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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:

\begin{equation} \boldsymbol{r_s}=\boldsymbol{r}-\boldsymbol{R} \end{equation}

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:

\begin{equation} \boldsymbol{v_s}=\boldsymbol{\omega}\times\boldsymbol{r}-\boldsymbol{\Omega}\times\boldsymbol{R} \end{equation}

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:

\begin{equation} v_s=0 \end{equation}

wich is false.

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  • $\begingroup$ 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. $\endgroup$ – garyp Mar 13 at 14:39
  • $\begingroup$ @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. $\endgroup$ – IchVerloren Mar 13 at 15:03
  • $\begingroup$ @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. $\endgroup$ – David White Mar 13 at 17:36
  • $\begingroup$ @DavidWhite that doesn't solve the problem $\endgroup$ – IchVerloren Mar 13 at 17:47
  • $\begingroup$ @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. $\endgroup$ – David White Mar 13 at 17:53

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