Correct Answer :   Helium

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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.

Correct Answer :   Hydrogen

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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.

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

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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.

Correct Answer :   False

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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.

Correct Answer :   Vertical variations of thermodynamic properties only

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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.

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

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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.

Correct Answer :   Nitrogen

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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.

Correct Answer :   Core of the planet

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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.

Correct Answer :   Mercury

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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.

Correct Answer :   78% and 21%

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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

Correct Answer :   Troposphere

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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.

Correct Answer :   216.65 K

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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.

Correct Answer :   98.46 km

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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

Correct Answer :   95.5% of gravity at MSL

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Explaination : Given,
Gravity at MSL (g₀) = 9.81 m/s²
Geopotential altitude (h) = 300 km
Radius of earth (R_E) = 6378 km
Gravity at 300 km above MSL (g) = g₀.(R_E/(R_E + h))²
= 9.81*(6378/(6378+300))²
= 9.718 m/s²
Percentage = (9.718/9.81) x 100% = 95.5% of MSL gravity

Correct Answer :   250.17 K

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Explaination : Given,
Temperature gradient (a) = 0.001 K/m = 1 K/km
At point 1, h₁ = 20 km, T₁ = 216.65 K
At point 2, h₂ = 32 km, T₂ =?
T₂ = a.(h₂ – h₁) + T₁
= 1*(32-20) + 216.65
= 228.65 K

Correct Answer :   0.365 kg/m³

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Explaination : Given,
At MSL, Temperature (T₁) = 288.16 K
Pressure (p₁) = 101325 Pa
Density (ρ₁) = 1.225 kg/m³
At 11 km, Temperature (T) = 216.66 K
Pressure (p) = 22700 Pa
Density (ρ) =?
From the equation of state,
p/p₁ = (ρT)/(ρ₁/T₁)
ρ = (pρ₁T₁)/(p₁T)
= (22700*1.225*288.16)/(101325*216.66)
= 0.365 kg/m³

Correct Answer :   False

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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.

Correct Answer :   Rocket Engine

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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.

Correct Answer :   Mesopause

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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.

Correct Answer :   1.54 x 10^-4 per meter

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Explaination : Given,
Acceleration due to gravity (g) = 9.81 m/s²
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

Correct Answer :   390 ⁰R

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Explaination : Given,
Pressure (p) = 629.66 lb/ft²
Density (ρ) = 0.9408 x 10^-3 slug/ft³
Specific gas constant (R) = 1716 lb.ft/slug.⁰R
Using equation of state,
Temperature, T = p/(ρR)
= 629.66/(0.9408*10^-3*1716)
= 390 ⁰R

Correct Answer :   Karman line

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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.

Correct Answer :   262 K

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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

Correct Answer :   Center of mass of the Earth

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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.

Correct Answer :   23.4°

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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.

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

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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.

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

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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.

Correct Answer :   Vernal equinox

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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.

Correct Answer :   Horizontal system

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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.

Correct Answer :   Celestial Horizon

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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.

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

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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.

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

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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.

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

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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).

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

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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.

Correct Answer :   False

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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.

Correct Answer :   Vertical circles

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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.

Correct Answer :   Altitude, Azimuth

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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.

Correct Answer :   180° = West point

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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.

Correct Answer :   True

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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.

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

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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.

Correct Answer :   0° = Celestial Horizon

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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.

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

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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.

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

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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°.

Correct Answer :   Heliocentric system

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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.

Correct Answer :   Center of the Sun

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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.

Correct Answer :   Ecliptic Plane

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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.

Correct Answer :   False

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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.

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

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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.

Correct Answer :   It is constant for a fixed object

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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.

Correct Answer :   Azimuth

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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.

Correct Answer :   In the direction of the Vernal Equinox

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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.

Correct Answer :   In the direction of the Winter Solstice

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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.

Correct Answer :   In direction of the Celestial North Pole

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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.

Correct Answer :   Angular rotation

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

Correct Answer :   (-x, y)

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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.

Correct Answer :   Orthogonal matrix

Correct Answer :   3

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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.

Correct Answer :   Fixed Wireless

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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.

Correct Answer :   5.85 km/s

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Explaination : Given,
Distance between Earth and satellite (r) = 5250 + 6378
= 11,628 km
Standard gravitational parameter of Earth (μ) = 398,600 km³/s²
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

Correct Answer :   42,647 km³/s²

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Explaination : Given,
Force of attraction between Earth and Mars (F) = 6.39 x 10²³ N
Distance between Earth and Mars (r) = 225 x 10⁶ km
Standard Gravitational Parameter of Earth (μ_E) = 398,600 km³/s²
Universal gravitational constant (G) = 6.67408 × 10^-11 m³/(kg.s²)
From Newton’s law of gravitation,
Mass of Mars (m) = (Fr²)/μ_E
= (6.39*10²³*(225*10⁶)²)/(398,600*10⁹)
= 6.39 x 10²³ kg
Standard Gravitational Parameter of Earth (μ_M) = G.m
= 6.67408*10^-11*6.39*10²³
= 42,647 km³/s²

Correct Answer :   1.771 x 10²⁶ N

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Explaination : Given,
Standard gravitational parameter of earth (μ) = 398,600 km³/s²
Mass of moon (m) = 7.348 x 10²² kg
Distance between earth and moon (r) = 384,400 km
Force of attraction (F) = (μ.m) / (r)²
= (398,600 x 7.348 x 10²²) / (384,400)²
= 1.982 x 10¹⁷ N

Correct Answer :   9.78 m/s²

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Explaination : Given,
Acceleration due to gravity at MSL (g₀) = 9.81 m/s²
Radius of Earth (R_E) = 6378 km
Height above the surface (h) = 10 km
Acceleration due to gravity at 10 km (g) = g₀ / (1 + h / R_E)²
= 9.81/(1+10/6378)²
= 9.78 m/s²

Correct Answer :   5.85 m/s²

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Explaination : Given,
Thrust (T) = 31 x 10⁶ N
Mass of rocket (m) = 1.98 x 10⁶ kg
Resultant Force (F) = T-mg
= 31*10⁶-(1.98*10⁶*9.81)
= 11.57 x 10⁶ N
From Newton’s second law,
Resultant instantaneous acceleration (a) = F/m
=(11.57*10⁶)/(1.98*10⁶)
= 5.85 m/s²

Correct Answer :   N-body problem

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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.

Correct Answer :   False

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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.

Correct Answer :   6N

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

Correct Answer :   3N

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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.

Correct Answer :   Mass of one of the bodies is negligible

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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.

Correct Answer :   10^-3 g’s

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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.

Correct Answer :   One-body problem

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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.

Correct Answer :   Closed form solution does not exist

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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.

Correct Answer :   7.08 km/s

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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

Correct Answer :   3.47 km/s

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Explaination : Standard gravitational parameter (μ) = 398,600 km³/s²
Specific angular momentum (h) = 57,112 km²/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

Correct Answer :   6.98 km/s

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Explaination : Given,
Standard gravitational parameter (μ) = 398,600 km³/s²
Specific angular momentum (h) = 57,112 km²/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

Correct Answer :   vis-viva

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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.

Correct Answer :   -30.3 km³/s²

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Explaination : Given,
Standard gravitational parameter (μ) = 398,600 km³/s²
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 km³/s²

Correct Answer :   True

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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.

Correct Answer :   22.62 x 10⁶ km²

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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 10⁶ km²

Correct Answer :   1.633 km/s

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Explaination : Standard gravitational parameter (μ) = 4903 km³/s²
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

Correct Answer :   2931.8 s

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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

Correct Answer :   72,876.32 km²/s

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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 km²/s

Correct Answer :   h and e

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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.

Correct Answer :   Curvilinear

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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.

Correct Answer :   Tends to zero

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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.

Correct Answer :   Eccentricity

Correct Answer :   True

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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.

Correct Answer :   0

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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.

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

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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.

Correct Answer :   Vernal equinox

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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.

Correct Answer :   Nadir

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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.

Correct Answer :   Zenith

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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.

Correct Answer :   Declination

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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.

Correct Answer :   Ephemeris

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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.

Correct Answer :   East to West

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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.

Correct Answer :   Retrograde

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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.

Correct Answer :   6

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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.

Correct Answer :   Circular

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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.

Correct Answer :   Satellite’s location in the orbit

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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).

Correct Answer :   Orbit’s orientation in the orbital plane

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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).

Correct Answer :   52°

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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.

Correct Answer :   1 ≤ e

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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.

Correct Answer :   Undefined

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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.

Correct Answer :   Time used as reference for mentioning orbital elements

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

Correct Answer :   Semi-major axis

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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.

Correct Answer :   Eccentricity

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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.

Correct Answer :   u₀ = ω + v₀

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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 u₀ at epoch is the angle between the ascending node and radius vector to the satellite at some time t and true anomaly at epoch v₀ is the angle between periapsis and the position of satellite at a particular time t called epoch.
They are related by:
u₀ = ω + v₀

Correct Answer :   False

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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.

Correct Answer :   4

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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.

Correct Answer :   0.44

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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

Correct Answer :   0.691

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Explaination : Specific energy (ε) = -10.29 km²/s²
Specific angular momentum (h) = 10,112 km²/s
Standard gravitational parameter (μ) = 398,600 km³/s²
Using the e-h-ε relation
e = (1+(2εh)/μ)^1/2
= (1+(2*-10.29*10,112)/398,600)^1/2
= 0.691

Correct Answer :   Parabolic

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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.

Correct Answer :   Elliptical

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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.

Correct Answer :   0.2098

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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

Correct Answer :   5250 km

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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

Correct Answer :   Heliocentric-Ecliptic System

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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.

Correct Answer :   Argument of periapsis

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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.

Correct Answer :   0.822

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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

Correct Answer :   True-anomaly-averaged radius

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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.

Correct Answer :   64.16°

Correct Answer :   52,910 km

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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

Correct Answer :   0.648 km/s

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Explaination : Gravitational Parameter (μ) = 398,600 km³/s²
Semi-major axis (a) = 50,000 km
Eccentricity (e) = 0.9
From orbit equation for ellipse,
Specific angular momentum (h) = (μa(1-e²))^1/2
= (398,600*50,000*(1-0.9²))^1/2
= 61,536 km²/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

Correct Answer :   40,450 km

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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

Correct Answer :   9.63 km/s

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Explaination : Specific angular momentum (h) = 72,000 km²/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

Correct Answer :   20.518 hours

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Explaination : Apogee radius (r_a) = 69,000 km
Perigee radius (r_p) = 7,100 km
Gravitational Parameter (μ) = 398,600 k³/s²
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

Correct Answer :   -4.927 km²/s²

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Explaination : Apogee radius (r_a) = 73,000 km
Perigee radius (r_p) = 7,900 km
Gravitational Parameter (μ) = 398,600 km³/s²
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 km²/s²

Correct Answer :   12.307 km/s

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Explanation : Gravitational Parameter (μ) = 398,600 km³/s²
Semi-major axis (a) = 50,000 km
Eccentricity (e) = 0.9
From orbit equation for ellipse,
Specific angular momentum (h) = (μa(1 – e²))^1/2
= (398,600*50,000*(1-0.9²))^1/2
= 61,536 km²/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

Correct Answer :   0.89

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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

Correct Answer :   equal to

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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.

Correct Answer :   0°

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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.

Correct Answer :   True

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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.

Correct Answer :   0.0

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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.

Correct Answer :   0°

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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.

Correct Answer :   24.41 km/s

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Explaination : Initial velocity (v_i) = 27 km/s
Gravitational Parameter (μ) = 42,828 km³/s²
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

Correct Answer :   9.599 km/s

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Explaination : Specific angular momentum (h) = 47,862.73 km²/s
Gravitational parameter (μ) = 324,859 km³/s²
Escape velocity (v_esc) = 2^1/2(μ/h)
= 2^1/2*(324,859/47,862.73)
= 9.599 km/s

Correct Answer :   inversely proportional to square root of

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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.

Correct Answer :   10.38 km/s

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Explaination : Specific angular momentum (h) = 87,899 km²/s
Eccentricity (e) = 2.5
Gravitational Parameter (μ) = 398,600 km³/s²
Hyperbolic excess velocity (v_∞) = (μ/h)(e² – 1)^1/2
= (398,600/87,989)(2.5² – 1)^1/2
= 10.38 km/s

Correct Answer :   10,984 km

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Explaination : Radius of satellite (r) = 36,613 km
Standard gravitational parameter (μ) = 398,600 km³/s²
Eccentricity (e) = 1.2
We know for hyperbolic trajectories,
(Satellite speed)² = (Hyperbolic Excess Speed)² + (Escape Velocity)²
v² = v_∞² + v_esc²
4v_∞² = v_∞² + 2μ/r
3v_∞² = 2μ/r
r = (2μ)/(3v_∞²)
We know,
v_∞² = (μ/h)²(e² – 1)
Substituting back into equation for r, we get,
r = (2/3)(h²/μ)(1/e² – 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

Correct Answer :   9.74 km/s

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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_p² – v_esc²)^1/2
= (14.3² – 10.466²)^1/2
= 9.744 km/s

Correct Answer :   145,560 km²/s

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Explaination : Specific angular momentum = Perigee radius x Perigee velocity
= 7278 x 20
= 145,560 km²/s

Correct Answer :   9.562 km/s

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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 km²/s
Radial velocity (v_r) = (μ/h)esin(θ)
= (398,600/94,422.69)*2.3*sin(100°)
= 9.562 km/s

Correct Answer :   Characteristic

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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.

Correct Answer :   56.44°

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Explanation : Eccentricity (e) = 1.2
Turn angle = 2*asin(1/e)
= 2*asin(1/1.2)
= 56.44°

Correct Answer :   20,000 km

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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

Correct Answer :   True

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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.

Correct Answer :   663.32 km

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Explaination : Eccentricity (e) = 1.2
Specific Energy (ε) = 199.3 km²/s²
Gravitational parameter (μ) = 398,600 km³/s²
We know,
Semi-major axis (a) = μ/(2ε)
= 398,600/(2*199.3)
= 1000 km
Aiming radius (Δ) = a*(e² – 1)^1/2
= 1,000*(1.2² – 1)^1/2
= 663.32 km

Correct Answer :   830.66 km

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Explaination : Eccentricity (e) = 1.3
Specific Energy (ε) = 199.3 km²/s²
Gravitational parameter (μ) = 398,600 km³/s²
We know,
Semi-major axis (a) = μ/(2ε)
= 398,600/(2*199.3)
= 1000 km
Aiming radius (Δ) = a*(e² – 1)^1/2
= 1,000*(1.3² – 1)^1/2
= 830.66 km

Correct Answer :   Low Earth Orbit

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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.

Correct Answer :   Equatorial Low Earth Orbit

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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.

Correct Answer :   Low Earth Orbit

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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.

Correct Answer :   Medium Earth Orbit

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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.

Correct Answer :   Tundra Orbit

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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.

Correct Answer :   Apparent Retrograde motion

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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.

Correct Answer :   Molniya Orbit

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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.

Correct Answer :   Geosynchronous Orbit

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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.

Correct Answer :   Medium Earth Orbit

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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.

Correct Answer :   Change of inclination of the orbits

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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.

Correct Answer :   Major-axis becomes twice

Correct Answer :   Hohmann transfer

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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.

Correct Answer :   Magnitude of angular momentum

Correct Answer :   Hohmann transfer

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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.

Correct Answer :   2

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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.

Correct Answer :   3.933

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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

Correct Answer :   3

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Explanation : In order to carry out Bi-Elliptical Hohmann tranfer 3 burns are required. 1^st burn is to convert the circular orbit at r₁ to elliptical orbit with perigee r_p = r₁ and apogee r_a = r_*. The second burn is carried out at the apogee to raise the perigee of elliptical orbit to r₂. Finally at perigee, the orbit is circulized by lowering the perigee to r₂.

Correct Answer :   True

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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.

Correct Answer :   Elliptical

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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.