# Astronautics Questions and Answers – How Spacecraft Move in Orbit

This set of Astronautics Multiple Choice Questions & Answers (MCQs) focuses on “How Spacecraft Move in Orbit”.

1. In orbit around a planet, the centripetal force of the satellite’s revolution around the planet is played by _______________
a) the satellite’s inertia
b) the satellite’s momentum
c) the planet’s gravity
d) the satellite’s momentum

Explanation: Centripetal force, in general, is the inward force that sustains an object’s circular motion. In this case, the planet’s gravity plays the role of centripetal force. This is what keeps the satellite in orbit around the planet.

2. For an orbit to be maintained, an object must have sufficient ___________
a) sideways velocity
b) altitude
c) mass
d) size

Explanation: A high enough sideways or ‘tangential’ velocity is required for an object to be kept in orbit around a central body.Say a person standing on the ground throws a ball in the forward direction. Normally, the ball would fall back down after covering some horizontal distance. If the person swings the ball with more ferocity, the ball would trace a longer path before coming to rest on the ground. In either case, the ball follows a curved trajectory, but one which intersects the Earth. If the ball is thrown with sufficient velocity, the curvature of its trajectory would then match the curvature of the Earth. The ball would still be falling, but at the same time always ‘missing’ the Earth (unlike the previous scenarios wherein the curvature of the ball’s trajectory was always greater than Earth’s). This is only possible if the ball has enough sideways or tangential momentum. After a while, the ball (now said to have ‘orbital velocity’) would come back and hit that person at the back of his/her head after one revolution.

3. Weightlessness in Earth orbit results from ___________
a) a lack of gravity at orbital altitudes
b) continuous free-fall of objects in orbit
c) tangential velocity
d) the gravitational attraction from other external bodies

Explanation: Objects in orbit are continuously in free-fall and at the same time ‘missing’ the central body due to their tangential velocity. To understand how a free-fall is associated with weightlessness, consider the following situation: Imagine a person standing in an elevator stationed at the highest floor and, for some reason, the elevator cables suddenly fail and the elevator plummets to the ground. The person inside now has nothing to stand against and is literally in a state of free fall, experiencing weightlessness until the elevator impacts the ground. This analogy may be used to explain why astronauts aboard spacecraft encounter weightlessnesswhile in orbit.
At orbital altitudes, Earth’s gravity is just negligibly smaller than that on its surface, so a lack of gravity is not the correct explanation.

4. For a satellite to speed up in orbit, it must ___________
a) fire its thrusters in a direction opposite to that of its motion
b) fire its thrusters in the same direction as that of its motion
c) fire its thrusters in a direction perpendicular to that of its motion
d) stay put

Explanation: When the satellite fires its thrusters in a direction opposite to that of its motion, it loses energy and hence altitude. A lower altitude corresponds to a stronger gravitational attraction, so the satellite moves faster in order to continuously miss the central body and maintain its orbit.
If the satellite fired its thrusters in the same direction as that of its motion, it would reach a higher altitude which corresponds to a slightly weaker gravitational pull, meaning that the satellite would slow down to maintain an orbit.

5. ‘Sub-orbital’ refers to an orbit _____________
a) situated at lower altitudes
b) that does not occur around a central body
c) that is incomplete
d) that is open

Explanation: ‘Sub-orbital’ means an orbit that intersects the central body (a planet, for example) and does not form a complete loop. The trajectory of launch vehicles may be considered as sub-orbital flight since their orbits are of very short duration and pass through Earth’s surface (due to the fact that the launcher falls back to the ground).

6. For a spacecraft to be classified as a satellite, it must _________
a) be launched from the surface of a planet
b) be small and compact
c) orbit a central body
d) be a sizeable object

Explanation: A satellite is something which is bound to and orbiting a planet. A spacecraft is anything which moves about in outer space (whether or not it is bound to a central body or traversing the vastness of deep space in order to intercept another planet). Satellites may be thought of as a subset of spacecraft.

7. What is the difference between a satellite and spacecraft?
a) Both are entities which are manmade
b) A spacecraft is compulsorily man-made while a satellite may not be
c) All satellites are spacecraft, but not all spacecraft are satellites
d) There is no difference

Explanation: A spacecraft is a general term for man-made exploratory vehicles. A satellite, however, is something which orbits a larger and more massive central body. The moon, for instance, is a natural satellite of Earth. The hundreds of spacecraft circling our planet on the other hand are all ‘artificial satellites’ (i.e., man-made).

8. Which of the following organizations is not dedicated to spacecraft?
a) Virgin Atlantic
b) Boeing
c) NASA
d) ISRO

Explanation: Virgin Atlantic is a subsidiary of ‘Virgin Group’ that only deals with air travel. Boeing does in fact specialize in the manufacture of spacecraft in addition to airliners.

9. Two objects ‘A’ and ‘B’ revolve in the exact same circular orbits along the same direction, but are spatially separated by a fixed angular distance. In order for one of the two objects to intercept the other, which of the following maneuvers needs to be executed?
a) Either one should fire its thrusters in order to catch up with the other
b) Both must wait for a close approach
c) It is not possible for both to approach one another without changing orbits
d)B oth need to fire their thrusters

Explanation: Since both ‘A’ and ‘B’ are stuck in the same circular orbits and revolve in the same direction, it is not possible for either ‘A’ or ‘B’ to approach the other due to identical orbital velocities (like how two cars on a highway running at the same speed and with one behind the other never seem to get close to each other). If one of them attempts to fires its own thrusters hoping to catch up with the other, it will then end up in a higher or lower (depending on the direction of firing) orbit with a different height compared to the original. Note that firing an on-board thruster causes a change in orbital altitude and not a change in speed within the same orbit (remember that each orbit corresponds to a unique orbital velocity).

10. The gravitational potential energy of a satellite of mass ‘m’ in an orbit of radius ‘r’ around a massive body of mass ‘M’ is given by ____________
a) $$-\frac{GMm}{r}$$
b) $$\frac{GMm}{r}$$
c) $$\frac{GMm}{r2}$$
d) $$-\frac{Mm}{r^2}$$

Explanation: Gravitational potential energy is the energy possessed by any object of finite mass due to its position in a gravitational field. In other words, it is the work done by a force ‘F’ in bringing the object from infinite to that position in the gravity field.
We know that work done by a force is given by force×displacement. Let’s say that a force ‘F’ displaces an object by an amount ‘r’. The work done by the force is give byW=F×r.
The gravitational potential energy ‘U’ of an object within a gravitational field of a larger mass ‘M’ is the total work done in bringing the object of mass ‘m’ from infinite to its position in the gravity field, which lies at a height ‘r’ from the center of the massive body. So ‘U’ is nothing but the difference between the potential energy of the object at infinite and at ‘r’.
U = Wr=∞ – Wr=r
U = (F * r)r=∞ – (F * r)r=r
Here, ‘F’ is the gravitational force between ‘m’ and ‘M’. So we have
U = $$(\frac{GMm}{r^2} * r)_{r=∞} – (\frac{GMm}{r^2} * r)_{r=r}$$
U = $$(\frac{GMm}{r})_{r=∞} – (\frac{GMm}{r})_{r=r}$$
Substituting r=∞ and r=r in the first and second terms, respectively, we have
U = $$-\frac{GMm}{r}$$
As is evident, the gravitational potential energy is a negative quantity. The minus sign signifies that the object is bound to the central body in an orbit around it. From the equation, we also see that as the altitude ‘r’ increases, the gravitational potential energy becomes less negative, i.e., increases. So a higher orbit corresponds to a higher energy (since more work is being done in order to reach greater heights) while a lower orbit corresponds to a more ‘negative’ or lower energy (since relatively less work needs to be done in this case).

11. When a satellite’s orbit is boosted to a higher altitude, then _____________
a) both the kinetic and potential energies of the satellite increase
b) both the kinetic and potential energies of the satellite decrease
c) the kinetic energy increases while the potential energy decreases
d) the kinetic energy decreases while the potential energy increases

Explanation: The total energy of a satellite in orbit is always conserved and remains constant. It consists of the kinetic energy of the satellite plus its gravitational potential energy (or simply, potential energy). If the satellite is boosted to a higher orbit, the velocity of the satellite will decrease (a higher altitude corresponds to a lower velocity). This means that the kinetic energy of the satellite decreases. Since the total energy must remain unchanged, the decrease in kinetic energy is accompanied by an increase in potential energy in a way that keeps the total energy conserved.

12. The velocity of a satellite’s revolution in a circular orbit is related to the orbital radius ‘r’ as _________
a) v∝$$\frac{1}{r^2}$$
b) v∝$$\frac{1}{r}$$
c) v∝$$\frac{1}{\sqrt{r}}$$
d) v∝$$\frac{1}{r^3}$$

Explanation: In an orbit, the centripetal force that maintains the satellite’s circular motion is provided by the force of gravity between the central body and the satellite. If ‘m’ is the mass of the satellite in an orbit of radius ‘r’ revolving with velocity ‘v’, then the centripetal force is given by((mv2)/r). This is provided by the gravitational attraction between the satellite and the central body of mass ‘M’, given by(GMm/r2 ). Equating the two expressions, we may obtain a relationship between the velocity ‘v’ and radius ‘r’:
$$\frac{mv^2}{r} = \frac{GMm}{r^2} = v= \sqrt{\frac{GM}{r}}$$
We see that ‘v’ is inversely proportional to the square root of ‘r’.

13. The total energy of a satellite of mass 800 kg in an orbit 500 km above Earth’s surface is _________ (assume Earth’s radius as 6370 kilometers and mass as 6×1024 kg; G =6.67×10-11 m3kg-1s-2).
a) 4.66 Giga Joules
b) 46.6 Giga Joules
c) 2.33 Giga Joules
d) 23.3 Giga Joules

Explanation: We know that the energy of a satellite consists of its kinetic energy as well as potential energy. So
E=K.E+P.E
If ‘v’ is the speed with which a satellite of mass ‘m’ revolves in an orbit of radius ‘r’ around a central body of a large mass ‘M’, then we have
$$E=(\frac{1}{2}mv^2)+(-\frac{GMm}{r})$$
or
$$E=\frac{1}{2}mv^2-\frac{GMm}{r}$$
We know that $$v^2=\frac{GM}{r}$$
Substituting for v2 in the first term of the equation for ‘E’, we have
$$E=\frac{1}{2}m(\frac{GM}{r})-\frac{GMm}{r}$$
$$E=\frac{1}{2} \frac{GM}{r} -\frac{GMm}{r}$$
$$E= -\frac{1}{2}\frac{GMm}{r}$$
Substituting the given values: M = 6×1024 kg; m = 800kg; r = (6370+500)x103 m, we get
E = 2.33 x 1010 Joules
OR
E = 23.3 Giga Joules

14. A satellite orbiting in a near-Earth orbit has some orbital velocity. If the mass of the satellite is doubles, then its orbital velocity __________
a) doubles
b) halves
c) triples
d) does not change

Explanation: We know that
$$v = \sqrt{\frac{GM}{r}}$$
Clearly, ‘v’ is not a function of the satellite’s mass ‘m’.

15. Assuming the Moon revolves around the Earth in a circular orbit of radius 385,000 kilometers, the velocity with which it orbits is approximately ___________ (assuming Earth’s mass to be 5.972 x 1024 kg).
a) 1 km/s
b) 1 m/s
c) 10 km/s
d) 10 m/s

Explanation: We know that
$$v = \sqrt{\frac{GM}{r}}$$
r = 385000000 meters; M = 5.972 x 1024 kg. Substituting the values, we get v≈ 1017.5 meters per second, which may be rounded off to 1 kilometer per second.

Sanfoundry Global Education & Learning Series – Astronautics.

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