# Satellite period and mass

1. Oct 15, 2015

### cyberdiver

Why is the period of a satellite's orbit independent of its mass? I understand that its mass cancels out mathematically, but I don't understand it intuitively. The way I'm seeing this, if a satellite has a greater mass, it would have a greater pull on the body it is orbiting, and hence would require a greater orbital velocity to counter centripetal force.

I find it difficult to comprehend that an asteroid and a neutron star would orbit a planet with the same period, especially as it would be more accurate to say that the planet would be orbiting the neutron star.

2. Oct 15, 2015

### TurtleMeister

The way you are seeing it is correct. Mass does affect orbital period. But it's not just the mass of the satellite, it's the combined mass of both bodies that determine the orbital period (and the distance between them). So if you're talking about a man made satellite in earth orbit, then doubling the mass of the satellite alone isn't going to make any noticeable difference in the combined mass of the satellite and the earth together. So there would be no detectable difference in the satellite's orbit. But if you doubled the mass of the moon, now that would make a noticeable difference in the orbital period. Does that help you understand it intuitively? By the way there is another thread about this subject: https://www.physicsforums.com/threads/orbital-velocity.836915/

3. Oct 15, 2015

### cyberdiver

I just looked through the thread. Why does the equation only take into account the mass of the primary body if it is affected by both the primary and the satellite masses?

4. Oct 15, 2015

### DaveC426913

That greater mass also means it takes a greater force to get it moving at the same speed.

5. Oct 15, 2015

### TurtleMeister

That's because the equation you are talking about is assuming that the primary body is much more massive than the sattellite. The equation you are looking for is here: https://en.wikipedia.org/wiki/Orbital_period#Two_bodies_orbiting_each_other

6. Oct 15, 2015

### DaveC426913

Because usually you are considering a secondary body that has a negligible mass compared to the primary body.
It is effectively zero.

[EDIT] Boy, for a turtle, you sure are fleet-of-finger.

7. Oct 15, 2015

### cyberdiver

The equations I was taught in class are completely different. I will need to investigate this more thoroughly.

8. Oct 15, 2015

### TurtleMeister

Well, the equation I linked to is the most accurate. But it's really unnecessary when the satellite is not a planetary body.

Well, that was very uncharacteristic of me. I think I've had too much caffeine.