# Deriving formula for surface velocity of a satellite in orbit.

1. Oct 27, 2013

### spiruel

I may be being stupid here... But I'm trying to derive such an equation to no avail.

Maybe such a formula requires more than one variable - but I would like to easily display this graphically.
To start out the derivation I stated that the Surface Velocity = (2pi/T)-(v/r) where v/r=angular velocity of satellite in orbit and 2pi/T=angular velocity of a planet.

Can anyone help?

Last edited: Oct 27, 2013
2. Oct 27, 2013

### spiruel

Think I got it (in terms of time, not velocity):

2pi*sqrt((radius of planet+orbital altitude)^(3)/(GM))-rotationalperiodofplanet

Last edited: Oct 27, 2013
3. Oct 27, 2013

### Staff: Mentor

You can work out the angular velocities separately to begin with. You stated the one for the planet as $2\pi/T$. Call that $\omega_p$, and make sure that T is the sidereal period for the planet. For the satellite you again made a true statement, namely that the angular velocity is given by $\omega_s = v/r$, with r the radius of the orbit and v the orbital speed.

Assuming a circular orbit the orbital speed will be $v = \sqrt{GM/r}$, where M is the mass of the planet.

Assuming that the satellite's orbit is in the plane of the equator, the difference in angular velocities is then:
$$\omega_p - \omega_s = \Delta \omega = \frac{2\pi}{T} - \frac{1}{r}\sqrt{\frac{GM}{r}}$$
You can move the 1/r into the square root if you want.

With Δω in hand you can work out the surface speed of the sub-satellite point easily enough.