Gravitation Questions and Answers

Answer: A

The theory is regarded as excellent because it explains larger number of observations by making fewer assumptions.

A single law of universal gravitation explains the fall of objects on Earth, the orbits of the Moon and planets, the tides and much more from one simple assumption about how masses attract. Explaining much with little (parsimony) is a hallmark of a good scientific theory.

Being mathematical, being developed over a long time or after experimentation does not by itself make a theory good; many such theories have been discarded.

Answer: A

The period of the radial oscillation is \(\frac{2\pi\ell^{3}}{mk^{2}}\).

With \(\ell\) fixed, the radial motion is one-dimensional motion in the effective potential:

\[ U_{\text{eff}}(r) = \frac{\ell^{2}}{2mr^{2}} - \frac{k}{r} \]

The circular orbit sits at its minimum, \(U'_{\text{eff}}(r_{0}) = 0\), which gives \(r_{0} = \frac{\ell^{2}}{mk}\). The curvature there sets the oscillation frequency:

\[ \begin{aligned} U''_{\text{eff}}(r_{0}) &= \frac{3\ell^{2}}{mr_{0}^{4}} - \frac{2k}{r_{0}^{3}} \\ &= \frac{3k - 2k}{r_{0}^{3}} = \frac{k}{r_{0}^{3}} \\ \omega^{2} &= \frac{U''_{\text{eff}}}{m} = \frac{k}{mr_{0}^{3}} \\ &= \frac{m^{2}k^{4}}{\ell^{6}} \end{aligned} \]

So \(\omega = \frac{mk^{2}}{\ell^{3}}\) and \(T = \frac{2\pi}{\omega} = \frac{2\pi\ell^{3}}{mk^{2}}\). This equals the period of the circular orbit itself, which is why inverse-square orbits close into ellipses.

\(2\pi\sqrt{m/k}\) is not even a time: \(k\) has units of N m², so \(m/k\) is not in s². The forms with 3 or 5 in the denominator do not follow from the curvature \(k/r_{0}^{3}\); they come from keeping the wrong terms of \(U''_{\text{eff}}\).

Answer: A

The mine has higher air pressure, lower acceleration due to gravity.

Air pressure increases as you go down, because there is a taller column of air above. Gravity decreases with depth: \(g_{d} = g\left(1 - \frac{d}{R}\right)\), since only the mass of the earth beneath you attracts you as if from the centre.

The change in \(g\) at 3 km is small (about 0.05%), but its direction is downward, which rules out the options with higher \(g\). Lower pressure would apply to a mountain top, not a mine.

Answer: C

\(x = \) 9.

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

\[ \begin{aligned} \left(\frac{r_{A}}{r_{B}}\right)^{3} &= \left(\frac{T_{A}}{T_{B}}\right)^{2} = 27^{2} = 729 \\ \frac{r_{A}}{r_{B}} &= \sqrt[3]{729} = 9 \end{aligned} \]

3 is the cube root of 27, which forgets to square the period ratio first; 8, 4 and 12 do not satisfy the law.

Answer: A

The field is zero at \(\frac{x}{3}\) from m.

Between the two masses the fields point in opposite directions. At distance d from m they balance when

\[ \begin{aligned} \frac{Gm}{d^{2}} &= \frac{G(4m)}{(x - d)^{2}} \\ x - d &= 2d \Rightarrow d = \frac{x}{3} \end{aligned} \]

The neutral point lies nearer the smaller mass. \(\frac{x}{2}\) would be the midpoint, correct only for equal masses, and \(\frac{x}{5}\) comes from equating \(m/d\) with \(4m/(x - d)\) without squaring.

Answer: D

The statement is Kepler's second law, the law of areas.

Kepler's second law says that the line joining a planet to the Sun sweeps out equal areas in equal intervals of time. It means a planet moves faster when it is near the Sun and slower when it is far away, and it follows from the conservation of the planet's angular momentum about the Sun.

Kepler's first law is the law of orbits: each planet moves in an ellipse with the Sun at one focus. The law of periods is his third law, T² proportional to a³. Newton's law of gravitation gives the force between masses, and Newton's third law concerns action and reaction.

Answer: C

The ratio is 3 : 2.

The potential energy of 1 kg on the surface is \(U = -\frac{GM}{R}\), so the ratio of magnitudes is

\[ \begin{aligned} \frac{U_{1}}{U_{2}} \\ &= \frac{M_{1}}{M_{2}}\cdot\frac{R_{2}}{R_{1}} \\ &= \frac{1}{2}\cdot\frac{3}{1} \\ &= \frac{3}{2} \end{aligned} \]

2 : 3 inverts it, and 9 : 4 and 4 : 9 use \(R^{2}\), which is right for gravitational field strength but not for potential energy.

Answer: D

The satellite does not have vibrational energy.

Treated as a rigid body, the satellite has a moment of inertia, rotational kinetic energy from spinning, angular momentum from both its spin and its orbit, and gravitational potential energy in the Earth's field. A rigid body has no vibrational motion of its parts, so it has no vibrational energy.

Every other option describes a quantity the spinning, orbiting satellite does have.

Answer: B

The ratio of time periods is \(1:3\sqrt{3}\).

By Kepler's third law \(T^{2} \propto r^{3}\), where r is measured from the centre of the earth. The question gives the distances from the centre directly: 2.5R and 7.5R, in the ratio 1 : 3.

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

\(3\sqrt{3}:1\) inverts the ratio; the nearer satellite has the shorter period. \(1:\sqrt{3}\) uses a power of \(\frac{1}{2}\) instead of \(\frac{3}{2}\).

Answer: A

The orbital velocity is \(\frac{v_{e}}{2}\).

At height R above the surface the orbit radius is 2R. The orbital speed there and the escape speed from the surface are

\[ \begin{aligned} v_{o} &= \sqrt{\frac{GM}{2R}} \\ v_{e} &= \sqrt{\frac{2GM}{R}} \end{aligned} \]

\[ \begin{aligned} \frac{v_{o}}{v_{e}} &= \sqrt{\frac{GM/2R}{2GM/R}} \\ &= \sqrt{\frac{1}{4}} = \frac{1}{2} \end{aligned} \]

\(\frac{v_{e}}{\sqrt{2}}\) is the orbital speed just above the surface (orbit radius R), which ignores the height. The options equal to or larger than \(v_{e}\) would let the satellite escape.