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.
Correct Answer : Inert 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.
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.
tb
m = m/tb
Correct Answer : Logarithmic
Explanation : Propellant mass fraction has a logarithmic effect on the vehicle velocity.up = ?u = -c ln(1-ζ).
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.
Correct Answer : 200
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.
Correct Answer : 20
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.
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.
Correct Answer : Increases
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.
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.
Correct Answer : 1.5
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.
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.
Correct Answer : less
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.
Correct Answer : 5-10%
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.
Correct Answer : vehicle mass
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.
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.
Correct Answer : Thrust
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.
Correct Answer : Two-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.
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.
Correct Answer : True
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.
Correct Answer : 1470 N
Explaination : "target-id662786c5290d1" class="collapseomatic_content " style="">Answer: cExplanation: Required component of gravitational force = mgsinθ.Fgrav comp = 300 x 9.8 x sin(30) = 1470 N.
Correct Answer : 50 m/s<sup>2</sup>
Explaination : Required acceleration perpendicular to flight path is atan = u2/R.atan = 1002/200 = 50 m/s2.
Correct Answer : udθ/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.
Correct Answer : perpendicular
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.
Correct Answer : 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.
Correct Answer : more energetic
Explanation : Using a more energetic propellant increases the specific impulse of the vehicle. Higher specific impulse results in higher final velocity increment.
Correct Answer : logarithmic
Explanation : R = m0/mf is the mass ratio, where m0 is the initial mass and mf is the burnout mass.∆u; is proportional to c ln(R) where c is the exhaust velocity.
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).
ηprop = 2u/(u+c).
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.
Correct Answer : Skin 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.
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.
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.
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.
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.
Correct Answer : Only I
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
Correct Answer : 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.
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.
Correct Answer : 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.
Correct Answer : 1322.21 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.
Correct Answer : perigee
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.
Correct Answer : 8311.38 m/s
Explaination : u = [μ(2/R – 1/a)]1/2u = [3.986 x 1014 x(2/6874000 – 1/8500000)]1/2= 8311.38 m/s.
Correct Answer : 1751.6 km
Explanation : From the relation us = Ro√(go/Ro+h), using us = 7000 m/s, Ro = 6374 x 103 m, we haveh = [go/us2/Ro2] – Roh = 1751.6 km.
Correct Answer : 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.
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.
Correct Answer : Less than 500 km
Explanation : LEO stands for Low Earth Orbits. They are typically less than 500 km in altitude.
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.
Correct Answer : coasting
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.
Correct Answer : Infinite
Explanation : No thrust is required for a truly rotational maneuver. Only torque is generated in this case using couples.
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.
Correct Answer : 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.
Correct Answer : Reverse thrust
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.
Correct Answer : orbit maintenance maneuvers
Explanation : Station keeping maneuvers are orbit maintenance maneuvers. It keeps the spacecraft in its intended orbit and orbital position.
Correct Answer : 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.
Correct Answer : Long duration
Explanation : Inflight correction maneuvers are typically short duration maneuvers. They are intermittent (pulsing) operations performed with low thrust.
Correct Answer : 2, 6
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.
Correct Answer : 1750 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 + …… .
Correct Answer : parallel
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.
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.
Correct Answer : Spacecraft
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.
Correct Answer : decreases
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.
Correct Answer : 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.
Correct Answer : Take-off
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.
Correct Answer : Low altitude
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.
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.
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.
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.
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.
Correct Answer : False
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.
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.
Correct Answer : 55.56 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.
Correct Answer : anti-sloshing setup
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.
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.
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.
Correct Answer : 0.004
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.
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.
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.
Correct Answer : 1.525 hrs
Explanation : T=2π(R0+h)3/2/√μ= 2×3.14x(6728000)3/2/(3.986×1014)= 1.525 hrs.
Correct Answer : 52.36 m/s
Explaination : Pitch rate q = V/RThen, V = qR = 20 x π/180 x 150 = 52.36 m/s.