Question

A 2400 kg satellite is in a circular orbit around a planet. The satellite travels with a constant speed of 6000 m/s. The radius of the circular orbit is 10 6 m

Answer 1

The magnitude of the gravitational force exerted on the satellite by the planet  will be F = 11970.1 N

What is gravitational force?

The gravitational force is a force that attracts any two objects with mass.It is directly proportional to the masses and inverse of the distance between the two objects.

It is given that,

Mass of the satellite, m = 2400 kg

Speed of the satellite, v = 6670 m/s

Radius of the circular path,

Let F is the magnitude of the gravitational force exerted on the satellite by the planet. The centripetal force is equal to the gravitational force. It is equal to :

$F=\dfrac{mv^2}{r}$

$F=\dfrac{2400\times (6670)^2}{8.92\times 10^6}$

F = 11970.1 N

So, the magnitude of the gravitational force exerted on the satellite by the planet is 11970.1 N. Hence, this is the required solution.

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