Question

A satellite is in a 91.0 min period circular orbit 330 km above Earth's surface. A)Find the satellite's speed.
B)Find its centripetal acceleration.

Answer 1

Explanation:

It is given that,

The satellite is placed 330 km above Earth's surface, r' = 330 km

The radius of the Earth, R = 6,371  km

Time period of the satellite, T = 91 min = 5460 seconds

So, the radius of satellite, $r=(330+6371)\ km=6701\times 10^3\ m$

(A) The speed of the satellite is given by :

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

$v=\dfrac{2\pi \times 6701\times 10^3}{5460}$

v = 7711.28 m/s

(B) The centripetal acceleration is given by :

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

$a=\dfrac{(7711.28)^2}{6701\times 10^3}$

$a=8.87\ m/s^2$

Hence, this is the required solution.