Question

A satellite is revolving in circular orbit at a height of 1000 km from the surface of the Earth. Calculate the orbital velocity and time of revolution. The radius of the Earth is 6400 km and the mass of the Earth is 6 × 10^24 kg.

Answer 1

let the mass of satellite be m and velocity be v.
Distance b/w satellite and center of earth,r=6400+1000=7400
The necessary centripetal force for satellite to revolve around the earth is  given by the gravitational attraction b/w earth and satellite  
i.e $\frac{GMm}{ r^{2} } = \frac{m v^{2} }{r}$
$\frac{GM}{r}= v^{2}$
$v^{2} = \frac{6.67 * 10^{-11} * 6 * 10^{24} }{7400}$
$v= \sqrt{400.2 * 10^{10} }$
v=2 *10⁶ ans .....

Answer 2

2×10^6

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