Hint 4

(a) In terms of the preigee distance $r_1$ and the apogee distance $r_2$, the semimajor axis is

$$a+{r_1+2_2\over 2}=6R$$

The eccentricity is

$$\epsilon={a-r_1\over a}={2\over 3}$$

(b) The energy in orbit is

$$E=-{GMm\over a}$$

(which the same as for a circular orbit of radius $a$). The energy at rest on the Earth's surface is purely potential,

$$E_0=-{GMm\over R}$$

Hence an energy

$$E-E_0={11\over 12}{GMm\over R}$$

is needed.

(c) At perigee the energy consists of kinetic and potential pieces,

$$E={1\over 2}mv_1^2-{GMm\over r_1}$$

Using the expression for the total energy above gives

$$v_1=\sqrt{{5\over 6}{GM\over R}}$$

(d) At perigee the velocity is perpendicular to the radius and so the angular momentum is

$$L=mr_1v_1=m\sqrt{{10\over 3}GMR}$$

(and points along the normal to the plane of the orbit).