A satellite is in a circular orbit H km above the earth surface.

In how manyhoursdoes 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

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- Oct 3rd 2009, 03:53 AMm_i_k_oHow to find the orbit time of this satellite?
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 - Oct 3rd 2009, 04:01 AMmathaddict

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

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

Rearranging ..

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

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 . - Oct 3rd 2009, 04:19 AMm_i_k_oConverting to hours
To convert to hours I will just divide T by 360?

- Oct 3rd 2009, 04:26 AMe^(i*pi)
Divide by $\displaystyle 60 \times 60 = 3600$

- Oct 4th 2009, 01:41 AMCaptainBlack
Kepler's Third law tells us that for a circular orbit:

$\displaystyle \tau^2=kr^3$

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

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

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

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

and the rest is arithmetic.

CB