Question

17. A 500 kg satellite revolves around the earth at a
height of 10 km above the earth's surface. Calculate,
assuming circular orbit :
(i) speed of satellite,
(ii) angular velocity of satellite,
(iii) gravitational force of the earth on the satellite.
Given :
radius of the earth 6.4 x 10^6m, g = 9.80 m/s.​

Answer 1

Answer:

(i) Speed of satellite=7925.78 m/s

(ii) Angular velocity of satellite =$\varpi=1.236\times10^{-3}$rad/sec

(iii) Gravitational force of the earth on the satellite=4900 Newton

Explanation:

A 500 kg satellite revolves around the earth at a  height of 10 km above the earth's surface

Given m=500kg,h=10km,$r=6.4\times10^{6}$

(i)we  know that speed of satellite is given by $v=\sqrt{g(r+h)}$

so $v=\sqrt{9.8\times(6.4\times10^{6}+10\times10^{3)} }$

v=7925.78m/s

(ii) we know that angular velocity is given by $\varpi=\frac{v}{(r+h)}$

$\varpi=1.236\times10^{-3}$rad/sec

(iii) Gravitational force of the earth on the satellite F=mg

$F=500\times9.8$

Hence F=4900 Newton