Unit 6: Energy and Momentum of Rotating Systems
Weightage : 10-15%
Two identical satellites orbit a planet of radius \(R\) in circular orbits A and C of radii \(3R\) and \(12R\), respectively, as shown above.
The speed of the satellite in orbit A is \(v_A\). The speed of the satellite in orbit C is \(v_C\). What is the ratio \(\dfrac{v_A}{v_C}\)?
Correct Answer: \( \boxed{\mathrm{C}} \)
For a satellite in a circular orbit, the gravitational force provides the required centripetal force:
\(\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}\)
Solving for the orbital speed,
\(v=\sqrt{\dfrac{GM}{r}}\)
Therefore,
\(v_A=\sqrt{\dfrac{GM}{3R}}\)
and
\(v_C=\sqrt{\dfrac{GM}{12R}}\)
Taking the ratio,
\[ \frac{v_A}{v_C} = \sqrt{\frac{GM/(3R)}{GM/(12R)}} = \sqrt{\frac{12R}{3R}} = \sqrt{4} =2 \]
Thus,
\(\boxed{\dfrac{v_A}{v_C}=\dfrac{2}{1}}\)
Therefore, the correct answer is (C).
A satellite of mass \(120\,\mathrm{kg}\) is in a circular orbit at a height \(h=R\) above the surface of Earth, where \(R\) is the radius of Earth.
The radius and mass of Earth are \(6.4\times10^6\,\mathrm{m}\) and \(6.0\times10^{24}\,\mathrm{kg}\), respectively.
The gravitational potential energy of the satellite-Earth system is most nearly
Correct Answer: \( \boxed{\mathrm{B}} \)
The gravitational potential energy of a satellite-Earth system is
\(U=-\dfrac{GMm}{r}\)
where \(r\) is the distance from the center of Earth to the satellite.
Since the satellite is at a height \(h=R\),
\(r=R+h=2R=2(6.4\times10^6\,\mathrm{m})=1.28\times10^7\,\mathrm{m}\)
Substituting the given values,
\(U=-\dfrac{(6.67\times10^{-11}\,\mathrm{N\cdot m^2/kg^2})(6.0\times10^{24}\,\mathrm{kg})(120\,\mathrm{kg})}{1.28\times10^7\,\mathrm{m}}\)
\(U\approx-3.75\times10^9\,\mathrm{J}\)
The negative sign indicates that the satellite is gravitationally bound to Earth.
Therefore, the correct answer is (B).
Identical satellites X and Y of mass \(m\) are in circular orbits around a planet of mass \(M\). The radius of the planet is \(R\). Satellite X has an orbital radius of \(3R\), and satellite Y has an orbital radius of \(4R\). The kinetic energy of satellite X is \(K_X\).
In terms of \(K_X\), the gravitational potential energy of the planet-satellite X system is
Correct Answer: \( \boxed{\mathrm{A}} \)
For a satellite in a circular orbit, the gravitational force provides the centripetal force:
\( \frac{GMm}{r^2}=\frac{mv^2}{r} \)
Hence,
\( v^2=\frac{GM}{r} \)
The kinetic energy of satellite X is
\( K_X=\frac{1}{2}mv^2=\frac{GMm}{2r} \)
The gravitational potential energy of the planet-satellite X system is
\( U=-\frac{GMm}{r} \)
Comparing the two expressions,
\( U=-2K_X \)
Thus, the gravitational potential energy of the planet-satellite X system is equal to \(-2K_X\). Therefore, the correct answer is (A).
