This question is about the mechanics of circular and elliptical orbits. Consider a satellite of mass m in orbit around a planet of mass M such that it is in free fall and sufficiently far from the planet to be able to ignore atmospheric effects. The point of closest approach (perigee) is at a distance rp from the centre of the planet and the highest point in the orbit (apogee) is at a distance ra from the centre of the planet.
(a) As the orbit is not circular, the equation F=mv2/r does not apply at any point in the orbit. However, at the apogee and perigee the velocity vector is perpendicular to the gravitational force. Hence the angular momentum (= mrω where ω is the angular velocity) is constant and at those two points is simply equal to mvara=mvbrb.
Use this with the conservation of energy (kinetic + gravitational potential) to derive Newton's vis viva equation (eq. 1) to find the velocity at any given radius of orbit on the ellipse:
v2=GM(r2−a1)(eq. 1)
where a=(ra+rp)/2, the semi-major axis of the ellipse, ra=a(1+e) and rp=a(1−e), where e is the eccentricity. Note: the largest diameter of an ellipse has a length 2a so that a is the length of the semimajor axis. [4]
(b) In order to escape from orbit, the satellite undergoes a single short burn at a specific point in the orbit which gives it a new orbit with infinite semi-major axis. That burn creates a velocity change Δv which is a figure of interest to engineers as it is essentially proportional to the amount of fuel burnt in a given manoeuvre.
(i) Find expressions for the necessary Δv to reach escape velocity from apogee (Δva) and perigee (Δvp). [5]
(ii) Hence show that the ratio:
ΔvpΔva=2(1+e)−(1+e)2(1−e)−(1−e)
[2]
(iii) Recalling that, for an ellipse, 0 < e < 1, show that it requires a greater velocity change to launch a satellite into escape orbit from apogee than from perigee - this may be counter-intuitive! [2]
In order to change from one circular orbital radius (r1) to another (r2) in the same orbital plane, a very efficient way is to use the two-burn Hohmann transfer. A brief burn is applied at one point in the orbit which makes the orbit elliptical. The first burn point is the perigee of the ellipse. On reaching the apogee, a second burn is applied to give the necessary velocity to make a circular orbit at the new height.
(c) Show that the velocity changes at perigee and apogee are given by:
Δvp=r1GM(r1+r22r2−1)
and
Δva=r2GM(1−r1+r22r1)
[5]
(d) For a given initial orbit r1, the total Δv=Δva+Δvp will vary with r2.
By differentiation with respect to r2, keeping r1 fixed, show that the maximum value of Δv occurs for the positive root of the equation
x3−15x2−9x−1=0
Where x=r2/r1. This root has the approximate value of 15.6.
In other words, it requires more energy to place the satellite in a higher orbit than it does to remove it to escape velocity, another counter-intuitive result! [4]