In: Physics
A spy satellite is in circular orbit around Earth. It makes one
revolution in 5.99 hours. How high above Earth's surface is the
satellite?
: 10368477 m
11.[2pt]
What is the satellite's acceleration?
Time period of a satellite orbiting at a distance r from the center of earth is given by,
G is the universal gravitational constant and M is the mass of earth.
Time period of the satellite is 5.99 hr.
Mass of earth is, 5.972X10^24 kg
Putting the values,
Radius of earth is r_e=6371 km
So, height of the satellite above the earth's surface is
11) Angular velocity of the satellite with time period T is given by,
Time period is 5.99 hours
So, angular velocity is,
So the satellite's acceleration is,