Physics Help Forum Problem about a geostationary satellite

 Sep 11th 2008, 08:08 PM #1 Senior Member     Join Date: Apr 2008 Posts: 815 Problem about a geostationary satellite We want to put a satellite of mass $\displaystyle m_s$ in a circular orbit around the Earth at a distance $\displaystyle h$ from the ground. What is the angular velocity required to keep it in its orbit? My guess was "the same angular velocity than the Earth has" but the answer given is $\displaystyle \omega=\sqrt{\frac{GM_E}{(R_E+h)^3}}$ where $\displaystyle M_E$ is the mass of the earth, $\displaystyle G$ is the universal gravitational constant and $\displaystyle R_E$ is the radius of the Earth. How can I reach to the answer? (Please give me the full answer if you can because my exam is tomorrow and I promise I'll try to understand the exercise). Thanks in advance. P.S.: I suppose I have to use the formula $\displaystyle F_g=\frac{Gm_sM_E}{R^2}$ and maybe also $\displaystyle \omega=\frac{v}{r}$ and probably $\displaystyle a_c=\frac{v^2}{r}$ with $\displaystyle F_c=m_sa_c$. __________________ Isaac If the problem is too hard just let the Universe solve it.
Sep 11th 2008, 09:43 PM   #2
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 Originally Posted by arbolis We want to put a satellite of mass $\displaystyle m_s$ in a circular orbit around the Earth at a distance $\displaystyle h$ from the ground. What is the angular velocity required to keep it in its orbit? My guess was "the same angular velocity than the Earth has" but the answer given is $\displaystyle \omega=\sqrt{\frac{GM_E}{(R_E+h)^3}}$ where $\displaystyle M_E$ is the mass of the earth, $\displaystyle G$ is the universal gravitational constant and $\displaystyle R_E$ is the radius of the Earth. How can I reach to the answer? (Please give me the full answer if you can because my exam is tomorrow and I promise I'll try to understand the exercise). Thanks in advance. P.S.: I suppose I have to use the formula $\displaystyle F_g=\frac{Gm_sM_E}{R^2}$ and maybe also $\displaystyle \omega=\frac{v}{r}$ and probably $\displaystyle a_c=\frac{v^2}{r}$ with $\displaystyle F_c=m_sa_c$.
$\displaystyle \frac{m v^2}{r} = \frac{GM_{E} \, m}{r^2}$ where $\displaystyle r = R_{E} + h$.

Therefore ....

Last edited by Mr Fantastic; Sep 11th 2008 at 09:45 PM.

Sep 12th 2008, 03:13 AM   #3
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 Originally Posted by Mr Fantastic $\displaystyle \frac{m v^2}{r} = \frac{GM_{E} \, m}{r^2}$ where $\displaystyle r = R_{E} + h$. Therefore ....
I meant to mention that $\displaystyle \omega = v \, r$ (but you already know that, right?)

Sep 12th 2008, 05:05 AM   #4
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 I meant to mention that (but you already know that, right?)
Yes I know that, even if I typed "$\displaystyle \omega =\frac{v}{r}$", I just made an error of typo.
Thanks for the answer, I'll respond later because I have no time now.
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