Unit 6: Energy and Momentum of Rotating Systems
Weightage : 10-15%
A satellite orbits the moon in a circle of radius \(R\). For the satellite to double its speed but maintain a circular orbit, what must the new radius of its orbit be?
Correct Answer: \( \boxed{\mathrm{C}} \)
For a satellite in a circular orbit, gravity provides the centripetal force:
\( \frac{mv^2}{r}=\frac{GMm}{r^2} \)
Canceling \(m\) and solving for \(r\):
\( v^2=\frac{GM}{r} \)
\( r=\frac{GM}{v^2} \)
This shows that the orbital radius is inversely proportional to the square of the orbital speed:
\( r\propto \frac{1}{v^2} \)
If the speed is doubled,
\( v_{\text{new}}=2v \)
then
\( r_{\text{new}}=\frac{GM}{(2v)^2} \)
\( r_{\text{new}}=\frac{GM}{4v^2} \)
\( r_{\text{new}}=\frac{1}{4}R \)
Therefore, to double its orbital speed while remaining in a circular orbit, the satellite must orbit at one-fourth the original radius.
Hence, the correct answer is \( \boxed{\mathrm{C}} \).
If the speed of a satellite orbiting Earth at a distance \(r\) from the center of Earth is \(v\), what is the speed of a second satellite orbiting Earth at a distance \(2r\) from the center of Earth?
Correct Answer: \( \boxed{\mathrm{C}} \)
For a satellite in a circular orbit, the gravitational force provides the centripetal force:
\( F_g = F_c \)
\( G\frac{Mm}{r^2}=\frac{mv^2}{r} \)
Canceling \(m\) and solving for \(v\):
\( v=\sqrt{\frac{GM}{r}} \)
For the second satellite, the orbital radius is \(2r\):
\( v_2=\sqrt{\frac{GM}{2r}} \)
Factoring out \(2\):
\( v_2=\frac{1}{\sqrt{2}}\sqrt{\frac{GM}{r}} \)
Since \(v=\sqrt{\frac{GM}{r}}\),
\( v_2=\frac{v}{\sqrt{2}} \)
Therefore, the speed of the second satellite is \( \boxed{\frac{1}{\sqrt{2}}v} \).
Hence, the correct answer is \( \boxed{\mathrm{C}} \).
A rocket has landed on Planet X, which has half the radius of Earth. An astronaut onboard the rocket weighs twice as much on Planet X as on Earth. If the escape velocity for the rocket taking off from Earth is \(u_0\), then its escape velocity on Planet X is
Correct Answer: \( \boxed{\mathrm{C}} \)
The escape velocity from a planet is given by
\( u=\sqrt{\frac{2GM}{R}} \)
We are told that Planet X has radius
\( R_X=\frac{R_E}{2} \)
and that an astronaut weighs twice as much on Planet X as on Earth.
Since weight is proportional to gravitational acceleration,
\( g_X=2g_E \)
Using
\( g=\frac{GM}{R^2} \),
\( \frac{G M_X}{R_X^2}=2\frac{G M_E}{R_E^2} \)
Substituting \(R_X=\frac{R_E}{2}\),
\( \frac{G M_X}{(R_E/2)^2}=2\frac{G M_E}{R_E^2} \)
\( \frac{4GM_X}{R_E^2}=2\frac{GM_E}{R_E^2} \)
\( M_X=\frac{M_E}{2} \)
Therefore,
\( u_X=\sqrt{\frac{2G(M_E/2)}{R_E/2}} \)
\( u_X=\sqrt{\frac{GM_E}{R_E/2}} \)
\( u_X=\sqrt{\frac{2GM_E}{R_E}} \)
But
\( u_0=\sqrt{\frac{2GM_E}{R_E}} \)
Hence,
\( u_X=u_0 \)
Therefore, the escape velocity on Planet X is the same as that on Earth.
Hence, the correct answer is (C).
