Gravitation Questions and Answers

Answer: D

The periods are in the ratio \(2\sqrt{2} : 3\sqrt{3}\).

Kepler's third law: \(T^{2} \propto a^{3}\), where a is the semi-major axis. The ratio of semi-major axes equals the ratio of major axes, 2 : 3.

\[ \begin{aligned} \frac{T_{1}}{T_{2}} &= \left(\frac{2}{3}\right)^{3/2} \\ &= \frac{2\sqrt{2}}{3\sqrt{3}} \end{aligned} \]

4 : 9 uses the power 2 instead of \(\frac{3}{2}\), and 2 : 3 assumes T is proportional to a. Option C mixes the roots: \(2\sqrt{3} : 3\sqrt{2}\) equals \(\sqrt{2} : \sqrt{3}\), the square root of the ratio.

Answer: D

Its gravitational potential energy decreases.

For a satellite of mass \(m\) in an orbit of radius \(r\), \(U = -\frac{GMm}{r}\). Moving to a smaller \(r\) makes \(U\) more negative, so the potential energy decreases.

This is the same as for any object falling closer to the Earth. It does not become zero (that would be at infinite distance) and does not stay constant. Its kinetic energy, by contrast, increases as it moves inward.

Answer: C

The speed at the farthest point is \(\frac{v}{3}\).

The gravitational force is central, so the satellite's angular momentum is conserved. At the nearest and farthest points the velocity is perpendicular to the radius, so

\[ \begin{aligned} m v r_{\min} &= m v' r_{\max} \\ v' &= v\,\frac{r_{\min}}{r_{\max}} = \frac{v}{3} \end{aligned} \]

\(3v\) inverts the ratio; the satellite moves fastest when closest. \(\frac{v}{6}\) and \(9v\) would follow from wrongly using energy or squared ratios.

Answer: B

The ratio is 2 : 1.

For a satellite in a circular orbit of radius \(r\), gravity provides the centripetal force, which gives \(\frac{1}{2}mv^{2} = \frac{GMm}{2r}\). The potential energy is \(-\frac{GMm}{r}\). So

\[ \frac{|U|}{K} = \frac{GMm/r}{GMm/2r} = 2 \]

The total energy is \(-\frac{GMm}{2r}\), equal in magnitude to the kinetic energy. 1 : 2 inverts the ratio.

Answer: E

The acceleration of the centre of mass is zero.

The gravitational forces between the two particles are internal forces: equal in size and opposite in direction (Newton's third law), so they add to zero for the system. The centre of mass accelerates only under an external force, and none acts here.

So the centre of mass stays at rest while the particles move towards it. The options in terms of G and m describe the attraction itself, not the motion of the centre of mass.

Answer: C

The angular speed is \(\frac{\pi}{12}\) rad h^-1.

A geostationary satellite turns once with the Earth, completing \(2\pi\) radians in 24 hours.

\[ \begin{aligned} \omega &= \frac{2\pi}{24\ \text{h}} \\ &= \frac{\pi}{12}\ \text{rad h}^{-1} \end{aligned} \]

\(\frac{\pi}{24}\) forgets that a full turn is \(2\pi\), not \(\pi\). \(\frac{\pi}{365}\) mixes in the Earth's year, which has nothing to do with the satellite's period.

Answer: C

The new period is \(\sqrt{27}t\).

By Kepler's third law \(T^{2} \propto R^{3}\), so \(T \propto R^{3/2}\):

\[ \begin{aligned} \frac{T'}{t} &= \left(\frac{3R}{R}\right)^{3/2} \\ &= 3^{3/2} \\ &= \sqrt{27} \end{aligned} \]

So \(T' = \sqrt{27}t = 3\sqrt{3}t\), about 5.2 times longer. \(\sqrt{3}t\) uses \(T \propto R^{1/2}\), and \(t\) would mean the period is independent of the radius.

Answer: C

The ratio of the weights is 9 : 4.

At a height \(h\) above the surface, \(g_{h} = g\frac{R^{2}}{(R + h)^{2}}\).

\[ \begin{aligned} h = R &: g_{1} = \frac{g}{4} \\ h = 2R &: g_{2} = \frac{g}{9} \\ \frac{W_{1}}{W_{2}} &= \frac{1/4}{1/9} = \frac{9}{4} \end{aligned} \]

4 : 9 inverts the ratio, and 4 : 1 uses the distances from the surface (R and 2R) instead of from the centre (2R and 3R).

Answer: A

The potential energy at the new height has magnitude \(5 \times 10^{6}\) J.

The magnitude of gravitational potential energy is \(\frac{GMm}{r}\), with \(r\) measured from the centre of the earth, so it is inversely proportional to \(r\).

\[ \begin{aligned} r_{1} &= 6.4 + 3.6 = 10 \times 10^{6} \text{ m} \\ r_{2} &= 6.4 + 5.6 = 12 \times 10^{6} \text{ m} \\ U_{2} &= 6 \times 10^{6} \times \frac{10}{12} = 5 \times 10^{6} \text{ J} \end{aligned} \]

Using the heights above the surface instead of the distances from the centre gives \(6 \times 10^{6} \times \frac{3.6}{5.6}\), about \(3.9 \times 10^{6}\) J, which matches none of the options.

Answer: B

The height is 0.732 R.

Gravity at height h is \(g_h = \frac{g}{(1 + h/R)^{2}}\). The weight is one third when this ratio is \(\frac{1}{3}\).

\[ \begin{aligned} \left(1 + \frac{h}{R}\right)^{2} &= 3 \\ \frac{h}{R} &= \sqrt{3} - 1 = 0.732 \end{aligned} \]

\(\frac{R}{3}\) uses the approximation \(g_h \approx g(1 - \frac{2h}{R})\), which only holds for small heights. 0.414R (\(\sqrt{2} - 1\)) is the height at which the weight is halved.