A 2130 kg satellite used in a cellular telephone network is in a circular orbit at a height of 800 km above the surface of the earth. What is the gravitational force on the satellite? What fraction is this force of the satellite's weight at the surface of the earth?
Given
The mass of the satellite (m1) = 2130 kg
The height above the Earth = 800 km
As we know that,
Force due to gravity, F = (G x m1 x m2 )/r^2
Where
G is the gravitational constant i.e. 6.673 x 10^(-11)
m1 is the mass of satellite
m2 is the mass of Earth i.e. 5.972 x 10^24 kg
r is the distance between the two.
r = 800 km + 6371 km = 7171 km
Now, F = (6.673 x 10^(-11) x 2130 x 5.972 x 10^24 )/(7171 x 10^3)² = 16506.731 N
the gravitational force on the surface of the Earth is given by, F" = mg
That implies, F" = 2130 x 9.81 = 20895.3
Now F/F" = 16506.731/20895.3 = 0.78
Get Answers For Free
Most questions answered within 1 hours.