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