# Rocket Propulsion Questions and Answers – Space Flight

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This set of Rocket Propulsion Multiple Choice Questions & Answers (MCQs) focuses on “Space Flight”.

1. Determine the velocity of a satellite in a circular orbit around the earth at a height of 112 km.
a) 3219.28 m/s
b) 2472.76 m/s
c) 1591.98 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.

2. 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) 30 mins
b) 25 mins
c) 35 mins
d) 40 mins

Explanation: Time 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.

3. 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) 25 MJ
b) 16 MJ
c) 32 MJ
d) 19 MJ

Explanation: 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.

4. What is the typical altitude of LEO?
a) more than 500 km
b) Less than 800 km
c) Less than 500 km
d) More than 800 km

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

5. Why are the orbits of satellites outside of the earth’s atmosphere?
a) The orbit needs to be elliptical
b) The orbit needs to be circular
c) Within the atmosphere, the satellite spirals down to the earth
d) 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.

6. For a circular trajectory of a satellite around the earth, the centrifugal forces must balance the ______
a) propulsive forces
b) gravitational forces
c) lift forces
d) drag 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.

7. Find the altitude of a satellite in a circular orbit around earth with a velocity of 7000 m/s.
a) 1751.6 km
b) 2134.3 km
c) 9053.1 km
d) 899.5 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.

8. 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) 8813.38 m/s
c) 8311.38 m/s
d) 8181.38 m/s

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

9. Satellite velocity is maximum at __________ for an elliptical orbit.
a) apogee
b) perigee
c) focal point
d) infinity

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.

10. 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) 1349.54 m/s
b) 1439.45 m/s
c) 1233.33 m/s
d) 1322.21 m/s

Explanation: 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.

Sanfoundry Global Education & Learning Series – Rocket Propulsion.

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