201. In a circular satellite orbit, the kinetic energy of a satellite of mass \(m\) around a planet of mass \(M\) at orbital radius \(r\) is
ⓐ. \(K=-\frac{GMm}{2r}\)
ⓑ. \(K=\frac{GMm}{2r}\)
ⓒ. \(K=\frac{GMm}{r^2}\)
ⓓ. \(K=\frac{GMm}{r}\)
Correct Answer: \(K=\frac{GMm}{2r}\)
Explanation: For a circular orbit, gravitational force supplies the required centripetal force. The force equation is \(G\frac{Mm}{r^2}=\frac{mv_o^2}{r}\). Multiplying both sides by \(r\) gives \(mv_o^2=\frac{GMm}{r}\). Kinetic energy is \(K=\frac{1}{2}mv_o^2\). Substituting \(mv_o^2=\frac{GMm}{r}\) gives \(K=\frac{GMm}{2r}\). The factor \(\frac{1}{2}\) comes from kinetic energy, not from the gravitational force formula.
202. For a satellite in a circular orbit, its gravitational potential energy is \(U=-\frac{GMm}{r}\) and its kinetic energy is \(K=\frac{GMm}{2r}\). Therefore,
ⓐ. \(U=+2K\)
ⓑ. \(U=\frac{K}{2}\)
ⓒ. \(U=-K\)
ⓓ. \(U=-2K\)
Correct Answer: \(U=-2K\)
Explanation: The potential energy of the satellite-planet system is \(U=-\frac{GMm}{r}\). The kinetic energy in a circular orbit is \(K=\frac{GMm}{2r}\). Comparing the two expressions shows that the magnitude of \(U\) is twice the kinetic energy. The negative sign is retained because gravitational potential energy is negative with zero at infinity. Thus \(U=-2K\), and this relation is specific to a circular gravitational orbit.
203. The total mechanical energy of a satellite in a circular orbit is
ⓐ. \(E=0\)
ⓑ. \(E=-\frac{GMm}{2r}\)
ⓒ. \(E=-\frac{GMm}{r^2}\)
ⓓ. \(E=+\frac{GMm}{2r}\)
Correct Answer: \(E=-\frac{GMm}{2r}\)
Explanation: The total mechanical energy is the sum of kinetic energy and potential energy. For a circular orbit, \(K=\frac{GMm}{2r}\) and \(U=-\frac{GMm}{r}\). So \(E=K+U=\frac{GMm}{2r}-\frac{GMm}{r}\). This simplifies to \(E=-\frac{GMm}{2r}\). The negative value shows that the satellite is in a bound orbit around the planet.
204. A satellite in a stable circular orbit has negative total mechanical energy mainly because
ⓐ. its potential energy is negative and larger in magnitude
ⓑ. its kinetic energy is negative and larger in magnitude
ⓒ. satellite mass becomes negative throughout the orbit
ⓓ. gravitational force is zero throughout the orbit
Correct Answer: its potential energy is negative and larger in magnitude
Explanation: In a circular gravitational orbit, the kinetic energy \(K=\frac{GMm}{2r}\) is positive. The gravitational potential energy \(U=-\frac{GMm}{r}\) is negative. Since \(|U|=2K\), the negative potential energy has a larger magnitude than the positive kinetic energy. Their sum is \(E=-\frac{GMm}{2r}\). The negative total energy means the satellite cannot escape to infinity unless extra energy is supplied.
205. Consider the following relations for a satellite in a circular orbit.
I. \(K=\frac{GMm}{2r}\)
II. \(U=-\frac{GMm}{r}\)
III. \(E=K+U=-\frac{GMm}{2r}\)
The correct set is
ⓐ. I only
ⓑ. I, II, and III
ⓒ. I and II only
ⓓ. II and III only
Correct Answer: I, II, and III
Explanation: Statement I follows from equating gravitational force to centripetal force and then using \(K=\frac{1}{2}mv_o^2\). Statement II is the gravitational potential energy of two masses with zero reference at infinity. Statement III is obtained by adding the kinetic and potential energies. The result is negative because the orbit is bound. These three equations are linked and should not be memorized as unrelated formulas.
206. A circular satellite has kinetic energy \(4.0\times10^9\,\text{J}\). Its gravitational potential energy and total mechanical energy are respectively
ⓐ. \(+8.0\times10^9\,\text{J}\), \(+12.0\times10^9\,\text{J}\)
ⓑ. \(+4.0\times10^9\,\text{J}\), \(-8.0\times10^9\,\text{J}\)
ⓒ. \(-8.0\times10^9\,\text{J}\), \(-4.0\times10^9\,\text{J}\)
ⓓ. \(-4.0\times10^9\,\text{J}\), \(0\)
Correct Answer: \(-8.0\times10^9\,\text{J}\), \(-4.0\times10^9\,\text{J}\)
Explanation: \( \textbf{Given kinetic energy:} \)
\[
K=4.0\times10^9\,\text{J}
\]
For a circular gravitational orbit,
\[
U=-2K
\]
So,
\[
U=-2(4.0\times10^9)
\]
\[
U=-8.0\times10^9\,\text{J}
\]
Total mechanical energy is
\[
E=K+U
\]
\[
E=4.0\times10^9-8.0\times10^9
\]
\[
E=-4.0\times10^9\,\text{J}
\]
\( \textbf{Final answer:} \) \(U=-8.0\times10^9\,\text{J}\) and \(E=-4.0\times10^9\,\text{J}\), with the negative total energy showing a bound orbit.
207. A satellite is shifted from a circular orbit of radius \(r\) to a circular orbit of radius \(4r\) around the same planet. Its total mechanical energy changes from \(E\) to
ⓐ. \(2E\)
ⓑ. \(4E\)
ⓒ. \(\frac{E}{4}\)
ⓓ. \(\frac{E}{2}\)
Correct Answer: \(\frac{E}{4}\)
Explanation: The total energy of a circular satellite orbit is
\[
E=-\frac{GMm}{2r}
\]
For fixed \(G\), \(M\), and \(m\), the total energy varies as
\[
E\propto-\frac{1}{r}
\]
If the radius becomes \(4r\), then
\[
E'=-\frac{GMm}{2(4r)}
\]
\[
E'=\frac{1}{4}\left(-\frac{GMm}{2r}\right)
\]
\[
E'=\frac{E}{4}
\]
Since \(E\) is negative, \(\frac{E}{4}\) is less negative and closer to zero.
\( \textbf{Final answer:} \) \(\frac{E}{4}\), meaning the higher circular orbit is less tightly bound.
208. A graph of total mechanical energy \(E\) of a circular satellite orbit against orbital radius \(r\) is described. The curve lies below the \(r\)-axis and approaches \(0\) as \(r\) increases. This matches
ⓐ. \(E=-\frac{GMm}{2r}\)
ⓑ. \(E=-\frac{GMm}{2r^2}\)
ⓒ. \(E=GMmr\)
ⓓ. \(E=+\frac{GMm}{2r}\)
Correct Answer: \(E=-\frac{GMm}{2r}\)
Explanation: The total energy of a circular gravitational orbit is \(E=-\frac{GMm}{2r}\). It is negative for any finite orbit radius, so the graph lies below the \(r\)-axis. As \(r\) increases, the magnitude of the negative energy decreases. Therefore the curve approaches \(0\) from below at very large radius. The negative sign and the inverse-distance dependence are both needed to describe the graph correctly.
209. A satellite in circular orbit has total energy \(E=-6.0\times10^8\,\text{J}\). The minimum external energy required to remove it from that orbit to infinity with zero final speed is
ⓐ. \(0\)
ⓑ. \(6.0\times10^8\,\text{J}\)
ⓒ. \(-6.0\times10^8\,\text{J}\)
ⓓ. \(3.0\times10^8\,\text{J}\)
Correct Answer: \(6.0\times10^8\,\text{J}\)
Explanation: \( \textbf{Initial total energy:} \)
\[
E_i=-6.0\times10^8\,\text{J}
\]
For removal to infinity with zero final speed,
\[
E_f=0
\]
The required external energy is the increase in mechanical energy:
\[
\Delta E=E_f-E_i
\]
Substitute:
\[
\Delta E=0-\left(-6.0\times10^8\right)
\]
\[
\Delta E=6.0\times10^8\,\text{J}
\]
This positive energy input is the binding energy of the satellite in that orbit.
\( \textbf{Final answer:} \) \(6.0\times10^8\,\text{J}\), because the satellite must be lifted from negative total energy to zero total energy.
210. Binding energy of a satellite in a circular orbit is best described as the energy required to
ⓐ. reduce its kinetic energy to zero at the same radius
ⓑ. keep it moving at constant speed for one full revolution
ⓒ. bring it from infinity to the planet with zero final speed
ⓓ. take it from orbit to infinity with zero final speed
Correct Answer: take it from orbit to infinity with zero final speed
Explanation: Binding energy is the positive energy that must be supplied to free a satellite from its bound orbit. A circular orbit has total energy \(E=-\frac{GMm}{2r}\). To reach infinity with zero speed, the final total energy must be \(0\). The energy required is \(0-E=\frac{GMm}{2r}\). It is not the energy needed to maintain the orbit, because ideal orbital motion under gravity does not require continuous fuel input.
211. The binding energy of a satellite in a circular orbit of radius \(r\) is
ⓐ. \(\frac{GMm}{r}\)
ⓑ. \(\frac{GMm}{r^2}\)
ⓒ. \(-\frac{GMm}{2r}\)
ⓓ. \(\frac{GMm}{2r}\)
Correct Answer: \(\frac{GMm}{2r}\)
Explanation: The total energy of a satellite in a circular orbit is \(E=-\frac{GMm}{2r}\). Binding energy is the amount of energy that must be supplied to make the total energy zero. Therefore binding energy is the positive magnitude of the total energy. It is \(\frac{GMm}{2r}\). The expression \(\frac{GMm}{r}\) is the magnitude of gravitational potential energy, not the binding energy of a circular orbit.
212. A satellite of mass \(500\,\text{kg}\) moves in a circular orbit where \(GM=4.0\times10^{14}\,\text{m}^3\text{s}^{-2}\) and \(r=2.0\times10^7\,\text{m}\). Its binding energy is
ⓐ. \(5.0\times10^9\,\text{J}\)
ⓑ. \(2.5\times10^9\,\text{J}\)
ⓒ. \(2.0\times10^{10}\,\text{J}\)
ⓓ. \(1.0\times10^{10}\,\text{J}\)
Correct Answer: \(5.0\times10^9\,\text{J}\)
Explanation: \( \textbf{Given data:} \)
\[
m=500\,\text{kg}
\]
\[
GM=4.0\times10^{14}\,\text{m}^3\text{s}^{-2}
\]
\[
r=2.0\times10^7\,\text{m}
\]
For a circular orbit, binding energy is
\[
E_b=\frac{GMm}{2r}
\]
Substitute:
\[
E_b=\frac{(4.0\times10^{14})(500)}{2(2.0\times10^7)}
\]
Calculate the numerator:
\[
(4.0\times10^{14})(500)=2.0\times10^{17}
\]
Calculate the denominator:
\[
2(2.0\times10^7)=4.0\times10^7
\]
So,
\[
E_b=\frac{2.0\times10^{17}}{4.0\times10^7}
\]
\[
E_b=5.0\times10^9\,\text{J}
\]
\( \textbf{Final answer:} \) \(5.0\times10^9\,\text{J}\), the positive energy needed to free the satellite from that circular orbit.
213. The energy required to make a satellite escape from a circular orbit of radius \(r\), without changing its position first, is equal to
ⓐ. twice its kinetic energy in that orbit
ⓑ. half its kinetic energy in that orbit
ⓒ. its kinetic energy in that orbit
ⓓ. zero because the satellite is already weightless
Correct Answer: its kinetic energy in that orbit
Explanation: In a circular orbit, the kinetic energy is \(K=\frac{GMm}{2r}\). The total energy is \(E=-\frac{GMm}{2r}\). To escape with zero final speed at infinity, the total energy must become \(0\). The required added energy is therefore \(\frac{GMm}{2r}\), which is equal to \(K\). Weightlessness does not mean absence of gravity or absence of binding energy.
214. A spacecraft in a circular orbit fires its engine briefly so that its speed becomes the local escape speed at the same radius. If the original orbital speed is \(v_o\), the required new speed is
ⓐ. \(\frac{v_o}{\sqrt{2}}\)
ⓑ. \(v_o\)
ⓒ. \(2v_o\)
ⓓ. \(\sqrt{2}v_o\)
Correct Answer: \(\sqrt{2}v_o\)
Explanation: At the same radius \(r\), circular orbital speed is
\[
v_o=\sqrt{\frac{GM}{r}}
\]
Escape speed at the same radius is
\[
v_e=\sqrt{\frac{2GM}{r}}
\]
Comparing these gives
\[
v_e=\sqrt{2}\sqrt{\frac{GM}{r}}
\]
\[
v_e=\sqrt{2}v_o
\]
The spacecraft must increase speed from \(v_o\) to \(\sqrt{2}v_o\), not to \(2v_o\).
\( \textbf{Final answer:} \) \(\sqrt{2}v_o\), because escape speed is based on energy while circular speed is based on centripetal force balance.
215. The following table gives energy quantities for a circular satellite orbit.
| Quantity | Expression | Sign |
|---|
| P. Kinetic energy | \(\frac{GMm}{2r}\) | Positive |
| Q. Potential energy | \(-\frac{GMm}{r}\) | Negative |
| R. Total energy | \(-\frac{GMm}{2r}\) | Negative |
| S. Binding energy | \(\frac{GMm}{2r}\) | Positive |
The table is
ⓐ. only P and Q are correct
ⓑ. only R and S are correct
ⓒ. only Q and R are correct
ⓓ. all four rows are correct
Correct Answer: all four rows are correct
Explanation: Kinetic energy is always positive for the moving satellite and equals \(\frac{GMm}{2r}\). Potential energy is negative because zero is chosen at infinity, so \(U=-\frac{GMm}{r}\). The total energy is the sum \(K+U=-\frac{GMm}{2r}\), making the circular orbit bound. Binding energy is the positive energy needed to raise the total energy to zero. The table correctly separates signed energy from energy that must be supplied.
216. Assertion: A satellite in a larger circular orbit has less negative total energy. Reason: For a circular orbit, \(E=-\frac{GMm}{2r}\).
ⓐ. Both assertion and reason are true, and the reason explains the assertion
ⓑ. The assertion is true, but the reason is false
ⓒ. Both assertion and reason are true, but the reason does not explain the assertion
ⓓ. The assertion is false, but the reason is true
Correct Answer: Both assertion and reason are true, and the reason explains the assertion
Explanation: The assertion is true because a larger orbital radius makes the magnitude \(\frac{GMm}{2r}\) smaller. Since the energy has a negative sign, the value moves closer to zero as \(r\) increases. The reason is the correct formula for total energy in a circular orbit. It directly explains why higher circular orbits are less tightly bound. A less negative total energy still represents a bound orbit as long as the energy remains below zero.
217. Two satellites of equal mass orbit the same planet in circular orbits. Satellite \(P\) has orbital radius \(r\), and satellite \(Q\) has orbital radius \(2r\). The ratio of their binding energies \(\frac{E_{bP}}{E_{bQ}}\) is
ⓐ. \(1\)
ⓑ. \(4\)
ⓒ. \(2\)
ⓓ. \(\frac{1}{2}\)
Correct Answer: \(2\)
Explanation: Binding energy for a circular orbit is
\[
E_b=\frac{GMm}{2r}
\]
For equal satellite masses around the same planet, \(GMm\) is the same for both.
Thus,
\[
E_b\propto\frac{1}{r}
\]
For satellite \(P\),
\[
E_{bP}=\frac{GMm}{2r}
\]
For satellite \(Q\), with radius \(2r\),
\[
E_{bQ}=\frac{GMm}{2(2r)}
\]
\[
E_{bQ}=\frac{GMm}{4r}
\]
Therefore,
\[
\frac{E_{bP}}{E_{bQ}}=\frac{GMm/(2r)}{GMm/(4r)}=2
\]
\( \textbf{Final answer:} \) \(2\), because the lower orbit is more strongly bound.
218. A claim says, “A satellite in orbit has zero total energy because it is not falling to Earth.” The best evaluation is that the claim is
ⓐ. correct, because gravitational potential energy vanishes in orbit
ⓑ. wrong, because a circular bound orbit has negative total energy
ⓒ. correct, because circular motion always has zero energy
ⓓ. wrong, because kinetic energy is always negative in orbit
Correct Answer: wrong, because a circular bound orbit has negative total energy
Explanation: A satellite in circular orbit is continuously falling around Earth, even though it does not hit the surface. Its kinetic energy is positive, but its gravitational potential energy is negative and twice as large in magnitude. Therefore the total energy is \(E=-\frac{GMm}{2r}\). Zero total energy corresponds to the minimum escape condition, not a circular bound orbit. Orbiting does not mean gravity or potential energy has disappeared.
219. A compact passage states: “A satellite is in a circular orbit of radius \(r\). Its speed is adjusted so that it just escapes from that same radius. No air resistance is present.” The extra kinetic energy needed is
ⓐ. \(\frac{GMm}{2r}\)
ⓑ. \(0\)
ⓒ. \(\frac{GMm}{r}\)
ⓓ. \(\frac{2GMm}{r}\)
Correct Answer: \(\frac{GMm}{2r}\)
Explanation: In the circular orbit, the total energy is
\[
E_i=-\frac{GMm}{2r}
\]
For just escaping to infinity with zero final speed, the final total energy must be
\[
E_f=0
\]
The extra energy needed is
\[
\Delta E=E_f-E_i
\]
\[
\Delta E=0-\left(-\frac{GMm}{2r}\right)
\]
\[
\Delta E=\frac{GMm}{2r}
\]
This is also equal to the kinetic energy of the original circular orbit.
\( \textbf{Final answer:} \) \(\frac{GMm}{2r}\), not \(\frac{GMm}{r}\), because the satellite already has orbital kinetic energy.
220. A circular satellite orbit has \(K=3.0\times10^{10}\,\text{J}\). After a maneuver, the satellite just escapes from the same radius. The minimum extra energy supplied by the engine is
ⓐ. \(3.0\times10^{10}\,\text{J}\)
ⓑ. \(1.5\times10^{10}\,\text{J}\)
ⓒ. \(9.0\times10^{10}\,\text{J}\)
ⓓ. \(6.0\times10^{10}\,\text{J}\)
Correct Answer: \(3.0\times10^{10}\,\text{J}\)
Explanation: \( \textbf{Initial circular-orbit kinetic energy:} \)
\[
K=3.0\times10^{10}\,\text{J}
\]
For a circular orbit,
\[
E=-K
\]
So the initial total energy is
\[
E_i=-3.0\times10^{10}\,\text{J}
\]
For just escape,
\[
E_f=0
\]
The minimum extra energy is
\[
\Delta E=E_f-E_i
\]
\[
\Delta E=0-\left(-3.0\times10^{10}\right)
\]
\[
\Delta E=3.0\times10^{10}\,\text{J}
\]
\( \textbf{Final answer:} \) \(3.0\times10^{10}\,\text{J}\), because raising the total energy from \(-K\) to \(0\) requires an energy input equal to \(K\).
