Central Forces and Kepler Problem Questions and Answers

Answer: C

The trajectory is a parabola.

The total energy per unit mass at closest approach is

\[ \begin{aligned} \varepsilon &= \frac{v^{2}}{2} - \frac{GM}{R} \\ &= \frac{GM}{R} - \frac{GM}{R} = 0 \end{aligned} \]

The sign of the energy fixes the conic: negative for bound orbits (circle or ellipse), zero for a parabola, positive for a hyperbola. Here \(v\) is exactly the escape speed \(\sqrt{2GM/R}\), so the energy is zero and the path is a parabola.

A circular orbit would need \(v^{2} = GM/R\), and any speed between that and escape speed gives an ellipse. A hyperbola needs \(v^{2} \gt 2GM/R\).

Answer: A

The correct statement is \(v_{0}b = v_{m}r_{m}\).

The Coulomb force is central, so angular momentum about the nucleus is conserved. Far away it is \(mv_{0}b\); at closest approach the velocity is perpendicular to the radius, so it is \(mv_{m}r_{m}\). Hence \(v_{0}b = v_{m}r_{m}\).

Energy conservation, with the alpha particle's charge 2e, gives \(v_{0}^{2} = v_{m}^{2} + 4r_{0}v_{0}^{2}/r_{m}\). Using \(v_{m} = v_{0}b/r_{m}\):

\[ \begin{aligned} r_{m}^{2} - 4r_{0}r_{m} - b^{2} &= 0 \\ r_{m} &= 2r_{0} + \sqrt{4r_{0}^{2} + b^{2}} \\ &\approx 4r_{0} + \frac{b^{2}}{4r_{0}} \end{aligned} \]

So the correction is \(b^{2}/4r_{0}\), not \(b^{2}/2r_{0}\) or \(b^{2}/8r_{0}\). And \(v_{0}b = 2v_{m}r_{0}\) would need \(r_{m} = 2r_{0}\), less than the head-on value \(4r_{0}\). Only the angular momentum statement holds.

Answer: C

The key gives 216T.

By Kepler's third law the period grows as the radius to the power 3/2:

\[ \begin{aligned} T_{4a} &= T(4)^{3/2} = 8T \\ T_{9a} &= T(9)^{3/2} = 27T \end{aligned} \]

All three planets are back where they started, in the same line as at \(t_{0}\), after a whole number of each period: the least common multiple of \(T\), \(8T\) and \(27T\), which is \(216T\).

The key reads 'the same straight line' as the same configuration, all planets on the same side as at \(t_{0}\). If a planet on the opposite side of the star also counted, an earlier alignment (at \(108T\)) would qualify, but that is not offered. \(8T\) and \(27T\) are single periods, and \(512T\) is \(8^{3}\), not a common multiple built from these periods.

Answer: C

The speed is \((gR/3)^{1/2}\).

At a height 2R above the surface the orbital radius is \(r = R + 2R = 3R\). Gravity provides the centripetal force, and \(GM = gR^{2}\) at the surface:

\[ \begin{aligned} \frac{mv^{2}}{r} &= \frac{GMm}{r^{2}} \\ v^{2} &= \frac{GM}{3R} = \frac{gR^{2}}{3R} = \frac{gR}{3} \\ v &= (gR/3)^{1/2} \end{aligned} \]

The option \((gR/2)^{1/2}\) takes the radius as 2R, forgetting to add the earth's radius to the height. The options gR/3 and gR/2 are values of \(v^{2}\), not of v, and do not have the dimensions of speed.

Answer: B

The period is \(2\pi\sqrt{R/g}\).

For a satellite close to the earth (\(H \ll R\)), gravity supplies the centripetal force, and \(GM = gR^{2}\):

\[ \begin{aligned} \frac{mv^{2}}{R} &= mg \Rightarrow v = \sqrt{gR} \\ T &= \frac{2\pi R}{v} = 2\pi\sqrt{\frac{R}{g}} \end{aligned} \]

In such an orbit the man and the satellite fall freely together, so he is weightless. This is about 84 minutes. The option \(2\pi\sqrt{g/R}\) has the ratio inverted and is not even a time, and the \(1/2\pi\) options are frequencies rather than periods.

Answer: C

For a repulsive inverse square force the orbit is Hyperbolic.

With a repulsive force, such as an alpha particle approaching a nucleus in Rutherford scattering, the effective potential is positive everywhere and the total energy is always positive. A positive-energy orbit in an inverse square field is a hyperbola, with the force centre at the outer focus.

Circular and elliptical orbits are bound orbits, possible only for an attractive force with negative total energy, and the parabola is the zero-energy boundary case of an attractive force.