Suppose that an Earth satellite of mass 4700 kg, originally in a circular orbit of radius 7.18 * 10 ^ 6 m, increases its orbital radius to 7.28 * 10 ^ 6 m.
At orbital distance r the gravitational field strength will be (r / rEarth) ^ -2 times as great as at the surface of the Earth (rEarth stands for the radius of the Earth, approximately 6.38 * 10 ^ 6 meters). At the orbital distances 7.18 * 10 ^ 6 meters and 7.28 * 10 ^ 6 meters, we therefore see that the gravitational field strengths are
At any point the field strength is equal to the centripetalacceleration v^2 / r; we therefore have
We obtain velocities
For the 4700 kg mass of the satellite we easily obtain kinetic energies of 130500 MJ (megaJoules) and 128700 MJ. The kinetic energy difference is therefore
The midpoint orbital radius is 7.229, which implies a midpoint gravitational field of
and gravitational force of
At the orbit is 'raised', the distance parallel to the gravitational field is the difference of the orbital radii:
The approximate work required to 'raise' the orbit against the gravitational pull will therefore be
Since the PE change is the work done by the satellite against gravity, this is the potential energy change of the satellite.
The ratio of this potential energy change to the kinetic energy change is thus
The ratio computed here is based on a midpoint approximation to an inverse square gravitational field. If the ratio is based on a more accurate approximation (e.g., on a subdivision of the distance into a sufficient number of subintervals, with a midpoint approximation on each), the approximate ratio will approach the precise ratio which is exactly -2.
- KE = .5 m v^2 with the velocity v = `sqrt(G M / r) obtained from setting centripetal acceleration equal to gravitational attraction, and
- `dPE = work required to change orbital distance, which is the integral of force * distance
- `dPE = int( G M m / r^2, r, r1, r2) = G M m (1 / r1 - 1 / r2).
The change of KE in a transition of circular orbits is always exactly half the magnitude of the potential energy change, and of the opposite sign. That is, the kinetic energy loss is always exactly half of the kinetic energy gain.
The KE of an orbit is .5 m * G M / r, which can be obtained from v = `sqrt( G M / r), obtained by setting centripetal force equal to gravitational force; however to illustrate the use of proportionality we use a different and perhaps more cumbersome approach. This approach will express the orbital velocity and KE in terms of g = 9.8 m/s^2 and the ratio rEarth / r of Earth radius to orbital radius. These end results involve easily visualized quantities are in many ways more intuitive than the equation .5 m * G M / r.
A circular orbit of radius r1 will have orbital velocity v1 such that v1^2 / r1 = g1, where g1 is the acceleration of gravity, or gravitational field strength, at distance r1 from the center of the Earth. Since the ratio of gravitational fields at two distances is the inverse square of the distance ratio, at radius r1 the field is
where g is the gravitational field strength at the surface of the Earth.
The squared velocity of the orbit whose radius is r1 is therefore
and the kinetic energy is
Similarly the kinetic energy in the second orbit is
The difference in the kinetic energies is
To find the approximate PE difference we use the midpoint gravitational field (the field at r halfway between r1 and r2) to estimate the average force required to 'raise' the satellite, then multiply approximate force by distance.
The gravitational field midway between orbital radii r1 and r2 is
where rMid = (r1 + r2) / 2. The force on the mass m at this distance is therefore
The work required to 'raise' the mass through the required distance (r2 - r1) parallel to the gravitational field is
Since rMid between r = r1 and r = r2 is (r1 + r2) / 2 we have
It can be shown that if r2 / r1 is nearly 1, the expression (r2 - r1) / [ (r2 + r1) / 2 ] ^2 is very nearly equal to 1 / r1 - 1 / r2, and we have
More generally for any planet of mass M and satellite of mass m we have `dPE = G M m ( 1 / r1 - 1 / r2 ) and `dKE = - .5 G M m ( 1 / r1 - 1 / r2 ).
Comparing the expressions for `dPE and `dKE, we see that `dPE / `dKE = -2. That is,
For increasing orbital distance PE increases by a certain amount and KE decreases by half that amount.
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