True or false on energy and orbits

Problem

True or false? Briefly justify your answer. (a) the gravitational potential energy $U = -GMm/r$ is always negative if $U(\infty) = 0$; (b) the escape velocity from a planet depends on the mass of the escaping object; (c) an elliptical orbit has negative total mechanical energy; (d) Kepler’s second law is equivalent to conservation of the planet’s angular momentum.

Solution

$\ev{(a)\ \text{T} \quad (b)\ \text{F} \quad (c)\ \text{T} \quad (d)\ \text{T}}$

(a) True. With the convention $U(\infty)=0$ we have $U(r) = -GMm/r$. Since $G$, $M$, $m$, $r$ are all positive, the fraction is positive and the minus sign makes $U<0$ for every finite $r$. $U$ increases (approaching zero) as the distance grows.

(b) False. From conservation of energy $v_f = \sqrt{2GM/R}$: it depends only on the planet’s mass $M$ and radius $R$, not on the mass of the escaping object (which cancels out). A stone and a rocket have the same escape velocity.

(c) True. An elliptical orbit is bound (closed): the object cannot move infinitely far away. This corresponds to total mechanical energy $E_\text{tot}<0$. If $E_\text{tot}=0$ the orbit is parabolic, if $E_\text{tot}>0$ it is hyperbolic: in both cases unbound.

(d) True. The gravitational force is central (directed along the line joining the planet and the star): its torque about the star is zero, so $\vec L$ is conserved. Conservation of $\vec L$ means the radius vector sweeps out equal areas in equal times: this is exactly Kepler’s second law.

Links

Topics: Gravitazione Concepts: Energia potenziale gravitazionale universale · Velocità di fuga · Orbite · Conservazione dell’energia meccanica · Conservazione del momento angolare · Leggi di Keplero