# Time dilation of an orbiting ship

• B
A.T.
Science Advisor
Using an apparent uniform (but constantly changing due to the "orbit") gravitational field with attendant time dilation effects is one way to solve the problem, but obviously not the only way.
You can use a frame that translates on a circle with the ship, but doesn't rotate. This will produce a uniform gravitational field with a time dependent direction. However since it's time dependent, you cannot assign a potential to it. You can just argue that the asteroid is always higher in the field.

The co-rotating common rest frame has a non-uniform but time independent gravitational field, so you can assign a potential it.

PeroK
Science Advisor
Homework Helper
Gold Member
With a small radius, which makes the potential difference low again.
How are you assigning a gravitational potential?

A.T.
Science Advisor
How are you assigning a gravitational potential?
The centrifugal potential is based on the conservative centrifugal force in the rotating frame.

Janus
Staff Emeritus
Science Advisor
Gold Member
While the centripetal acceleration is found by;
$$A_c = \omega^2 r$$

and
$$A_c = \frac{v2}{}$$
How are you assigning a gravitational potential?
It is basically related to the work needed to move a test particle from one radius to another in a rotating frame.

The centripetal potential difference between a point at a given radius and the axis is found by:

$$\phi = -\frac{\omega^2 r^2}{2}$$
(moving towards the center mean moving to a higher potential)

And since tangential velocity at r is found by:

$$v= \omega r$$

we then get:
$$\phi = -\frac{v^2}{2}$$

Which means that the potential difference between the Axis and r depends only on the velocity at r and is independent of r.

Contrast this to centripetal acceleration which depends on both v and r.

This is not much different than clock placed in the nose and tail of a rocket ship and undergoing an equal proper acceleration, the nose clock will tick faster than the tail clock. Now imagine instead that your rocket is continually pointing its nose towards the Asteroid as it circles under rocket power. The nose clock will now have a slightly lower proper acceleration than the tail clock. It will still run faster than the tail clock, but not by as much as it did when the nose and tail had the same proper acceleration. If you extend the nose of the rocket all the way to the axis of the forced orbit, the difference in rate between nose and tail clock will be the same as the time dilation for the tangential speed of the tail of the rocket.

Nugatory, A.T. and Ibix