Homework Problems
Orbital Energy and Angular Momentum

1. Starting with the general expression for total energy of any orbit:

E = 1/2 m(GMm / L)2 (e2 - 1)      

show that this reduces to the simpler expression for an ellipse,

E = - GMm/2a

by inserting the value of L at an appropriate place in the elliptical orbit (either at perihelion or aphelion), and simplifying the resulting expression. You'll need the value of rperi or rap from the general ellipse equation, and the expression for vperi or vap, given in Lecture 6.

2. To send a space probe directly to the Sun, we would have to remove the angular momentum from its orbital motion.  Consider a space probe of mass 2500 kg, in circular orbit at 1 AU from the Sun.  The orbital speed of such an orbit (the same as Earth's speed) is about 30 km/s.
(a) What is the probe's orbital angular momentum, in kg m2/s?
(b) What is the minimum kinetic energy needed to lose this angular momentum, in J?
(c) Show that this is the same as the kinetic energy that would have to be added in order for the probe initially at 1 AU to escape the solar system. [Hint: The energy needed to escape the solar system is that needed to change its total energy to zero.]

3. Investigate the energy of circular orbits as follows: Start with the expression for the effective potential, Ueff = L2 / 2mr2 - GMm / r. A circular orbit is one where the total energy is a minimum, so E = Ueff. Find the value of r (the radius of the circular orbit, call it r0) where Ueff is a minimum. Then calculate the ratio Ueff(r0)/U(r0) and show that it is equal to 1/2. This is a general result--the minimum energy orbit is one in which the total energy is 1/2 the potential energy (i.e. E = U/2).