Question

A 2510 kg satellite used in a cellular telephone network is in a circular orbit at a height of 800 kmabove the surface of the earth.

a. What is the gravitational force on the satellite?

b. What fraction is this force of the satellite's weight at the surface of the earth?

Answer 1

The mass of the satellite (m₁) = 2510 kg

The height above the Earth = 800 km

You need to find the gravitational force on the satellite first, using this equation:

Where F is the froce due to gravity, G is the gravitational constant, m₁ is the mass of the first object, m₂ is the mass of the second object, and r is the distance between the two.

Here are those constants:

G = 6.673 * 10^-11
Mass of the Earth = 5.9742 * 10^24 kg

We need to find the distance between the two objects. We know the satellite is 800 km above the surface of the Earth, and we know the radius of the Earth to be 6371 km. The distande between the centers of gravity for both objects is simply adding the distances:

r = 800 km + 6371 km = 7171 km

Now that we have all the necessary components, all we have to do is plug the values into the gravitational force formula:

F = [(6.67*10^-11*2510*5.9742*10^24)/(7171*1000)^2]

Calulated out:

F =19.45 kN

To find the gravitational force on the surface of the Earth, use Newton's Second Law:

Where m is the mass of the object and a is its acceleration.

Because we are on the surface of the Earth, we know that the acceleration of the satellite is 9.81 m/s².

F_Earth = 2510*9.81

F_earth = 24.623 kN

The fraction of the two is as simple as it sounds:

F_space/F_earth = [19.45/24.623]

= 0.7899

I hope this helps!