Answer:
Clock on the satellite is slower than the one present on the earth = 29.376 s
Given:
Distance of satellite from the surface, d = 250 km
Explanation:
Here, the satellite orbits the earth in circular motion, thus the necessary centripetal force is provided by the gravitation force and is given by:
[tex]\frac{mv^{2}}{R} = \frac{GMm}{R^{2}}[/tex]
where
v = velocity of the satellite
R = radius of the earth = 6350 km = 6350000 m
G = gravitational constant = [tex]6.674\times 10^{- 11} m^{3}/ks-s^{2}[/tex]
M = mass of earth = [tex]5.972\times 10^{24} kg[/tex]
Therefore, the above eqn can be written as:
[tex]v = \sqrt{\frac{GM}{R}}[/tex]
Now, for relativistic effects:
[tex]\frac{v}{c} = \sqrt{\frac{GM}{Rc^{2}}} = 26.41\times 10^{- 6}[/tex]
Now,
r = R + 250
[tex]\frac{v_{surface}}{c} = {\frac{1}{c}\frac{2\pi R}{24} = 1.54\times 10^{-6}[/tex]
Ratio of rate of satellite clock to surface clock:
[tex]\frac{\sqrt{1 - \frac{v^{2}}{c^{2}}}}{\sqrt{1 - \frac{v_{surface}^{2}}{c^{2}}}} = 3.43\times 10^{-10}[/tex]
Clock on the satellite is slower than the one present on the earth:
[tex]3.43\times 10^{-10}\times 24\times 3600 = 29.376 s[/tex]
