Answer:
The centripetal acceleration of the satellite is [tex]a=0.22\ m/s^2[/tex].
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 :
[tex]v=\dfrac{d}{t}[/tex]
[tex]v=\dfrac{2\pi d}{t}[/tex]
[tex]v=\dfrac{2\pi\times 42250 \times 10^3}{86400 }[/tex]
v = 3072.5 m/s
The centripetal acceleration of the satellite is given by :
[tex]a=\dfrac{v^2}{d}[/tex]
[tex]a=\dfrac{(3072.5)^2}{42250 \times 10^3}[/tex]
[tex]a=0.22\ m/s^2[/tex]
So, the centripetal acceleration of the satellite is [tex]a=0.22\ m/s^2[/tex]. Hence, this is the required solution.
