Question

A satellite is put into a circular orbit 200 km above
the earth's surface where the acceleration due to
gravity is 9.2 ms 2. Determine (i) satellite's speed
(ii) Time period (time to take one revolution). The
radius of earth is 6400 km.​

Answer 1

Explanation:

It is given that,

Radius of the Earth, R = 6400 km

A satellite is put into a circular orbit 200 km above  the earth's surface, r = 200 km

Value of acceleration due to gravity, $a=9.2\ m/s^2$

Total distance, d = R + r

d = 6400 km + 200 km = 6600 km

$d=66\times 10^5\ m$

1. The centripetal acceleration acts on the satellite which is given by :

$a=\dfrac{v^2}{d}$

$v=\sqrt{ad}$

$v=\sqrt{9.2\ m/s^2\times 66\times 10^5\ m}$

$v=7792.30\ m/s$

2. The period to taken one revolution is given by :

$T=\dfrac{2\pi d}{v}$

$T=\dfrac{2\pi \times 66\times 10^5\ m}{7792.30\ m/s}$

T = 5321.79 seconds

or

T = 1.47 hours

Hence, this is the required solution.

Answer 2

Answer:

hope it helps you. it's a tricky question anyhow.