Show that period of a satellite revolving around the Earth depends upon mass of the Earth.

Explanation

Consider a satellite of mass m projected in a circular orbit round the earth from a height h above the earth's surface with radius of orbit `r=R+h.`
Let, in one revolution the distance covered by the satellite is equal to the circumference of its circular orbit and the time taken is the periodic time, T.
Cirtical speed `(v_(c))=("Circumference of the orbit")/("Period time") `
`v_(c)=(2pi r)/(T)`
But, `v_(c)=(2pi r)/(T)`
But, `v_(c)=sqrt((GM)/(r))`
On squaring both sides, we get
`(GM)/(r)=(4pi^(2)r^(2))/(T^(2))`
`therefore T^(2)=(4pi^(2)r^(3))/(GM)=((4pi^(2))/(GM))r^(3)`
`therefore T^(2) prop r^(3) " " ( because (4pi^(2))/(GM) " is constant")`
Thus, the square of the time period of revolution of the stalite is directly proportional to the cube of the radius of its orbit.
Since, `T^(2)=(4pi^(2)r^(3))/(GM)`
`therefore T=2pi sqrt((r^(3))/(GM))`
Which is the expression for the time period of the satellite revolving around the earth. Time period of the satellite depends on mass of the earth.