If the total mechanical energy in a satellite's orbit (assuming circular) is greater when it is closer to the earth, and hence smaller when it is farther from the earth, then we can say that as the moon drifts from the earth, the moon loses energy in translational speed and gravitational potential energy. If only those two are taken into consideration, then there is a net energy loss from the moon.

I had first thought that the energy a satellite has increases as it goes on a larger orbit, but I ran some numbers and it didn't appear so. If I went wrong somewhere, please someone, correct me. Here are my numbers:

For a geostationary satellite (r = 42 164 km, v = 11 068 km/s, m = 1 kg), its total energy is PE + KE. PE = mgh, but g = 0.22416 m/s^2. The result is PE = 9 451.650 kJ, KE = 4 726.582 kJ

For a satellite at r = 45 000 km , m = 1kg, then v = sqrt(GM/r) = 2 976.06 km/s. g at that height is g = 0.19680 The result is PE = 8 856.094 kJ, KE = 4 428.047 kJ

At the larger orbit, both PE and KE are lower than if it was at a lower orbit. Is this right?

Now, the earth slows down its rotation, which allows the moon to go into a larger orbit by conservation of angular momentum. Since the moon goes into a larger orbit, it loses energy. But, since the spin of the earth has slowed down, it also loses energy. Moreover, the moon is still tidally locked with the earth, so its rotational speed isn't increasing.

All in all, there seems to be an energy loss that's going on. How is this being compensated? Is it in the translational speed of the moon (so that the moon is actually moving faster than it should be to maintain a stable orbit)? That seems reasonable - there could be an increase in translational and rotational speed to compensate for the energy loss, maintaining the moon to be tidally locked.

But that's just me. What really happens? How does the energy transfer occur, and are there mathematical equations describing this exchange?


1 Answer 1


It appears you made a few mistakes.

The formula $E_P = mgh$ is only an approximation for objects near the ground. The more complete formula is

$ E_P = -\frac{\mu m}{r} $

where $\mu = 398600.44$ is Earth's standard gravitational parameter, and $r$ is the distance between the object and the Earth's center of gravity.

Especially note the negative sign; this has to do with the definition of potential energy in the context of orbits. This is where I think you went wrong.

Also, where did you find $V = 11.068$ km/s for a geostationary orbit? That looks more like an escape speed than a normal orbital speed...Indeed, if you look up the altitude for a geostationary orbit you see that it is $35768$ km above the equator. That means the total pathlength traversed by the satellite in one stellar day is

$2\pi \cdot ( 35768+R_E) \approx 264,811 $km

making the speed

$264,811 \mathrm{\ km} / 86164 \mathrm{\ seconds} \approx 3.07 \mathrm{\ km/s}$

so much much slower than the ~11 km/s you stated. Lumping all this together:

$E_P^{GEO} \approx -\frac{398600.44}{42164} = -9.45 $ kJ/kg

$E_K^{GEO} \approx \frac{3.07^2}{2} = 4.71 $ kJ/kg

$E_{tot}^{GEO} = 4.71-9.45 = -4.74 $ kJ/kg

while for the other orbit

$E_P^{alt} \approx -\frac{398600.44}{45000} = -8.86 $ kJ/kg

$E_K^{alt} \approx \frac{2.98^2}{2} = 4.44 $ kJ/kg

$E_{tot}^{alt} = 4.44-8.86 = -4.42$ kJ/kg

which is indeed higher than the GEO orbit.

This makes sense -- you need to input a lot more energy to let anything escape from Earth's gravity than, say, an apple falling to the ground (which is also in an "orbit", albeit one far closer to the Earth, and not exactly on an escape trajectory).

If what you say would be true, everything would simply fall up and escape the Earth. There are a few experiments that will show that that is not actually what happens :)

With regard to your statement about the moon: the moon is indeed slowly escaping from the Earth. The mechanism here is that the Moon is gaining orbital speed at the expense of Earth's rotational momentum, through tidal interaction.

Roughly translated: as Earth's rotation slows down, the Moon speeds up, making the Moon progress farther away from the Earth, towards a lower speed.

The total energy in that higher orbit is higher, because the drop in speed is disproportionally small in relation to the gain in potential energy. Eventually, after a few million years of repeating the above, the moon will have gained enough energy to escape the Earth and orbit the Sun on its own.

  • $\begingroup$ Thank you. The numbers I got were counterintuitive, so I asked. I laughed when you pointed out that things would fall up if what I said was right - I didn't think about that. Also, I think I didn't put the correct units there, too. It should be 11 km/h, but apparently I didn't use the incorrect value for getting the kinetic energy. The KE I got was 4 726.582, which uses v = 3.07 and not v ~ 11. $\endgroup$ Jan 22, 2013 at 10:03

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