satellites orbit earth at distance of 42,000,000 m from earths center. Their angular velocity at this height is the same as the rotation of the earth, So they appear stationary at certain locations in the sky. what is the period of the satellite? The mass of the earth is 5.97x10^24 kg
electron1
The period is the time for the satellite to move a distance that is equal to the circumference of a circle. Since the satellite is moving at a constant speed, the centripetal force is equal to the Universal gravitational force.
m * v^2 ÷ r = G * M * m ÷ r^2
v = √(G * M ÷ r
G * m = 6.67 * 10^-11 * 5.97 * 10^24 = 3.98199 * 10^14
r = 4.2 * 10^7
v = √(3.98199 * 10^14 ÷ 4.2 * 10^7)
This is approximately 3,079 m/s
C = 2 * π * 4.2 * 10^7 = π * 8.4 * 10^7 meters
To determine the time for the satellite to orbit the earth, divide this distance by its velocity.
t = π * 8.4 * 10^7 ÷ √(3.98199 * 10^14 ÷ 4.2 * 10^7)
This is approximately 85,704.5 seconds
One day = 24 hours
One hour = 3600 seconds
One day = 24 * 3600 = 86,400 seconds.
If we round by answer, it will be one day. This proves that the satellite and earth have the same angular velocity.
Andrew Smith
24 hours ( approximately)
If it appears stationary then it must turn at the same rate that the earth does.
To be quite precise the earth turns one full turn in slightly less than 24 hours.
It must turn a little more than a full turn so that it faces the sun at the same moment each day.
Hence 24 hours = T * ( 1+1/365.3 )
T = 24 / ( 1+1/365.3) = 23.93 hours
or 8.6 * 10 ^ 4 s