Sweeping satellite
Problem
All satellites moving under the gravitational attraction of a single body travel in the path of an ellipse, where the body being orbited is at one focus of the ellipse. This is a consequence of the attractive force being inversely proportional to the square of the distance. For example the Moon's orbit around Earth is close to circlular, but Mars' orbit around the Sun is quite a bit more
elliptical.
Moment of momentum (sometimes called angular momentum) $H = mvr\sin(\theta)$, where $m$, $v$, $r$, and $\theta$ are the mass, velocity, radius from the axis of rotation, and angle between the radius and the velocity, of the satellite. When $dH/dt = 0$, clearly $H$ is not changing, so we can say it is "conserved". $dH/dt = Fr\sin(\theta)$, i.e. moment (where $\theta$ is the angle between the
radius and the force). So when there is no moment about the axis of rotation, which is always the case for a satellite since the gravitational force is parallel to the radius, moment of momentum is conserved. When a satellite is at the nearest point of its orbit ("perigee"), and the furthest point ("apogee"), the direction of motion and the radius are perpendicular, so $\theta = 90^{\circ}$, so
we can say that $mr_pv_p = mr_av_a$ where subscript $a$ denotes apogee, and subscript $p$ denotes perigee.
The sum of kinetic and potential Energy can also be assumed to be conserved for a satellite. This being the case, show that for those two positions, $v_av_p(r_a + r_p) = 2gR^2$, where $R$ and $g$ are the radius and gravitational field strength at the surface of the body being orbited.
Getting Started
Hint 1: Gravitational potential energy is the integral of the gravitational force, and evaluates to a negative number, theoretically being zero when infinitely far away.
Hint 2: In the Newtonian gravitational field equation, it can be shown that $GM = gR^2$. That assumes a spherical body of course.
Student Solutions
Conserving energy, $(1/2)mv_a^2 - GMm/r_a = (1/2)mv_p^2 - GMm/r_p$.
Conserving moment of momentum, $r_av_a = r_pv_p$.
And since $GM = gR^2$,
$v_a^2 - 2gR^2/r_a = v_p^2 - 2gR^2/r_p$.
And since $v_ar_a = v_pr_p$,
$$v_a^2 - v_p^2 = 2gR^2(1/r_a - 1/r_p)\;,$$
and
$$v_a^2 - v_p^2 = 2gR^2(v_a/v_pr_p - v_p/v_pr_p)\;.$$
And applying the difference between 2 squares,
$$(v_a + v_p)(v_a - v_p) = 2gR^2(v_a/v_pr_p - v_p/v_pr_p)\;,$$
$$v_pr_p(v_a + v_p) = 2gR^2\;.$$
Applying the moment of momentum relationship again,
$$v_pr_p(v_a + v_ar_a/r_p) = 2gR^2\;,$$ so
$$v_av_p(r_a + r_p) = 2gR^2\;,$$ as required.
I found this simple formula that combines conservation of energy and moment of momentum while studying 1st year engineering, and decided to remember it in case a relevant question came up in the exam, rather than deriving the conservation from first principles, to save time!