}

D)

120.5 x 10^{3} km^{2}

Correct Answer : 22.62 x 10^{6} km^{2}

Explaination : Semi-major axis (a) = 9,000 km

Semi-minor axis (b) = 4,000 km

Area swept (at t = T/5) (A)=?

Due to similarity,

A/(T/3) = πab/T || Area of ellipse = πab

A = πab/5

= π*9,000*4,000/5

= 22.62 x 10^{6} km^{2}

A)

1.015 km/s

B)

1.633 km/s

C)

2.089 km/s

D)

3.211 km/s

Correct Answer : 1.633 km/s

Explaination : Standard gravitational parameter (μ) = 4903 km^{3}/s^{2}

Radius of moon (R_{M}) = 1738 km

Altitude from surface of moon (z) = 100 km

Velocity of satellite (v) = (μ/(R_{M} + z))^{1/2}

= (4903/(1738+100))^{1/2}

= 1.633 km/s

A)

2931.8 s

B)

4211.11 s

C)

5863.53 s

D)

7983.13 s

Correct Answer : 2931.8 s

Explaination : Perigee distance (r_{p}) = 6378 + 400 = 6778 km

Apogee distance (r_{a}) = 6378 + 900 = 7278 km

Semi-major axis (a) = (r_{p} + r_{a})/2

= (6778 + 7278)/2

= 7028 km

Time period of the orbit (T) = 2πa^{3/2}/μ^{1/2}

= 2π*7028^{3/2}/398,600^{1/2}

= 5863.53 s

Time taken to travel from perigee to apogee = T/2 = 5863.53/2 = 2931.8 s

A)

4,536.32 km^{2}/s

B)

61,456.23 km^{2}/s

C)

70,234.12 km^{2}/s

D)

72,876.32 km^{2}/s

Correct Answer : 72,876.32 km^{2}/s

Explaination : Velocity of satellite (v) = 10.7 km/s

Flight path angle (γ) =15°

Satellite distance (r) = 7048 km

Tangential velocity (v_{t}) = vcos(γ) = 10.7*cos(15°) = 10.34 km/s

Specific angular momentum (h) = rv_{t} = 7048*10.34 = 72,876.32 km^{2}/s

A)

Only μ

B)

h, μ and e

C)

h and e

D)

h, μ and θ

Correct Answer : h and e

Explaination : h and e (specific angular momentum and eccentricity) are the orbital constants among all the letters in the orbit equation. μ (standard gravitational parameter) is a constant that is independent of the orbit and depends on the mass of the larger body. Whereas, true anomaly (θ) is not constant and is dependent on the spacecraft/satellite position relative to perigee line.

A)

Rotary

B)

Curvilinear

C)

Brownian

D)

Rectilinear

Correct Answer : Curvilinear

Explanation : Curvilinear motion because all orbits/trajectories have a radius of curvature. Or in other words, all orbits are conic section and therefore curved. Rectilinear motion could be useful if a satellite moved in a straight line. Rotary motion also rotates, but it rotates along its own axis. Though most celestial bodies have rotary motion, it also does not apply to point masses (which is how bodies are assumed in two body problems). Brownian motion is a random motion of particles and has application in quantum world.

A)

Tends to zero

B)

Tends to infinity

C)

Remains constant

D)

Increases proportionally

Correct Answer : Tends to zero

Explanation : The satellite while orbiting an attracting body has negative potential energy and its maximum potential energy tends to zero when the radius tends to infinity.

A)

Radius

B)

Inclination

C)

Eccentricity

D)

True anomaly

Correct Answer : Eccentricity

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : In a system with two point masses m1, m2 having mutual gravitational force acting on each other, the center of mass lies somewhere along the line joining the two masses. The center of mass of this system is non accelerating having a constant velocity.

A)

0

B)

1

C)

Infinite

D)

Negligible

Correct Answer : 0

Explanation : When the equation of motion is derived for a to-body problem, the bodies experience gravitational force only. All the other forces and torques are assumed to be zero.

A)

The point masses never touch

B)

The bodies ae considered to be point masses

C)

The center of mass of two bodies does not lie in the center

D)

No external force acts on the bodies apart from gravitational force

Correct Answer : The center of mass of two bodies does not lie in the center

Explanation : While deriving the equation of motion, there are three main assumptions made as follows :

• Bodies of mass **m1** and **m2** are spherical point masses.

• The point masses never touch.

• No external force other than gravitational force acts on the two bodies.

A)

Ephemeris

B)

Greenwich

C)

Vernal equinox

D)

Intersection of ecliptic and equatorial planes

Correct Answer : Vernal equinox

Explanation : Vernal equinox lies on the celestial equator which is the outward projection of Earth’s equator on the celestial sphere. This is the origin for measuring the longitude which is done is degrees from east to west.

A)

Zenith

B)

Nadir

C)

Horizon

D)

Azimuth

Correct Answer : Nadir

Explanation : Nadir is the point on the imaginary celestial sphere which is directly below the observer. It is directly opposite to the Zenith. Usually when the astronauts carry out spacewalks, they refer to nadir which is the downward view of the satellite in the orbit.

A)

Nadir

B)

Horizon

C)

Azimuth

D)

Zenith

Correct Answer : Zenith

Explanation : Zenith is the point on the imaginary celestial sphere which is directly overhead the observer. It is the highest point on the sphere and this is the point when the shadows appear to be at its smallest when Sun is at the zenith.

A)

Declination

B)

Meridian

C)

Inclination

D)

Right Ascension

Correct Answer : Declination

Explanation : On the celestial sphere, latitude are nothing but the declination since the positive latitudes run from equator to the north pole, and the negative values run from equator to the south pole. These are measure in degrees.

A)

Zenith

B)

Sidereal

C)

Ephemeris

D)

Celestial coordinates

Correct Answer : Ephemeris

Explanation : The coordinates of celestial body which includes stars, comets, asteroid, planets etc. are known as ephemeris when its coordinates are expressed as a function of time. The ephemeris depends on epoch or vernal equinox at that time.

A)

West to East

B)

East to West

C)

North to South

D)

South to North

Correct Answer : East to West

Explanation : Due to the Earth’s rotation which is in the direction of East to West, the celestial sphere also rotates East to West as it is just a fictious sphere with Earth’s centre as its own.

A)

Polar

B)

Prograde

C)

Hohmann

D)

Retrograde

Correct Answer : Retrograde

Explanation : Inclination is the measure of angle between the orbital plane of the satellite and the equatorial plane of the orbit it revolves around. If the inclination angle is 0 degrees, it’s called a prograde orbit. When the inclination angle is 90 degrees, it is a polar orbit and when angle is 180 degrees it is known as retrograde orbit. In this particular orbit, the satellite moves in the opposite direction to the rotation of the planet around which it is orbiting.

A)

2

B)

3

C)

4

D)

6

Correct Answer : 6

Explanation : In total, there are six orbital elements defined. Out of these 2 parameters: eccentricity and angular momentum are used to define the orbit. True anomaly is used to find the location of the object in the orbit. Three additional Euler angles are used to describe the orientation of the orbit. These are- inclination, argument of perigee and right ascension of the ascending node.

A)

Circular

B)

Elliptical

C)

Parabolic

D)

Hyperbolic

Correct Answer : Circular

Explanation : For the circular orbits, the eccentricity is zero and there is no defined perigee. Thus, true anomaly and argument of perigee are not defined for the circular orbits.

A)

Orbit’s shape

B)

Orbit’s orientation

C)

Satellite’s location in the orbit

D)

Orbit’s plane’s rotation about the Earth

Correct Answer : Satellite’s location in the orbit

Explanation : True anomaly is the angle measured in the direction of satellite motion, from perigee to the satellite’s location. The value varies from 0° to 360° and it is undetermined when the eccentricity is zero (i.e. for circular orbits).

A)

Orbit’s shape

B)

Orbit’s orientation in the orbital plane

C)

Satellite’s location in the orbit

D)

Orbit’s plane’s rotation about the Earth

Correct Answer : Orbit’s orientation in the orbital plane

Explanation : Argument of perigee is the angle measured in the direction of satellite motion, from the ascending node to the perigee. The value varies from 0° to 360° and it is undetermined when i = 0° or 180° or when the eccentricity is zero (i.e. for circular orbits).

A)

0°

B)

52°

C)

98°

D)

100°

Correct Answer : 52°

Explanation : The orbital inclination of ISS is 52° and it orbits around the Earth from west to east. 98° is the orbital inclination of Mapping and 0° is for Geostationary satellites.

A)

0 ≤ e ≤ 1

B)

1 ≤ e

C)

e = 1

D)

e ≤ 0

Correct Answer : 1 ≤ e

Explaination : Eccentricity is the ratio of half the foci separation to the semi-major axis. For closed orbits, the eccentricity varies from 0 ≤ e ≤ 1 and foe the open orbits 1 ≤ e. Parabolic and Hyperbolic orbits are considered as open orbits as the body flies off to infinity eventually and does not return to the same angular position.

A)

0

B)

90

C)

360

D)

Undefined

Correct Answer : Undefined

Explanation : Argument of perigee is the angle measured in the direction of the satellite’s motion from the ascending node to the perigee. When inclination is zero, there are no nodes thus its value is undefined.

A)

Eclipse

B)

Particular position at time t

C)

Angle swept by satellite in time t

D)

Time used as reference for mentioning orbital elements

Correct Answer : Time used as reference for mentioning orbital elements

Explanation : Epoch is used to refer to a particular time at the moment when the astronomical quantities such as the orbital elements are defined.

A)

Inclination

B)

Eccentricity

C)

Semi-major axis

D)

Argument of periapsis

Correct Answer : Semi-major axis

Explanation : Out of the six Keplerian orbital elements, semi-major axis (a) helps in determining the size of the conic orbit. It is defined as the half of the long axis of the ellipse. The orbital period and energy depend on the orbit size.

A)

Eccentricity

B)

Inclination

C)

Semi-major axis

D)

Argument of periapsis

Correct Answer : Eccentricity

Explanation : Eccentricity helps in determining the shape of the conic orbit. It is defined as the ratio of half the foci separation to the semi-major axis. For closed orbit, eccentricity ranges from 0 to 1 and for open orbits it is greater than 1.

A)

u_{0} = ω – v_{0}

B)

u_{0} = ωv_{0}

C)

u_{0} = ω + v_{0}

D)

ω = u_{0} + v_{0}

Correct Answer : u_{0} = ω + v_{0}

Explaination : Argument of periapsis ω is the angle formed between the ascending node and periapsis point measured in the direction of the object’s motion. Whereas, argument of latitude u_{0} at epoch is the angle between the ascending node and radius vector to the satellite at some time t and true anomaly at epoch v_{0} is the angle between periapsis and the position of satellite at a particular time t called epoch.

They are related by:

u_{0} = ω + v_{0}

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : False

Explanation : Keplerian orbits have a fundamental assumption that the only influenece on the orbit is gravitational force of the attracting body and it ha s aspherical potential field. Due to this the orbital elements do not change as a function of time.

Although in real life, potentional field is not precisely sphrecial and there may be few changes in the orbital elements with time which is often neglected.

A)

1

B)

2

C)

3

D)

4

Correct Answer : 4

Explanation : The angle between the earth’s equitorial plane and orbital plane is known as inclination. Based on the different inclination angles, there are four types of orbits present.***** **Equitorial orbit :** Angle of inclination is 0 or 180 deg.***** **Polar orbit :** Angle of inclination is 90 deg.***** **Prograde orbit :** Angle of inclincation is between 0 and 90 deg.***** **Retrograde orbit :** Angle of inclination is between 90 and 180 deg.

A)

0.25

B)

0.44

C)

0.64

D)

0.89

Correct Answer : 0.44

Explanation : Semi-major axis (a) = 6580 km

Semi-minor axis (b) = 5910 km

Eccentricity (e) = (1 – (b/a)^{2})^{1/2}

= (1-(5910/6580)^{2})^{1/2}

= 0.44

A)

0.691

B)

1.21

C)

1.45

D)

1.97

Correct Answer : 0.691

Explaination : Specific energy (ε) = -10.29 km^{2}/s^{2}

Specific angular momentum (h) = 10,112 km^{2}/s

Standard gravitational parameter (μ) = 398,600 km^{3}/s^{2}

Using the e-h-ε relation

e = (1+(2εh)/μ)^{1/2}

= (1+(2*-10.29*10,112)/398,600)^{1/2}

= 0.691

A)

Circular

B)

Elliptical

C)

Parabolic

D)

Hyperbolic

Correct Answer : Parabolic

Explanation : Parabolic orbits are formed at the exact escape velocity of a planet. Elliptical and circular orbits are formed when the satellite’s velocity is less than the escape velocity. Whereas, at velocities higher than the escape velocity, hyperbolic orbits are formed.

A)

Elliptical

B)

Circular

C)

Parabolic

D)

Hyperbolic

Correct Answer : Elliptical

Explanation : Elliptical orbits have an eccentricity greater than zero and less than one. Most planets and orbits orbiting around the sun are elliptical in shape.

A)

0.2098

B)

0.5

C)

0.7051

D)

0.901

Correct Answer : 0.2098

Explanation : The eccentricity of an orbit can be determined as follows:

e = (Apoapsis distance – Periapsis distance) / (Apoapsis distance + Periapsis distance)

e = (10,378 – 6778) / (10,378 + 6778)

e = 0.2098

A)

4950 km

B)

5250 km

C)

5000 km

D)

5150 km

Correct Answer : 5250 km

Explanation : Semi-major axis of an ellipse is half of the major axis. Major axis can be represented as a sum of perigee and apogee distances. Therefore,

Semi-major axis = (2500 + 8000) / 2 = 5,250 km

A)

Geocentric-Ecliptic System

B)

Geocentric-Equatorial System

C)

Right Ascension-declination System

D)

Heliocentric-Ecliptic System

Correct Answer : Heliocentric-Ecliptic System

Explanation : Heliocentric-Ecliptic system is a system which has sun as its center or origin. Fundamental plane of operation of this coordinate system is ecliptic plane of our solar system. Most of the planets along with the sun lie on this plane.

A)

Inclination

B)

True anomaly

C)

Argument of periapsis

D)

Longitude of the ascending node

Correct Answer : Argument of periapsis

Explanation : Argument of periapsis is defined as the angle between ascending node and periapsis. The angle between vernal equinox and ascending node is longitude of the ascending node. The angle between earth’s equatorial plane unit vector and satellite’s angular momentum vector is inclination. True anomaly is the angle between perigee and spacecraft position relative to center of the earth.

A)

0.251

B)

0.412

C)

0.822

D)

0.99

Correct Answer : 0.822

Explanation : Apogee radius (r_{a}) = 77,000 km

Perigee radius (r_{p}) = 7,500 km

Eccentricity (e) = (r_{a} – r_{p}) / (r_{a} + r_{p})

= (77,000 – 7,500) / (77,000 + 7,500)

= 0.822

A)

Apogee radius

B)

Perigee radius

C)

Satellite radius

D)

True-anomaly-averaged radius

Correct Answer : True-anomaly-averaged radius

Explanation : True-anomaly-averaged radius is the average distance between satellite and earth for one complete orbit. To find the average distance between the earth and a satellite, the total true anomaly angle of 2π is divided into n equal segments with Δθ, and a sum of satellite radiuses is calculated from every nth interval. The sum is then divided by n. To make the solution easier, the sum is written as an integral and solved.

A)

43.24°

B)

45.32°

C)

50.44°

D)

64.16°

Correct Answer : 64.16°

A)

12,012 km

B)

52,910 km

C)

55,000 km

D)

71,000 km

Correct Answer : 52,910 km

Explanation : Apogee radius (r_{a}) = 100,000 km

Eccentricity (e) = 0.89

Semi-major axis (a) = r_{a} / (1 + e)

= 100,000 / (1 + 0.89)

= 52,910 km

A)

0.648 km/s

B)

1.293 km/s

C)

2.791 km/s

D)

12.307 km/s

Correct Answer : 0.648 km/s

Explaination : Gravitational Parameter (μ) = 398,600 km^{3}/s^{2}

Semi-major axis (a) = 50,000 km

Eccentricity (e) = 0.9

From orbit equation for ellipse,

Specific angular momentum (h) = (μa(1-e^{2}))^{1/2}

= (398,600*50,000*(1-0.9^{2}))^{1/2}

= 61,536 km^{2}/s

Apogee radius (r_{a}) = a(1 + e)

= 50,000 (1 + 0.9)

= 95,000 km

Apogee velocity (v_{a}) = h/r_{a}

= 61,536/95,000

= 0.6477 km/s

A)

8,900 km

B)

12,000 km

C)

15,060 km

D)

40,450 km

Correct Answer : 40,450 km

Explanation : Apogee radius (r_{a}) = 73,000 km

Perigee radius (r_{p}) = 7,900 km

Semi-major axis (a) = (r_{a} + r_{p}) / 2

= (73,000 + 7,900) / 2

= 40,450 km

A)

7.71 km/s

B)

7.89 km/s

C)

9.63 km/s

D)

11.01 km/s

Correct Answer : 9.63 km/s

Explaination : Specific angular momentum (h) = 72,000 km^{2}/s

Satellite radius (r) = Earth’s radius + distance from Earth surface

= 6378 + 1,100

= 7478 km

Tangential velocity (v_{T}) = h/r

= 72,000/7,478

= 9.63 km/s

A)

11.214 hours

B)

11.25 hours

C)

20 hours

D)

20.518 hours

Correct Answer : 20.518 hours

Explaination : Apogee radius (r_{a}) = 69,000 km

Perigee radius (r_{p}) = 7,100 km

Gravitational Parameter (μ) = 398,600 k^{3}/s^{2}

Semi-major axis (a) = (r_{a} + r_{p})/2

= (69,000 + 7,100)/2

= 38,050 km

Time period (T) = (2π/μ^{1/2})a^{3/2}

= (2π/398,600^{1/2})*38,050^{3/2}

= 73,865.77 s

= 20.518 hours

A)

-2.257 km^{2}/s^{2}

B)

-4.927 km^{2}/s^{2}

C)

-5.238 km^{2}/s^{2}

D)

-6.231 km^{2}/s^{2}

Correct Answer : -4.927 km^{2}/s^{2}

Explaination : Apogee radius (r_{a}) = 73,000 km

Perigee radius (r_{p}) = 7,900 km

Gravitational Parameter (μ) = 398,600 km^{3}/s^{2}

Semi-major axis (a) = (r_{a} + r_{p}) / 2

= (73,000 + 7,900) / 2

= 40,450 km

Specific Energy (ε) = -μ/(2a) = -398,600*(2*40,450)

= -4.927 km^{2}/s^{2}

A)

0.648 km/s

B)

1.293 km/s

C)

2.791 km/s

D)

12.307 km/s

Correct Answer : 12.307 km/s

Explanation : Gravitational Parameter (μ) = 398,600 km^{3}/s^{2}

Semi-major axis (a) = 50,000 km

Eccentricity (e) = 0.9

From orbit equation for ellipse,

Specific angular momentum (h) = (μa(1 – e^{2}))^{1/2}

= (398,600*50,000*(1-0.9^{2}))^{1/2}

= 61,536 km^{2}/s

Perigee radius (r_{p}) = a(1-e)

= 50,000 (1-0.9)

= 5,000 km

Perigee velocity (v_{p}) = h/r_{p}

= 61,536/5,000

= 12.307 km/s

A)

0.11

B)

0.41

C)

0.64

D)

0.89

Correct Answer : 0.89

Explanation : Perigee radius (r_{p}) = 6,000 km

Semi-major axis (a) = 52,910 km

Eccentricity (e) = 1 – r_{p} / a

= 1 – 6,000 / 52,910

= 0.89

A)

equal to

B)

less than

C)

more than

D)

equal to or less than

Correct Answer : equal to

Explanation : Tangential velocity is exactly equal to the spacecraft’s velocity at apogee. True anomaly at the apogee is 180° and as the radial velocity is dependent on the true anomaly as a multiple of sin θ, it should be zero. Therefore, at this instant the tangential velocity and spacecraft’s velocity are the same.

A)

0°

B)

90°

C)

180°

D)

270°

Correct Answer : 0°

Explanation : Flight path angle (γ) = atan (esinθ/(1 + ecosθ))

At perigee, true anomaly (θ) = 0°, this implies sinθ = 0

γ = atan (0) = 0°, therefore 0° is the correct answer.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : Tangential velocity is equal to the satellite velocity in a circular orbit since both the velocities travel along the same direction throughout the orbit. Therefore, there is no radial component. There is acceleration along the radial direction though which keeps the satellite in orbit without falling back to earth.

A)

1.0

B)

1.01

C)

0.0

D)

0.87

Correct Answer : 0.0

Explanation : A geostationary orbit (GEO) is an orbit over a planet in which a satellite has a fixed position over the planet. This makes the orbital shape of GEO exactly as the shape of the planet. Since, all planets are spherical in shape their cross-sections are circular. Hence, the GEO is a circular orbit with eccentricity of 0. It might be slightly elliptical but for the sake of simplicity it can be assumed to circular.

A)

0°

B)

22.5°

C)

27°

D)

45°

Correct Answer : 0°

Explanation : Since, the tangential velocity equals to the satellite velocity at any point across a circular orbit, the flight path angle is always zero. Regardless, of what the true anomaly or height of the satellite, in a circular orbit the flight path angle is zero.

A)

15.8 km/s

B)

21.11 km/s

C)

24.41 km/s

D)

22.67 km/s

Correct Answer : 24.41 km/s

Explaination : Initial velocity (v_{i}) = 27 km/s

Gravitational Parameter (μ) = 42,828 km^{3}/s^{2}

Radius of Satellite (r) = 3,389.5 + 3,000

= 6,389.5 km

Orbit velocity (v) = (μ/r)^{1/2}

= (42,828/6,389.5)^{1/2}

= 2.589 km/s

Delta-v = v_{i} – v

= 27 – 2.589

= 24.41 km/s

A)

11.11 km/s

B)

10.36 km/s

C)

9.599 km/s

D)

9.45 km/s

Correct Answer : 9.599 km/s

Explaination : Specific angular momentum (h) = 47,862.73 km^{2}/s

Gravitational parameter (μ) = 324,859 km^{3}/s^{2}

Escape velocity (v_{esc}) = 2^{1/2}(μ/h)

= 2^{1/2}*(324,859/47,862.73)

= 9.599 km/s

A)

directly proportional to

B)

inversely proportional to

C)

inversely proportional to square of

D)

inversely proportional to square root of

Correct Answer : inversely proportional to square root of

Explanation : Only for circular orbits is the velocity of a satellite inversely proportional to square root of the radius of orbit. In all the other cases, the relationship is not as straightforward due to the tangential and radial velocity components. It cannot be inversely proportional either, though v = h/r. Since h, is dependent on r. Only equation that best describes the relationship is v = (μ/r)^{1/2} and μ is a constant.

A)

10.22 km/s

B)

10.38 km/s

C)

11.19 km/s

D)

12.56 km/s

Correct Answer : 10.38 km/s

Explaination : Specific angular momentum (h) = 87,899 km^{2}/s

Eccentricity (e) = 2.5

Gravitational Parameter (μ) = 398,600 km^{3}/s^{2}

Hyperbolic excess velocity (v_{∞}) = (μ/h)(e^{2} – 1)^{1/2}

= (398,600/87,989)(2.5^{2} – 1)^{1/2}

= 10.38 km/s

A)

8,900 km

B)

10,984 km

C)

12,000 km

D)

15,060 km

Correct Answer : 10,984 km

Explaination : Radius of satellite (r) = 36,613 km

Standard gravitational parameter (μ) = 398,600 km^{3}/s^{2}

Eccentricity (e) = 1.2

We know for hyperbolic trajectories,

(Satellite speed)^{2} = (Hyperbolic Excess Speed)^{2} + (Escape Velocity)^{2}

v^{2} = v_{∞}^{2} + v_{esc}^{2}

4v_{∞}^{2} = v_{∞}^{2} + 2μ/r

3v_{∞}^{2} = 2μ/r

r = (2μ)/(3v_{∞}^{2})

We know,

v_{∞}^{2} = (μ/h)^{2}(e^{2} – 1)

Substituting back into equation for r, we get,

r = (2/3)(h^{2}/μ)(1/e^{2} – 1)

Substituting orbit equation for perigee radius, we get,

r = (2/3)(r_{p}/(e – 1))

Therefore, rearranging the equation to get r_{p}

r_{p} = 3r(e-1)/2

= 3*36,613*(1.2-1)/2

= 10,984 km

A)

9.74 km/s

B)

9.98 km/s

C)

10.02 km/s

D)

10.22 km/s

Correct Answer : 9.74 km/s

Explaination : Perigee speed (v_{p}) = 14.3 km/s

Satellite radius (r) = 6378 + 900

= 7278 km

Escape velocity (v_{esc}) = (2μ/r)^{1/2}

= (2*398,600/7278)^{1/2}

= 10.466 km/s

Hyperbolic excess speed (v_{∞}) = (v_{p}^{2} – v_{esc}^{2})^{1/2}

= (14.3^{2} – 10.466^{2})^{1/2}

= 9.744 km/s

A)

110,156 km^{2}/s

B)

112,134 km^{2}/s

C)

125,981 km^{2}/s

D)

145,560 km^{2}/s

Correct Answer : 145,560 km^{2}/s

Explaination : Specific angular momentum = Perigee radius x Perigee velocity

= 7278 x 20

= 145,560 km^{2}/s

A)

8.221 km/s

B)

9.122 km/s

C)

9.562 km/s

D)

10.24 km/s

Correct Answer : 9.562 km/s

Explanation : Perigee radius (r_{p}) = 6778 km

Eccentricity (e) = 2.3

True anomaly (θ) = 100°

From the orbit equation,

Specific angular momentum (h) = [μr_{p}(1 + e)]^{1/2}

= [398,600*6778*(1 + 2.3)]^{1/2}

= 94,422.69 km^{2}/s

Radial velocity (v_{r}) = (μ/h)esin(θ)

= (398,600/94,422.69)*2.3*sin(100°)

= 9.562 km/s

A)

Excess

B)

Specific

C)

Dynamic

D)

Characteristic

Correct Answer : Characteristic

Explanation : Characteristic energy or specific excess energy of a hyperbolic orbit is the energy required to escape. Specific energy on its own is a measure of a spacecraft’s total energy divided by its mass. Characteristic energy times mass is excess energy of a spacecraft, but the question asks for energy divided by mass. While, dynamic energy is not a term used in orbital mechanics.

A)

12.45°

B)

41.24°

C)

45.32°

D)

56.44°

Correct Answer : 56.44°

Explanation : Eccentricity (e) = 1.2

Turn angle = 2*asin(1/e)

= 2*asin(1/1.2)

= 56.44°

A)

18,000 km

B)

19,000 km

C)

20,000 km

D)

21,000 km

Correct Answer : 20,000 km

Explanation : Perigee radius (r_{p}) = 9,800 km

Perigee radius (r_{a}) = -49,800 km

We know, for hyperbolas,

r_{p} = a(e – 1)

r_{a} = -a(e + 1)

where, a and e are semi-major axis and eccentricity, respectively

Dividing equations mentioned above and substituting the given values, we get,

9800/49,800 = (e – 1)/(e + 1)

0.1968(e + 1) = e – 1

0.8032e = 1.1968

e = 1.49

Substituting back into perigee radius equation, we get,

Semi-major axis (a) = r_{p}/(e – 1)

= 9800/(1.49 – 1)

= 20,000 km

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : True. Hyperbolic excess velocity is the relative velocity between helio-centered ellipse and earth’s orbit. It is achieved when a spacecraft accelerates to a speed more than escape velocity. Maintaining a proper hyperbolic excess is important to perform inter-planetary transfers.

A)

487.54 km

B)

532.90 km

C)

663.32 km

D)

789.97 km

Correct Answer : 663.32 km

Explaination : Eccentricity (e) = 1.2

Specific Energy (ε) = 199.3 km^{2}/s^{2}

Gravitational parameter (μ) = 398,600 km^{3}/s^{2}

We know,

Semi-major axis (a) = μ/(2ε)

= 398,600/(2*199.3)

= 1000 km

Aiming radius (Δ) = a*(e^{2} – 1)^{1/2}

= 1,000*(1.2^{2} – 1)^{1/2}

= 663.32 km

A)

704.31 km

B)

830.66 km

C)

1000 km

D)

1200 km

Correct Answer : 830.66 km

Explaination : Eccentricity (e) = 1.3

Specific Energy (ε) = 199.3 km^{2}/s^{2}

Gravitational parameter (μ) = 398,600 km^{3}/s^{2}

We know,

Semi-major axis (a) = μ/(2ε)

= 398,600/(2*199.3)

= 1000 km

Aiming radius (Δ) = a*(e^{2} – 1)^{1/2}

= 1,000*(1.3^{2} – 1)^{1/2}

= 830.66 km

A)

High Earth Orbit

B)

Low Earth Orbit

C)

Medium Earth Orbit

D)

Geosynchronous Orbit

Correct Answer : Low Earth Orbit

Explanation : Low Earth orbits are the orbits with its center as the Earth’s center. It has an altitude of 2000 km which is approximately one-third the Earth’s Radius, which is 6,371 km.

A)

Low Earth Orbit

B)

High Earth Orbit

C)

Medium Earth Orbit

D)

Equatorial Low Earth Orbit

Correct Answer : Equatorial Low Earth Orbit

Explaination : Equatorial Low Earth Orbit has the lowest ?v requirement because the spacecraft is launched due east which leads to least ?v as Earth rotates from east to west with a surface velocity of 1600 km per hour.

A)

Low Earth Orbit

B)

High Earth Orbit

C)

Medium Earth Orbit

D)

Geosynchronous Orbit

Correct Answer : Low Earth Orbit

Explanation : Most of the weather satellites and space stations are located in the circular or elliptical low-earth orbit. Post Apollo mission, there have been no spacecraft orbiting beyond low earth orbit.

A)

Low Earth Orbit

B)

High Earth Orbit

C)

Medium Earth Orbit

D)

Geosynchronous Orbit

Correct Answer : Medium Earth Orbit

Explanation : Van Allen Belts is the belt of high energy protons that lie within the altitude range of 20,000 to 20,650 km which is the Medium Earth orbit. The orbital period of this orbit range is 12 hours.

A)

Polar Orbit

B)

Tundra Orbit

C)

Molniya Orbit

D)

Sun-Synchronous Orbit

Correct Answer : Tundra Orbit

Explanation : Tundra orbit is a type of geosynchronous orbit which has a very high inclination of 63.4 degrees with a low eccentricity of range 0.2-0.3. The orbit is of figure 8 form with one of the smaller loops in either of the hemisphere. Compared to Molniya orbit, this has half the orbital time period.

A)

Prograde motion

B)

Retrograde motion

C)

Apparent Prograde motion

D)

Apparent Retrograde motion

Correct Answer : Apparent Retrograde motion

Explanation : At a high earth orbit of an altitude of approximately 35,000 km, the satellite has a apparent retrograde motion. In this the rotation speed of the earth is higher compared to the orbital velocity which leads to Earth’s surface moving westward.

A)

Polar Orbit

B)

Molniya Orbit

C)

Low-Earth Orbit

D)

Sun-synchronous Orbit

Correct Answer : Molniya Orbit

Explanation : Molniya orbit is a highly eccentric medium earth orbit. It has an eccentricity of 0.722 in order to increase the viewing time of the latitudes as it is an extremely elliptical orbit. The orbit takes 12 hours to complete with Earth rotating underneath it.

A)

High Earth Orbit

B)

Medium Earth Orbit

C)

Geosynchronous Orbit

D)

Equatorial Low Earth Orbit

Correct Answer : Geosynchronous Orbit

Explanation : The Communications satellite is always placed in the Geosynchronous orbit, as the satellite’s position remains same in this orbit and the fixed antennas are able to communicate with them.

A)

Low Earth Orbit

B)

High Earth Orbit

C)

Medium Earth Orbit

D)

Equatorial Low Earth Orbit

Correct Answer : Medium Earth Orbit

Explanation : GPS is placed at an altitude of 20,200 km which is the medium-earth orbit. The orbital time-period of GPS is about 12 hours. Most of the navigation satellites are placed in the medium- earth orbit.

A)

Change of inclination of the orbits

B)

Hohmann transfer

C)

Adjustment of perigee an apogee height

D)

General coplanar transfer between circular orbits

Correct Answer : Change of inclination of the orbits

Explanation : When the satellite is launches, there’s often error in burnout altitude, speed, flight path angle and the exact orbit that was desired is not achieved. This is corrected by applying a small delta-v in the same plane to adjust the orbital parameters. The in-plane corrections include- Hohmann transfer, adjustment of perigee and apogee height, general coplanar transfer between circular orbits.

A)

Major-axis becomes half

B)

Major-axis becomes twice

C)

Major-axis is unaffected

D)

Major-axis becomes one-fourth

Correct Answer : Major-axis becomes twice

A)

Bi-elliptical transfer

B)

Phasing maneuver

C)

One-tangent transfer

D)

Hohmann transfer

Correct Answer : Hohmann transfer

Explanation : The transfer between two circular coplanar orbits is carries out using Hohmann transfer. It is a way of moving the satellite to a high-altitude orbit from the parking orbit or the other way round. The transfer takes place using a transfer elliptical orbit in the same plane.

A)

Direction of velocity

B)

Magnitude of velocity

C)

Direction of angular momentum

D)

Magnitude of angular momentum

Correct Answer : Magnitude of angular momentum

A)

Bi-elliptic transfer

B)

Hohmann transfer

C)

Non-Hohmann transfer

D)

One tangent-burn transfer

Correct Answer : Hohmann transfer

Explanation : Hohmann transfer is considered at the most energy efficient two-impulse transfer. It does not truly exist. In this, both the initial and final orbital planes are coplanar and the change in velocity magnitude is tangential to both the initial and final orbits.

A)

1

B)

2

C)

3

D)

4

Correct Answer : 2

Explanation : Two velocity increments in total are required to carry out Hohmann transfer. The first velocity increment is done at the inner circular orbit in the direction of the spacecraft. This places it in a higher energy elliptical trajectory. Once reaching the perigee of the elliptical orbit, the second velocity increment is carries out using thrusters to place it in the outer circular orbit.

A)

1.874

B)

2.894

C)

3.933

D)

4.784

Correct Answer : 3.933

Explaination : Given values are: r_{A} = (6,378 + 200)km = 6,578 km (Where radius of Earth of 6,578 km has to be added)

r_{B} = 35,786 + 6,328 = 42,164 km

A)

1

B)

2

C)

3

D)

4

Correct Answer : 3

Explanation : In order to carry out Bi-Elliptical Hohmann tranfer 3 burns are required. 1^{st} burn is to convert the circular orbit at r_{1} to elliptical orbit with perigee r_{p} = r_{1} and apogee r_{a} = r_{*}. The second burn is carried out at the apogee to raise the perigee of elliptical orbit to r_{2}. Finally at perigee, the orbit is circulized by lowering the perigee to r_{2}.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : r* is a free parameter and can have any value. Usually as r* tends to infinity, the bi-elliptic tranfer becomes more efficient. As r* tends to infinity, time tends to infinity so there is trade-off between time and efficiency.

A)

Circular

B)

Elliptical

C)

Parabolic

D)

Hyperbolic

Correct Answer : Elliptical

Explanation : The Hohmann transfer involves tranfer of the object/satellite between two circular orbits using a transfer elliptical orbit. When transfering from a smaller orbit to a larger one, the change in velocity is applied in the direction of motion. Transfer orbits cannot take the shapes of Parabolic or Hyperbolic as they are usually escapy trajectories.

A)

Same as the period of orbit

B)

Twice of the period of orbit

C)

One-third of the period of orbit

D)

Half of the period of orbit

Correct Answer : Half of the period of orbit

Explanation : The transfer time is half of the period of orbit.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : False

Explanation : The Hohmann transfer is carried out between two circular co-planer orbits. The tranfer orbit is the elliptical orbit. The co-planer orbits result in optimising energy.

A)

Parallel to the plane of orbit

B)

Tangent to the plane of orbit

C)

At an angle to the plane of orbit

D)

Perpendicular to the plane of orbit

Correct Answer : Perpendicular to the plane of orbit

Explanation : The velocity change delta-v component perpendicular to the plane of orbit changes the orientation of the orbital plane. The velocity can change the size, shape or even rotate the line of apsides.

A)

Altitude

B)

Flight path angle

C)

Speed and Flight path angle

D)

Speed of the satellite

Correct Answer : Speed and Flight path angle

Explanation : After applying a finite delta-v, the speed and flight-path angle of the satellite are unchanged, then only the plane of the orbit has been altered. This is called a simple plane change

A)

Inclination change

B)

In-plane orbit change

C)

Out-of-plane orbit change

D)

Three-dimensional orbit change

Correct Answer : Out-of-plane orbit change

Explanation : Changing an inclined orbit to an equatorial orbit involves change in plane which is an example of a simple plane change. This is a part of out- of-plane change as it involves change in the orientation of the orbital plane.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : For a satellite orbiting in a Molniya orbit, which is a highly elliptical orbit, the speed is high at the nodal crossing. Due to this, even a small inclination change requires a significantly larger impulse as delta-v is directly proportional to the velocity at the node.

A)

Focus

B)

Apogee

C)

Equator

D)

Perigee

Correct Answer : Apogee

Explanation : The maximum efficiency of inclination is achieved at the apogee where the orbital velocity is the lowest thus the delta-v required is the lowest. This results in high efficiency to change the satellite’s orbital plane.

A)

Eccentricity

B)

Inclination

C)

True anomaly

D)

Right of ascension

Correct Answer : Inclination

Explanation : Orbital inclination change is a maneuver carried out to change the inclination of the orbital plane. This requires a change in the orbital velocity vector by applying a small impulse (delta-v) at the orbital node.

A)

At an angle i

B)

Perpendicular

C)

In the same direction

D)

In the opposite direction

Correct Answer : In the opposite direction

Explanation : The orbit normal component of the delta-v that is applied to change the orbital plane is always in the opposite direction to the angular momentum vector h? when it is applied at the ascending node. At descending node, both the vectors are in the same direction.

A)

Rockets with instantaneous bursts

B)

Rocket fire short bursts for required delta-v

C)

Maneuvers which are analyzed by Lambert’s equation

D)

Rocket fires long bursts for long time for the required delta-v

Correct Answer : Rocket fire short bursts for required delta-v

Explanation : Maneuvers are required by the spacecrafts to move from one orbit to another. Impulsive maneuvers are the ones in which the rockets fire in short bursts for small duration to produce the required velocity change. This changes the magnitude and direction of the velocity vector.

A)

Pumping maneuver

B)

Cranking maneuver

C)

Orbital raising maneuver

D)

Orbital lowering maneuver

Correct Answer : Pumping maneuver

Explanation : Impulsive maneuver is carried out by firing brief on-board rocket motor. This results in a velocity shift of the spacecraft. If Δv reflects a magnitude change, it is referred to as a pumping maneuver.

A)

Pumping maneuver

B)

Orbital raising maneuver

C)

Cranking maneuver

D)

Orbital lowering maneuver

Correct Answer : Cranking maneuver

Explanation : Each impulsive maneuver results in a velocity change of the spacecraft. This velocity change yield both magnitude and direction change of the velocity vector. When Δv represents a change in direction it is referred as a cranking maneuver.

A)

Rockets with instantaneous bursts

B)

Rocket fire short bursts for required delta-v

C)

Maneuvers which are analyzed by Lambert’s equation

D)

Rocket fires long bursts for long time for the required delta-v

Correct Answer : Rocket fires long bursts for long time for the required delta-v

Explanation : When the thrust applied by the rocket is done over a longer period to obtain the required delta-v, it is known as the non-impulsive maneuver. Unlike impulsive maneuver it does not involve applying impulse over a short period.

A)

Energy change

B)

Simple plane change

C)

Adjustment of perigee and apogee height

D)

Combined change of apsis altitude and plane orientation

Correct Answer : Energy change

Explanation : There are four types of single impulse maneuver-Adjustment of perigee and apogee height, simple rotation of the line of apsides, simple plane change and combined change of apsis altitude with plane orientation. There is no change in energy involved by applying single impulse.

A)

One-Tangent burn

B)

One-Impulse transfer

C)

Two-Impulse Hohmann transfer

D)

Three-impulse Bi-elliptical transfer

Correct Answer :

Explanation : Phasing maneuver is a used when a spacecraft in a given orbit is moved to a different location within the same orbit. This is carried out using two-impulse Hohmann transfer. The first burn is carried out in the original orbit to shrink or expand the orbit. On completing the orbit and returning to the original burn point, it performs second maneuver.

A)

To save fuel

B)

To adjust the inclination angle of the orbit

C)

For the spacecraft to move ahead of another spacecraft

D)

For the spacecraft to move with another spacecraft behind it

Correct Answer : To adjust the inclination angle of the orbit

Explanation : The phasing maneuver is usually carried out when the spacecraft needs to rendezvous ahead or behind another spacecraft. Also, sometimes this is carried out to save the fuel by entering a phasing orbit closer the original orbit and performing multiple orbits before returning back to the orbital burn point.

A)

ΔV = v_{2} – v_{1}

B)

ΔV = v_{2} + v_{1}

C)

ΔV = v_{2}^{2} – v_{1}^{2}

D)

ΔV = v_{2}^{2} + v_{1}^{2}

Correct Answer : ΔV = v_{2} – v_{1}

Explaination : In order to calculate the total impulse which is required by the space shuttle/ satellite to change its position from initial orbit to phasing orbit, ΔV = v_{2} – v_{1} is used.

Where, v_{1} is the spacecraft’s velocity at point of intersection in original orbit

v_{2} is the spacecraft’s velocity at point of intersection in phasing orbit.

A)

Speed up

B)

Speed down

C)

Change its inclination

D)

Change its true anomaly

Correct Answer : Speed up

Explanation : If the target is behind the interceptor in the same orbit, the interceptor must speed up to enter a higher, slower orbit thereby allowing the target to catch up.

A)

Hohmann transfer

B)

High energy orbit

C)

Electric propulsion transfer

D)

Low-thrust chemical transfer

Correct Answer : High energy orbit

Explanation : For military operation mission, more energy is required than what is required for the Hohmann transfer. High energy orbit is used when transfer time is of the essence. In this operation, an impulse is given at the initial circular orbit which pushes the spacecraft into an elliptical orbit. After reaching the desired point, another small impulse is given to circularize the orbit and place the spacecraft at the desired altitude.

A)

High energy orbit

B)

Hohmann transfer

C)

Low-thrust chemical transfer

D)

Electric propulsion transfer

Correct Answer : Electric propulsion transfer

Explanation : In case of electric propulsion transfer, the final orbit is reached by spiraling outward as there is less consumption of propellant due to the high I_{sp} value of the electric propulsion.

A)

Speed up

B)

Speed down

C)

Change its inclination

D)

Change its true ascension

Correct Answer : Speed down

Explanation : To catch a spacecraft ahead of it in the same orbit, the interceptor slows down, entering a smaller phasing orbit with a shorter period. This allows it to catch the target and go ahead of it.

A)

1

B)

2

C)

3

D)

4

Correct Answer : 2

Explanation : Non-Hohmann transfer makes use of two impulses. The first impulse is made at the initial orbit which helps in entering the transfer orbit. After that, a second impulse is applied to enter the final orbit.

188 .

A)

Minor axis

B)

Imaginary line connecting the apsides of circular orbit

C)

Imaginary line connecting the apsides of ellipse

D)

Line tracing the direction of movement of satellite in the orbit

Correct Answer : Imaginary line connecting the apsides of ellipse

Explanation : Apse line is the imaginary line that connects the two extreme apside points in the orbit. It is only defined for elliptic parabolic or hyperbolic orbits as the circular orbit has an eccentricity of 0. For the Non-Hohmann transfer, the two orbits share a common apse line.

A)

η = θ_{1} + θ_{2}

B)

η = γ_{1} – γ_{2}

C)

η = γ_{1} + γ_{2}

D)

η = θ_{1} – θ_{2}

Correct Answer : η = θ_{1} – θ_{2}

Explaination : The rotation of the line of apsides is given as the difference between the true anomalies of the point of intersection as measured from the periapsis of the orbit. This can ben seen in the figure below.

A)

Common plane

B)

Transfer between elliptical orbit

C)

Transfer is tangent to both the orbit

D)

Transfer shares the apse line to both the orbits

Correct Answer : Transfer shares the apse line to both the orbits

Explanation : In both Hohmann and Non-Hohmann transfers, there’s a transfer between the two co-axial elliptical orbits. The main difference is that in Hohmann transfer, the transfer trajectory is tangent to both the initial and final orbits, whereas in Non-Hohmann that is not true. In the later case, both the orbits share a common apse line.

A)

Lambert’s problem

B)

Jacobi problem

C)

Euler’s problem

D)

Kepler’s problem

Correct Answer : Lambert’s problem

Explanation : Chase maneuvers are solved using Lambert’s problems. In chase maneuver, there are two spacecrafts which orbit in two non-coplanar elliptic orbits. One of the spacecraft tries to intercept the another one. This intercept trajectory is bi-elliptic orbital transfer.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : False

Explanation : Chase maneuvers are considered to be energy demanding maneuvers which involves one spacrcraft trying to intercept the other spacecraft using bi-elliptic transfer orbit.

A)

LEO

B)

GEO

C)

HEO

D)

MEO

Correct Answer : LEO

Explanation : Chase maneuvers are considred to be impractical for Low Earth orbits as they are very energy demanding with high delta-v requirements. They are usually carried out for interplanetary missions.

A)

Active

B)

Target

C)

Chase

D)

Passive

Correct Answer : Chase

Explanation : In case of chase maneuver, there are two spacecrafts involvolved in which one spacecraft chases the other. The term used for this spacecraft is ‘Chase’.

A)

Active

B)

Chaser

C)

Target

D)

Passive

Correct Answer : Target

Explanation : When two spacecrafts are orbiting in a rendezvous, and one spacecraft tries to intercept the other spcecraft, the maneuvering spacecraft is known as ‘Chase’ and the one that is being chased is called ‘Target’.

A)

For deorbiting

B)

Perform docking operation

C)

To target another spacecraft

D)

To catch up spacecraft for maintenance

Correct Answer : To target another spacecraft

Explanation : Chase maneuver is performed for docking operation. This is done in otder to carry out maintenance task, refuel another spacecrafts or deorbiting.

A)

0

B)

1

C)

2

D)

3

Correct Answer : 3

Explanation : There are three orbital maneuvers carried out in a bi-elliptical orbit. The first maneuver moves spacecraft into an elliptical transfer orbit, second maneuver raises periapsis of first transfer ellipse equal to radius of outer circle. The final maneuver reduces the radius at apogee of the second transfer ellipse to equal the radius of the outer circle.

A)

1

B)

2

C)

3

D)

4

Correct Answer : 2

Explanation : In the bi-elliptic orbit, there are two delta-v burn provided. The first burn is at the apogee of the initial circular orbit whichmakes it enter into first transfer orbit. After this, the second burn is at the perigee of the transfer orbit 1 which makes the object enter the second transfer orbit.

A)

First transfer orbit

B)

Final orbit

C)

Parking orbit

D)

Second transfer orbit

Correct Answer : First transfer orbit

Explanation : In the first maneuver of the bi-elliptic transfer, the object is moves to the first transfer ellipse that has a radius much larger than the radius of the final circular orbit.

A)

First burn

B)

Second burn

C)

Third burn

D)

Fourth burn

Correct Answer : Third burn

Explanation : The third burn is retrograde i.e. the burn is opposite to the direction of the motion. This is done to slow down the object and circularize the orbit by lowering the perigee to the final orbit’s radius.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : False

Explanation : Despite making use of more than one engine burn, bi-elliptic maneuver is more efficient than the Hohmann transfer. It has a greater travel time, but requires less delta-v in total.

A)

Final orbit

B)

Parking orbit

C)

First transfer orbit

D)

Second transfer orbit

Correct Answer : Second transfer orbit

Explanation : In the first maneuver of the bi-elliptic transfer, the object is moves to the second transfer ellipse. This raises the perigee of the first transfer ellipse equal to the radius of the outer circle.

A)

Energy change

B)

Simple plane change

C)

Adjustment of perigee and apogee height

D)

Combined change of apsis altitude and plane orientation

Correct Answer : Energy change

Explanation : There are four types of single impulse maneuver-Adjustment of perigee and apogee height, simple rotation of the line of apsides, simple plane change and combined change of apsis altitude with plane orientation. There is no change in energy involved by applying single impulse.

A)

0.1 km/s

B)

0.25 km/s