How to find the orbit time of this satellite?

**m_i_k_o** · October 3rd 2009, 03:53 AM

A satellite is in a circular orbit H km above the earth surface.
In how many hours does the satellite complete one orbit?

The radius of the earth is 6370 km.

H[km] = 289;

Could someone please show me how to solve this

**mathaddict** · October 3rd 2009, 04:01 AM

Originally Posted by m_i_k_o

A satellite is in a circular orbit H km above the earth surface.
In how many hours does the satellite complete one orbit?

The radius of the earth is 6370 km.

H[km] = 289;

Could someone please show me how to solve this

HI

The gravitational attraction between the satelite and the earth is the same as the centripetal force of the satelite so

Let M be the mass of earth and small m be the mass of satelite .
Note that omega=2pi/T

$\frac{GMm}{(R+H)^2}=m(R+H)(\frac{2\pi}{T})^2$

Rearranging ..

$T=\sqrt{\frac{4\pi^2(R+H)^3}{GM}}$

Now you know R , H , G (Gravitational constant) , Mass of the earth is not given but you can check it out in your book since i hv forgotten .

Then substitute .

**m_i_k_o** · October 3rd 2009, 04:19 AM

To convert to hours I will just divide T by 360?

**e^(i*pi)** · October 3rd 2009, 04:26 AM

Divide by $60 \times 60 = 3600$

**CaptainBlack** · October 4th 2009, 01:41 AM

Originally Posted by m_i_k_o

A satellite is in a circular orbit H km above the earth surface.
In how many hours does the satellite complete one orbit?

The radius of the earth is 6370 km.

H[km] = 289;

Could someone please show me how to solve this

Kepler's Third law tells us that for a circular orbit:

$\tau^2=kr^3$

where $\tau$ is the period of the orbit and $r$ its' radius.

For Geostationary orbit $\tau_{geo}=24$ hours, and $r_{geo}=35786$ km.

So the period $\tau_1$ that we seek satisfies:

$\frac{\tau_1^2}{\tau_{geo}^2}=\frac{6659^3}{r_{geo }^3}$

and the rest is arithmetic.

CB

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