Question

A 2400-kg satellite is in a circular orbit around a planet. the satellite travels with a constant speed of 6670 m/s. the radius of the circular orbit is 8.92 × 106 m. calculate the magnitude of the gravitational force exerted on the satellite by the planet.

Answer 1

Answer:

Gravitational force, F = 11970.1 N

Explanation:

It is given that,

Mass of the satellite, m = 2400 kg

Speed of the satellite, v = 6670 m/s

Radius of the circular path, $r=8.92\times 10^6\ m$

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.

Answer 2

The gravitational force exerted on the satellite is called the centrifugal force, the force keeping it orbiting to the planet. Its formula is F= mass times the square of the velocity all over the radius.Thus,

F = 2400 * 6670^2 * (1/8.92x10^6) 
F = 11,970 N

I hope I was able to help you. Have a good day.