Rocket Propulsion - Flight Performance Quiz(MCQ)
A)
One dimensional, curved path
B)
Two-dimensional, curved path
C)
One dimensional, straight line path
D)
Two-dimensional, straight line path

Correct Answer :   One dimensional, straight line path

Explanation : One dimensional, straight-line acceleration path is followed in such an environment as the only force acting on the rocket is its thrust and it acts in the flight direction. Under the influence of gravity, the flight path will become curved.

A)
Inert mass
B)
C)
Structural mass
D)
Propellant mass

Explanation : We account for residual propellant mass in inert mass section. Inert mass includes the masses of the engine system, like that of the nozzles, tanks, cases or unused, residual propellants. It doesn’t come under the useful propellant mass consumed for propulsion.

A)
Mass flow decreases, thrust increases
B)
Mass flow and thrust remains constant
C)
Mass flow increases, thrust decreases
D)
Mass flow decreases, thrust decreases

Correct Answer :   Mass flow and thrust remains constant

Explanation : Both mass flow and thrust remain constant in such an environment. Then for a given burning time `tb`, total mass expelled from the rocket can be determined from mass flow rate `m = m/tb` equation.

A)
Linear
B)
Parabolic
C)
Logarithmic
D)
Exponential

Explanation : Propellant mass fraction has a logarithmic effect on the vehicle velocity.
up = ?u = -c ln(1-ζ).

A)
Better nozzle cooling
B)
Higher chamber pressure
C)
Use of better propellants
D)
A more favorable nozzle area ratio

Correct Answer :   Better nozzle cooling

Explanation : Because doing any of these except better nozzle cooling will result in better Isp. Velocity increment is directly proportional to Isp, so more Isp means more velocity increment.

A)
20
B)
100
C)
200
D)
1000

Explanation : Multistage vehicles can have mass ratios that exceed 200. In this case, the rocket is segmented into multiple segments in which each of the segment is a separate propulsive system.

A)
20
B)
100
C)
200
D)
1000

Explanation : Single-stage rocket vehicles can have mass ratio up to about 20. To minimize weights and lateral loads, spherical shape of the rocket is desirable, although it may not be the easiest to manufacture.

A)
Higher chamber pressure
B)
Higher system inert mass
C)
More favorable nozzle area ratio
D)
Higher temperature at the inlet of the nozzle

Correct Answer :   Higher system inert mass

Explanation : This is because the effective propellant fraction is the ratio of the mass of propellants to the initial mass of the system. If inert mass(mf) increases, then the total initial mass(mo) will also increase as mo = mp + mf.

9 .
What is the effect of increasing mass ratio (mo/mb, where mo is the initial mass and mb is the burnout mass) on Isp?
A)
Increases
B)
Decreases
C)
No effect
D)
Increases only after a limit

Explaination : Isp increases because of the relation up = c ln(mo/mb). Here c is the exhaust velocity of the vehicle and ln denotes natural logarithm.

A)
Thrust
B)
Gravitational forces
C)
Aerodynamic forces
D)
Forces due to thermal stress

Correct Answer :   Forces due to thermal stress

Explanation : For a vehicle flying in the earth’s atmosphere, thrust, aerodynamic (lift and drag) and gravitational forces are the major external forces. Thermal stress inside an engine leads to internal forces only.

A)
1.5
B)
2.1
C)
3.2
D)
4.5

Explanation : The value of c can increase by a small factor between 1.2 to 1.6 as altitude is increased. Rocket motors produce greater thrust at higher altitudes due to the lower atmospheric pressure.

A)
Sum of all the fin areas
B)
Minimum cross-sectional area normal to the missile axis
C)
Maximum cross-sectional area normal to the missile axis
D)
The total surface area of the missile/rocket including the fins

Correct Answer :   Maximum cross-sectional area normal to the missile axis

Explanation : Area A denotes the maximum cross-sectional area perpendicular to the longitudinal axis of the missile. So for a cylindrical missile body having a radius r, A = πr2.

A)
less
B)
more
C)
equal
D)
any random value

Explanation : At lower atmospheric altitudes, the dynamic pressure experienced will be low because of the lower velocity of the rocket even though the air density is high. Dynamic pressure Pdyn is proportional to ρV2.

14 .
What is the loss of velocity increment experienced by a rocket as a result of drag in the atmosphere?
A)
5-10%
B)
15-20%
C)
25-30%
D)
35-40%

Explaination : For space launch vehicles and ballistic missiles, the drag loss is about 5 to 10% of the final vehicle velocity increment. It will be lesser for space vehicles than atmospheric vehicles like fighter aircraft because the time spent in the atmosphere is lesser for the space vehicles.

A)
angle of attack
B)
vehicle mass
C)
flight Mach number
D)
vehicle configuration

Explanation : lift and drag coefficients are primarily a function of vehicle configuration, flight Mach number, and angle of attack. CL depends on lift force and not vehicle mass.

A)
The angle between the local horizon and the missile axis
B)
The angle between the missile axis and its flight path
C)
The angle between the local horizon and the missile flight path
D)
The angle between the relative wind and the missile longitudinal axis

Correct Answer :   The angle between the missile axis and its flight path

Explanation : Angle of attack for a missile is the angle between its flight path and the missile axis. For the pitching movement of a cruise missile, a range of angle of attack of 15° to 24° is found to be effective.

A)
Thrust
B)
Moments
C)
Lateral forces
D)
Control forces

Explanation : Control forces, Lateral forces and moments are taken to be zero to ensure that the vehicle doesn’t turn or slip in flight. However, the vehicle may need thrust to propel forward.

A)
One-dimensional
B)
Two-dimensional
C)
Three-dimensional
D)
Four-dimensional

Explanation : The trajectory for such a vehicle is considered to be two-dimensional and contained within a plane. In real cases, the flight is much more complex and three-dimensional and requires computational usage for the trajectory analysis.

A)
pitch angle
B)
roll angle
C)
angle of attack
D)
flight path angle

Correct Answer :   angle of attack

Explanation : The vehicle wings are at some angle with the flight path. This angle is called the angle of attack. It plays a significant role in determining the lift generated under atmospheric conditions in the vehicle.

A)
True
B)
False
C)
Can Not Say
D)
None of the above

Explanation : If the flight is not along a vertical path, then there is a component of gravitational force on the vehicle that tries to slow it down. For an angle θ between the flight direction and the local horizontal, this force component is mgsinθ and it acts towards the center of the earth.

21 .
Determine the component of gravitational force in the direction of the flight path if the angle of the flight path with the horizontal is 30°, angle of direction of thrust with the horizontal is 20° and mass of the vehicle is 300 kg. Assume the acceleration due to gravity to be 9.8 m/s2.
A)
1470 N
B)
2250 N
C)
2980 N
D)
3343 N

Explaination : "target-id662786c5290d1" class="collapseomatic_content " style="">Answer: c
Explanation: Required component of gravitational force = mgsinθ.
Fgrav comp = 300 x 9.8 x sin(30) = 1470 N.

22 .
Determine the acceleration perpendicular to flight path for constant value of flight speed u = 100 m/s and instantaneous radius R = 200 m.
A)
50 m/s2
B)
100 m/s2
C)
125 m/s2
D)
200 m/s2

Explaination : Required acceleration perpendicular to flight path is atan = u2/R.
atan = 1002/200 = 50 m/s2.

23 .
Which of the following is acceleration perpendicular to the flight path?
A)
udθ/dt
B)
θdu/dt
C)
d(uθ)/dt
D)
d(u/θ)/dt

Explaination : udθ/dt is the correct representation of the acceleration perpendicular to the flight path. du/dt is the acceleration in the direction of the flight path. It has a contribution from the lift, propulsive forces, and the gravitational forces.

A)
parallel
B)
perpendicular
C)
at an acute angle
D)
at an obtuse angle

Explanation : Lift generation is normal to the direction of the flight path for such a vehicle. Lift is normal to the relative wind, while drag acts parallel to it and in the opposite direction.

A)
Propulsive
B)
Gravitational
C)
Aerodynamic
D)
Electromagnetic

Explanation : Propulsive, aerodynamic and gravitational forces are the major forces that contribute to the flight of an airborne vehicle. Propulsive forces include the thrust produced by the vehicle’s engine, while the aerodynamic forces include lift and drag forces.

A)
highly stable
B)
less energetic
C)
more energetic
D)
highly unstable

Explanation : Using a more energetic propellant increases the specific impulse of the vehicle. Higher specific impulse results in higher final velocity increment.

A)
linear
B)
parabolic
C)
logarithmic
D)
exponential

Explanation : R = m0/mf is the mass ratio, where m0 is the initial mass and mf is the burnout mass.
&#8710u; is proportional to c ln(R) where c is the exhaust velocity.

28 .
If flight velocity is u and the vehicle exhaust velocity is c, when is the propulsive efficiency maximum?
A)
u ≅ c
B)
u ? c
C)
u ? c
D)
c → ∞

Correct Answer :   u ≅ c

Explaination : Propulsive efficiency is maximum when the flight velocity and the exhaust velocity of the vehicle are closer to each other. This can be seen from the formula `ηprop = 2u/(u+c).`

A)
lighter vehicle structure
B)
greater vehicle mass ratio
C)
more vehicle structural mass
D)
greater complexity of the vehicle design

Correct Answer :   lighter vehicle structure

Explanation : A shorter rocket length means that the overall vehicle’s structural mass will be lesser. This is because of the reduction in the amount of material required in the construction of the vehicle. It will also mean more mass ratio compared to longer rockets because of lesser inert mass in shorter vehicles.

A)
Base drag
B)
Skin drag
C)
Form drag
D)
Shock drag

Explanation : Skin friction drag is the drag caused by the friction of flowing air over the vehicle outer surface. A polished and smooth outer surface will help in its reduction.

A)
lower propellant density
B)
higher propellant density
C)
higher propellant flammability
D)
lower propellant flammability

Correct Answer :   higher propellant density

Explanation : Higher propellant density will allow lesser tank volume for the storage of the same mass of the propellant. Hence the frontal area requirement will be less. Flammability has no role in controlling rocket frontal area.

A)
Using lighter structures
B)
C)
Using lighter guidance/ control devices
D)
Using a lesser portion of propellant for flight

Correct Answer :   Using a lesser portion of propellant for flight

Explanation : If a lesser portion of propellant is used for the flight of the vehicle, then the residual propellant mass will be more. This means that the final mass mf will be more than usual.

A)
Increasing the thrust
B)
C)
Increase in structure
D)
Increase in specific impulse

Correct Answer :   Increase in structure

Explanation : The increase in structure or propulsion system masses is kept minimum for increasing the initial mass of the rocket. The effective mass ratio will be more if the structural mass is not kept minimum.

A)
aerodynamic shape of the body
B)
aeroelastic stresses
C)
solid structures far upstream
D)
solid structures far downstream

Correct Answer :   aerodynamic shape of the body

Explanation : Form drag depends on the aerodynamic shape of the body in the flow field. A streamlined body like a slender body with a pointed nose has lesser drag compared to a blunt object.

35 .
Which of the following can the gravitational losses?
I) Reducing the burning time.
II) Increasing the angle between the normal to the local horizon and the flight path angle.
A)
Only I
B)
Only II
C)
Both I and II
D)
Neither I nor II

Explaination : Reducing the burning time will only reduce the gravitational losses. Keeping the flight path normal to the local horizon will also help in reducing gravitational losses

A)
1591.98 m/s
B)
2472.76 m/s
C)
3219.28 m/s
D)
7834.95 m/s

Explanation : Satellite velocity us can be obtained from mus2 = mg. Here g = gox(Ro/Ro+h)2
⇒ us = Ro√(go/Ro+h)
⇒ us = 6374 x 103 x √(9.8/6486000) = 7834.95 m/s.

37 .
Determine the time period of revolution of a satellite in a circular orbit around a planet of radius 6374 m at a height of 600 m. Assume the acceleration due to gravity at sea level to be 9.8 m/s2.
A)
25 mins
B)
30 mins
C)
35 mins
D)
40 mins

Correct Answer :   30 minsTime period τ = 2π(Ro+h)/us = 2π(Ro+h)3/2/(Ro/√go)
⇒ τ = 2 x 3.14 x 69743/2/(6374/√9.8) = 1797.2 s or ∼ 30 mins.

38 .
Determine the energy necessary to bring a unit mass into a circular orbit around the earth having a height of 170 km. Neglect the effects of drag.
A)
16 MJ
B)
19 MJ
C)
25 MJ
D)
32 MJ

Explaination : The required energy E = 1/2 Rogo x (Ro + 2h)/(Ro + h)
⇒ E = 0.5 x 6374 x 103 x 9.8 x (6714/6544) ≅ 32 MJ.

39 .
For an eccentricity of 0.92 and semi major axis length of 9500 km, determine the apogee velocity for a satellite in elliptical orbit. Take μ = 3.986 x 1014 m3/sec2.
A)
1233.33 m/s
B)
1322.21 m/s
C)
1349.54 m/s
D)
1439.45 m/s

Explaination : Apogee velocity ua = √(μ(1-e)/a(1+e))
ua = √(3.986 x 1014 x (0.08/1.92)/(9500 x 103)) = 1322.21 m/s.

A)
infinity
B)
apogee
C)
perigee
D)
focal point

Explanation : The point on the elliptical orbit closest to the focal point (where the planet is situated) is the perigee. At perigee, the satellite will have its maximum velocity and at apogee it will have its minimum velocity. It is evident from the principles of angular momentum conservation.

41 .
At an instant of time, a satellite in an elliptical orbit is at a distance of 500 km from the surface of the earth. If the semi major axis of the orbit is 8500 km in length, determine the magnitude of the satellite velocity. Take μ = 3.986 x 1014 m3/sec2
A)
8118.33 m/s
B)
8181.38 m/s
C)
8311.38 m/s
D)
8813.38 m/s

Explaination : u = [μ(2/R – 1/a)]1/2
u = [3.986 x 1014 x(2/6874000 – 1/8500000)]1/2
= 8311.38 m/s.

A)
899.5 km
B)
1751.6 km
C)
2134.3 km
D)
9053.1 km

Explanation : From the relation us = Ro√(go/Ro+h), using us = 7000 m/s, Ro = 6374 x 103 m, we have
h = [go/us2/Ro2] – Ro
h = 1751.6 km.

A)
lift forces
B)
drag forces
C)
propulsive forces
D)
gravitational forces

Explanation : The centrifugal forces should be equal to the gravitational forces for a satellite revolving around the earth in a circular orbit. If m denotes the satellite mass, us denotes the satellite velocity, and R represents the radius of the orbit, then mus2/R = mg.

A)
The orbit needs to be circular
B)
The orbit needs to be elliptical
C)
Drag losses are absent outside the atmosphere
D)
Within the atmosphere, the satellite spirals down to the earth

Correct Answer :   Drag losses are absent outside the atmosphere

Explanation : Outside the atmosphere, due to the absence of air, aerodynamic effects like lift and drag are non-existent. Within the atmosphere, additional energy is required to make up for the drag losses.

A)
Less than 500 km
B)
Less than 800 km
C)
More than 500 km
D)
More than 800 km

Correct Answer :   Less than 500 km

Explanation : LEO stands for Low Earth Orbits. They are typically less than 500 km in altitude.

A)
neutral point
B)
center of gravity
C)
center of pressure
D)
maneuvering point

Correct Answer :   center of gravity

Explanation : The thrust vector passes through the center of gravity for a rocket propulsion system. Because of that, no residual moments will be generated about the center of gravity.

A)
coasting
B)
cruising
C)
freefall
D)
translational maneuver

Explanation : Coasting is an example for an unpowered maneuver. Cruising involves maintaining a steady level horizontal flight at a constant altitude. Several powered and unpowered segments of the flight trajectory can be grouped together under translational maneuvers.

A)
Zero
B)
Infinite
C)
In the order of milli-Newtons
D)
In the order of mega-Newtons

Explanation : No thrust is required for a truly rotational maneuver. Only torque is generated in this case using couples.

A)
Circular maneuvers
B)
Translational maneuvers
C)
Truly rotational maneuvers
D)
Misaligned thrust maneuvers

Correct Answer :   Truly rotational maneuvers

Explanation : Reaction control systems typically help in rotational maneuvers. It requires a set of 4 thrusters for to rotate the vehicle about any one axis in either of the two directions.

A)
Lock-on
B)
Spacecrafts
C)
Rendezvous maneuver
D)
Heat shield

Explanation : Docking is the process in which two spacecrafts gradually approach each other and link together. It is a kind of rendezvous maneuver involving pulse node thrusters and low thrust. It is sometimes called lock-on.

A)
Reverse thrust
B)
Deploying heat shields
C)
The atmospheric drag
D)
Deploying landing parachutes

Explanation : In those locations where there is little to no atmosphere (like moon or Mercury), drag is not a reliable factor for slowing down the rocket. Reverse thrust is usually employed in such cases for the same purpose during the descent and the touchdown.

A)
orbit transfer maneuvers
B)
orbit maintenance maneuvers
C)
first stage propulsion control
D)

Correct Answer :   orbit maintenance maneuvers

Explanation : Station keeping maneuvers are orbit maintenance maneuvers. It keeps the spacecraft in its intended orbit and orbital position.

A)
Coasting
B)
Re-entry
C)
Launching
D)
Orbit Injection

Explanation : Orbital injection or transferring the rocket vehicle from one orbit to another requires a predetermined, accurate amount of specific impulse. Here the thrust levels are lower in comparison to the launch stage.

A)
Pulsing
B)
Low thrust
C)
Long duration
D)
Uses reaction control systems

Explanation : Inflight correction maneuvers are typically short duration maneuvers. They are intermittent (pulsing) operations performed with low thrust.

55 .
Depending on the mission profile, the most economical number of steps / expended stages for a multistage rocket is between _____ and ______
A)
2, 6
B)
2, 10
C)
4, 7
D)
4, 9

Explaination : Generally, the number of stages for a multistage rocket is between two and six stages. Using more than one stage permits higher vehicle velocities, higher payload fractions and better performance of long ranged ballistic missiles. But having too many stages will limit the initial take-off mass and the mechanisms and machinery involved become too complex.

56 .
If the velocity increments of the first, second, and third stages are 1000 m/s, 500 m/s and 250 m/s respectively, determine the total velocity increment.
A)
1375 m/s
B)
1750 m/s
C)
2750 m/s
D)
3500 m/s

Explaination : For an n-staged vehicle, the total velocity increment is the sum of the velocity increments of all the stages. So the final velocity increment is δuf = δu1 + δu2 + δu3 + …… .

A)
parallel
B)
partial
C)
tandem
D)
piggyback

Explanation : When two or more booster stages are attached to the lower stage of the vertical configuration of a rocket vehicle, such a configuration is called as parallel staging. In this case, the booster stages are “strapped” on to the lower stage and this allows an increase in the vehicle performance.

A)
True
B)
False
C)
Can Not Say
D)
None of the above

Explanation : The payload that can be carried to an orbit will decrease with an increase in the altitude of the orbit. Assuming the total mission energy to be constant, a smaller mass can be taken to a higher altitude. Earlier launch vehicles are upgraded to allow an increase in mission energy so that a larger mass can be carried as payload.

A)
Spacecraft
B)
Booster stage
C)
Sustainer stage
D)
Control and guidance system

Explanation : Of the total launch vehicle, the only part that goes into the orbit or deep space is the spacecraft. It is that part which carries the payload. Some of the spacecraft are designed to return back to the earth.

A)
increases
B)
decreases
C)
first increases and then decreases
D)
first decreases and then increases

Explanation : The initial stage of the rocket is usually the stage that requires the largest thrust and largest total impulse. The lower stage is often called as the booster stage. The thrust required becomes smaller for the upper stages or the sustainer stages.

A)
Zero stage
B)
Half stage
C)
Booster stage
D)
Sustainer stage

Explanation : The upper stages are usually called as the sustainer stages. Booster stages usually include the strap-on motors which after their use drop off from the attached lower stage. They are also called as half or zero stages. Typically, there are 2 to 6 strap-on motors used in a rocket engine.

A)
Docking
B)
Take-off
C)
Attitude control
D)
Small trajectory corrections

Explanation : Same engines can be used for different purposes. An example for one such engine is the small reaction control thruster for the Space Shuttle which serves the main purpose of attitude control, but in addition to it plays a role in docking and small trajectory corrections as well. It is however not used for take-off as it is a low thrust, high specific impulse delivering device.

A)
Outer space
B)
During launch
C)
Low altitude
D)
Above atmosphere

Explanation : In general, rocket exhaust plumes are pencil shaped when they are at low altitudes. This is because the exhaust gases are restricted from expanding against ambient air pressure.

A)
Shock layer at the nozzle exit
B)
Rapid combustion of injector fuel
C)
De-excitation of internal modes of gas molecules
D)
Excitation of molecules in the combustion chamber

Correct Answer :   De-excitation of internal modes of gas molecules

Explanation : The molecules are excited in the combustion chamber. But these excited molecules de-excite in their internal modes causing radiation to be emitted.

A)
near sea level
B)
Outer space
C)
Stratospheric region
D)
Just outside the atmosphere

Correct Answer :   near sea level

Explanation : Afterburning in mixing zones lead to higher temperatures and higher infrared emissions. Afterburning is dependent on altitude as it depends upon the amount of oxygen readily available in the atmosphere. Oxygen concentration decreases at higher altitudes. So higher infrared emissions are observed near sea level.

A)
Viscous inner core flow
B)
Inviscid outer core flow
C)
Inviscid inner core flow
D)
Viscous outer core flow

Correct Answer :   Viscous outer core flow

Explanation : Nozzle core flow can be divided into inviscid inner core flow and viscous outer core flow. The viscous outer core flow is also called the mixing layer. No chemical reactions are assumed to take place in the nozzle inner core flow.

67 .
If A is the pre-exponential factor, n is the temperature exponent, Ea is the activation energy, then the expression for forward rate constant according to Arrhenius law is ______
A)
kf = AT exp(-Ea/nRT)
B)
kf = ATn exp(-Ea/RT)
C)
kf = T exp(-EaA/RTn)
D)
kf = ATn exp(-Ea/RTn)

Correct Answer :   kf = ATn exp(-Ea/RT)

Explaination : The expression for the forward reaction is given by kf = ATn exp(-Ea/RT). This is given by the Arrhenius law, which plays an important role in the calculation of the rate of chemical reactions and activation energy.

A)
True
B)
False
C)
Can Not Say
D)
None of the above

Explanation : Exhaust plumes with strong emission signatures easily detectable. Afterburning will enhance such an emission. For stealth aircraft, this isn’t an ideal mode of operation, unless the emission signature can be well suppressed.

69 .
Which of the following correspond to the spectral bands of CO2 and H2O respectively in the IR signature of rocket exhaust plumes?
A)
2.7μm and 4.3μm
B)
3.4 μm and 7.2μm
C)
4.3 μm and 2.7 μm
D)
7.2 μm and 3.4μm

Correct Answer :   4.3 μm and 2.7 μm

Explaination : Spectral band of H2O is in the 2.7 μm region, while for CO2, it is 4.3 μm. In missile launches, these are the two principal bands observed in the atmospheric absorption.

70 .
Determine the chamber pressure for a rocket engine if the thrust coefficient is 1.8, the throat area is 0.2 m2 and the total thrust is 20000 N.
A)
45.67 kPa
B)
55.56 kPa
C)
61.33 kPa
D)
67.98 kPa

Explaination : Thrust force F = Pc At Cf, where Pc is the chamber pressure, At is the throat area and Cf is the thrust coefficient.
Then, Pc = F/(At Cf) = 20000/(0.2×1.8) = 55.56 kPa.

A)
anti-sloshing setup
B)
control surfaces
C)
reaction control system
D)
hinged multiple rocket nozzles

Explanation : Sloshing of liquid propellants in the tank can be prevented by anti-sloshing setup. But when the vehicle moves into an unstable state, it cannot control the vehicle and help it return back to the stable regime.

72 .
Which of the following statements is true for the center of gravity (cog) and the center of pressure (cop) location along the axis of a fuel powered aircraft?
A)
cog and cop both changes
B)
cog and cop both remain fixed
C)
cog remains fixed but cop changes
D)
cog changes but cop remains fixed

Correct Answer :   cog and cop both changes

Explaination : Center of gravity changes due to fuel consumption of the aircraft. Since aerodynamics moments vary with Mach number, center of pressure location is also not fixed.

73 .
Suppose an aircraft trimmed at a 10° angle of attack faces a gust that changes the angle of attack to 15°. What is the nature of the moment required to bring it back to the trimmed condition?
A)
Positive yawing moment
B)
Positive pitching moment
C)
Negative yawing moment
D)
Negative pitching moment

Correct Answer :   Negative pitching moment

Explaination : Negative pitching moment is a nose-down pitching moment which will act as the restoring moment in this situation. Yawing moment happens about the vehicle’s z-axis.

74 .
Determine the normal force coefficient of a rocket having a maximum frontal area of 0.2 m2, flying at sea level with a velocity of 100 m/s. Consider the center of gravity to be ahead of the center of pressure by 0.5 m and the corrective moment coefficient to be 2.5.
A)
0.004
B)
0.008
C)
0.012
D)
0.016

Explaination : C1 = 1/2 ρ V2 Ar Cna [Z – W], where C1 is the corrective moment coefficient, ρ is the ambient density, V is the velocity, Cna is the normal force coefficient, Z-W is the location of center of gravity ahead of center of pressure, and Ar is the reference area.
Then Cna = 2.5/(0.5×1.225×1002x0.2×0.5) = 0.004.

A)
neutral point
B)
center of gravity
C)
center of pressure
D)
maneuvering point

Correct Answer :   center of pressure

Explanation : Center of pressure is the point where all the aerodynamic forces balance. Lift and drag are the two aerodynamic forces acting on the rocket.

A)
Bullet trains
B)
Artificial satellites
C)
Commercial aircrafts
D)
Unguided rocket projectiles

Correct Answer :   Unguided rocket projectiles

Explanation : Unguided rocket projectiles can be given a rotation about its longitudinal axis by means of inclined fins or smaller rocket nozzles to stabilize it. This is an example of spin stabilization.

A)
1.266 hrs
B)
1.395 hrs
C)
1.525 hrs
D)
1.738 hrs

Explanation : T=2π(R0+h)3/2/√μ
= 2×3.14x(6728000)3/2/(3.986×1014)
= 1.525 hrs.

78 .
For an airplane undergoing a steady pull-up maneuver along a curved flight path of the constant angle of attack, constant pitch rate of 20°/s, and having a radius of curvature of 150 m, determine the flight velocity.
A)
38.33 m/s
B)
47.66 m/s
C)
52.36 m/s
D)
69.42 m/s