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Old Sep 11th 2008, 07:08 PM   #1
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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 $\displaystyle F_c=m_sa_c$.
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Old Sep 11th 2008, 08:43 PM   #2
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Originally Posted by arbolis View Post
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 08:45 PM.
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Old Sep 12th 2008, 02:13 AM   #3
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Originally Posted by Mr Fantastic View Post
$\displaystyle \frac{m v^2}{r} = \frac{GM_{E} \, m}{r^2}$ where $\displaystyle r = R_{E} + h$.

Therefore ....
I meant to mention that $\displaystyle \omega = v \, r$ (but you already know that, right?)
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Old Sep 12th 2008, 04:05 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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