Why does geopotential not take into account the rotation of the Earth?

[baxter]

Hi

Geopotential U is the gravitational potential of the Earth without taking into account the rotation of the Earth whereas gravity's potential $U_T$ considers the rotation of the Earth such as :

$U_T = U + \frac{1}{2}\omega ^2r^2cos(\psi)^2$ (with common spherical coordinates)

However, for a satellite, the potential of the Earth undergone by the satellite is $U$ and not $U_T$.

My question : why?

In fact, I agree the satellite does not belong the the surface of the Earth but it seems that the geopotential $U$ does not consider the rotation of the Earth ? It would say that the rotation of the Earth does not affect the gravitational force of the Earth on the satellite ?

 

[D H]

The Newtonian answer is simple: The satellite isn't sitting on the surface of the Earth. The Earth's rotation rate has zero direct effect on the satellite.

The rotation of the Earth does have an indirect effect on the satellite because the Earth is not spherically symmetric. The Earth instead has a somewhat lumpy density distribution. The equatorial bulge is by far the biggest non-spherical effect of the Earth's rotation on satellites. This bulge is a direct consequence of the Earth's rotation. It is what makes Sun synchronous orbits possible. A prolate ellipsoid model yields a good but not perfect model of the Earth's gravity field. There are other deviations from a spherically distributed density. For example, the northern hemisphere has considerably more land mass than does the southern hemisphere, and the Pacific is much bigger than is the Atlantic.

In a Newtonian universe, there would be nothing left after removing all of these effects that result from the Earth's orientation. The Earth's angular velocity itself has no effect on the satellite's motion.

That's not quite true in general relativity. There is a frame dragging effect in general relativity that is tiny (very, very tiny!) in the case of the Earth. This frame dragging effect is much more sizable for an object closely orbiting a rapidly rotating neutron star.

 

[BobG]

Hi

Geopotential U is the gravitational potential of the Earth without taking into account the rotation of the Earth whereas gravity's potential $U_T$ considers the rotation of the Earth such as :

$U_T = U + \frac{1}{2}\omega ^2r^2cos(\psi)^2$ (with common spherical coordinates)

However, for a satellite, the potential of the Earth undergone by the satellite is $U$ and not $U_T$.

My question : why?

In fact, I agree the satellite does not belong the the surface of the Earth but it seems that the geopotential $U$ does not consider the rotation of the Earth ? It would say that the rotation of the Earth does not affect the gravitational force of the Earth on the satellite ?

It depends on what you want to know.

U relates to gravitational force solely due to the Earth (or the potential energy since the mass of the second object isn't actually known yet).

U_t relates to the net gravitational force felt by a person on Earth. The rotation of the Earth is relevant only because that rotation is imparting motion to the person and the motion of the person (or object) creates a centrifugal force that opposes gravity.

Both would be useful bits of information on Earth.

U_t would depend on the satellite's motion; not the rotation of the Earth. U_t wouldn't yield much useful information for a satellite in a circular orbit, as the net "weight" felt by the satellite would have to be zero in order for the satellite to be in a circular orbit.

"Weightlessness" doesn't mean the lack of a gravitational force. It means the centrifugal force created by the person's (or satellite's) velocity cancels out the gravitational force.

 

[baxter]

Thanks for your explanations, they are very useful ;)