Question

What is the period of revolution of a satellite with mass m

that orbits the earth in a circular path of radius 9480 km
(about 3100 km
above the surface of the earth)?
Express your answer in seconds.

Answer 1

Step-by-step explanation:

T = 2π√((9,480,000)³/(6.67430 × 10^(-11) * 5.972 × 10^24))
METHOD:-

Using the formula to determine period of revolution of a satellite

T = 2π√(r³/GM)

where

T=period of revolution (in seconds)

π=3.14159

r=radius of the orbit (in m)

G=6.67430 × 10^(-11) m³/(kg·s²(approx)

M =5.972 × 10^24 kg(approx)

Now to calculate the period of revolution, we will have to convert the radius from km(kilometers) to m(meters):

multiply the radius by 1000

r = 9480 km * 1000 = 9,480,000 meters

Substitute the values into the formula:

T = 2π√((9,480,000)³/(6.67430 × 10^(-11) * 5.972 × 10^24))

calculating this equation we will get our answer