Question

A 213.7 kg satellite is in a circular orbit of 22,236 miles (35,768 km) in radius. The force keeping the satellite in orbit is 44.1 N. What is the velocity (speed) of the satellite

Answer 1

Answer:

v = 3.08 k ms^-1

Explanation:

As we know that, $v=\sqrt{GM/r}$

where r = R + h = 35768 + 6400 = 42168 km = 4.2168 x 10^7 m

G = 6.673 x 10^-11 N m^-2 kg^-2 and M = 6.0 x 10^24 kg

By substituting the required values, we get...

v = $\sqrt{}$6.673 x 10^-11 x 6.0 x 10^24 / 4.2168 x 10^7 = $\sqrt{}$9.494 x 10^6 ms^-1 = 3081.2 ms^-1 = 3.08 k ms^-1

Answer 2

Answer:

2716.84 m/s

Explanation:

Using,

F = mv²/r................. Equation 1

Where F = The force keeping the satellite in the orbit, m = mass of the satellite, v = velocity of the satellite, r = radius of the circular orbit.

make v the subject of equation 1

v = √(Fr/m).................. Equation 2

Given: F = 44.1 N, r = 35768 km = 35768000 m, m = 213.7 kg.

Substitute into equation 2

v = √(44.1×35768000/213.7)

v = √(7381229.761)

v = 2716.84 m/s

Hence the velocity of the satellite = 2716.84 m/s