# Velocity of a satellite with a polar orbit with respect to the surface of the earth

Let's say I'm tryng to derive the expression for the velocity of a really low orbit satellite. Let $$S$$ be the frame of reference of the earth's center (in rest). Let $$S'$$ be the frame of an observer on the surface of the earth.

The first step should be writing:

$$$$\boldsymbol{r_s}=\boldsymbol{r}-\boldsymbol{R}$$$$

where $$\boldsymbol{r_s}$$ is the position vector of the satellite from the surface of the earth, $$\boldsymbol{r}$$ is the position vector of the satellite from the center, and $$\boldsymbol{R}$$ is the position vector of the surface frame from the center frame.

The next thing should be deriving the equality with respect to time.

Since $$\boldsymbol{R}$$ and $$\boldsymbol{r}$$ are vectors of constant magnitude, the result should be:

$$$$\boldsymbol{v_s}=\boldsymbol{\omega}\times\boldsymbol{r}-\boldsymbol{\Omega}\times\boldsymbol{R}$$$$

Where $$\boldsymbol{\omega}$$ is the angular velocity of the satellite and $$\boldsymbol{\Omega}$$ is the angular velocity of the earth. But I feel like something missing there. The "derivative operator" in rotating systems is defined as $$\frac{d}{dt}=[(\frac{d}{dt})_r+\boldsymbol{\omega}\times$$], where $$(\frac{d}{dt})_r$$ is the velocity in the rotating system.

This might be really simple, but I've always had a really hard time with rotating frames of reference.

EDIT:

My answer is wrong. If I suppose that the satellite has a polar orbit with the same angular velocity as the arth, and that the observer in the surface is on the equator, then:

$$$$v_s=0$$$$

wich is false.

• It's very unclear (to me) what you are trying to do. I suspect that you don't mean polar orbit, but rather geostationary? Otherwise why would $\dot{v}=0$. Is that correct? On the other hand, you do specify low earth orbit, for which $v'$ would be very complicated, and not geostationary. Maybe I'm misreading your post. – garyp Mar 13 at 14:39
• @garyp I've changed my notation. $v'$ wasn't a derivative but the speed of the satellite on the surface frame of reference. When I said " low orbit" I was thinking that the orbit radius of the satellite would be equal to the Earth's radius. You are rignt, for the result $v_s=0$ I considered a polar orbit. In that case, my velocity expression yields a zero speed. That means I'm wrong. – IchVerloren Mar 13 at 15:03
• @IchVerloren, set the force of gravity equal to the centripetal force on the satellite, and solve for velocity. The starting equation is $Gm_1m_2/r^2=m_2v^2/r$, where $m_2$ is the mass of the satellite. – David White Mar 13 at 17:36
• @DavidWhite that doesn't solve the problem – IchVerloren Mar 13 at 17:47
• @IchVerloren, if you could give a bit of detail regarding why you need to solve the velocity problem for your stated reference frame, that would be somewhat helpful. – David White Mar 13 at 17:53