**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
}(*e*^{2
}-
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 *r*_{peri}** or ***r*_{ap}** from the general ellipse equation, and the expression for ***v*_{peri}** or ***v*_{ap}**, 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 m**^{2}/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, ***U*_{eff}* = L*^{2}* / *2*mr*^{2}* - GMm / r*. **A circular orbit is one where the total energy is a minimum, so ***E* = *U*_{eff}**. Find the value of ***r*** (the radius of the circular orbit, call it ***r*_{0}**) where ***U*_{eff}** is a minimum. Then calculate the ratio ***U*_{eff}(*r*_{0})/*U*(*r*_{0})** 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**). **