A satellite orbits the earth near its surface. By what amount does the satellite's clock fall behind the earth's clock in one revolution ? Assume that non-relativistic analysis can be made to compute the speed of the satellite and only the time dilation is to be taken into account for calculation of clock speeds.

Explanation

The speed of the satellite may be obtained from the equation,
` GMm/ R^(2) = mv^(2) / R `
or, ` v = (sqrt GM / R) `
`= [ (6.67 xx 10^(11) Nm ^(2) kg)]/6400 xx 10^(3) m ]^(1/2)`
`= 7910 m s(^-1).
Thus . `v /c = 7910 / 3 xx 10 ^(5) `
or, (sqrt 1 - (v/c)^(2)) = [1 - 6.95 xx 10^(-10)] ^(1/2)`
`~~ 1 - 3.48 xx 10^(10) `
The time taken by the satellite to complete one revolution (this is proper time and 5080 s is improper time. ),
`t = (1- 3.48 xx 10^(-10)) xx (5080 s) `
or, `t / 5080 s = 1 - 3.48 xx 10^(-10) `
or, `(t - 5080 s )/ 5080 s = 3.48 xx 10^(-6) s .`
The satellite's clock falls behind by 1.77 xx 10^(-6)s. ` in one revolution .