What is the time period of satellite near the earth's surface? (neglect height of the orbit of satellite from the surface of the earth)?

Explanation

The force on the satellite due to earth is given by `F=(GmM)/(R^(2))`
M-Mass of earth `=6xx10^(24) kg`
m-mass of satellite,
R-radius of earth `=6.4xx10^(6) m`
Let v be the speed of the satellite
`v=(2pi R)/(T) rArr T=(2 pi R)/(v)`
Required centripetal force is provided to satellite by the gravitational force hence
`F_(C )=(mv^(2))/(R)`
But `F_(C)=(GMm)/(R^(2))` according to Newton.s law of gravitation.
i.e., `(GMm)/(R^(2))=(m(2pi R)^(2))/(T^(2)R)`
`rArr T^(2) =(4 pi^(2) R^(3))/(GM)`, as mass of the earth (M) and G are constants the value of T depends only on the radius of the earth.
`rArr T^(2) alpha R^(3)`
Substituting the values of M, R and G in above equation we get, T = 84.75 minutes. Thus the satellite revolving around the earth in a circular path near to the earth.s surface takes 1Hour and 24.7 minutes approximately to complete one revolution around earth.