Question

A geosynchronous satellite orbits Earth at a distance of 42,250 km from the center of Earth and has a period of 1 day. What is the centripetal acceleration of the satellite (in m/s2)?

Answer 1

Answer:

The centripetal acceleration of the satellite is $a=0.22\ m/s^2$.

Explanation:

Given that,

The distance covered by a geosynchronous satellite, d = 42250 km

The time taken by the satellite to covered distance, t = 1 day = 24 hours

Since, 24 hours = 86400 seconds

Let v is the speed of the satellite. It is given by the total distance divided by total time taken such that :

$v=\dfrac{d}{t}$

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

$v=\dfrac{2\pi\times 42250 \times 10^3}{86400 }$

v = 3072.5 m/s

The centripetal acceleration of the satellite is given by :

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

$a=\dfrac{(3072.5)^2}{42250 \times 10^3}$

$a=0.22\ m/s^2$

So, the centripetal acceleration of the satellite is $a=0.22\ m/s^2$. Hence, this is the required solution.