Question

A satellite orbits the Earth in the same direction it rotates in a circular orbit above the equator a distance of 250 km from the surface. By how many seconds per day will a clock on such a satellite run slow compared to a clock on the Earth? (Compute just the special relativistic effects.)

Answer 1

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:

$\frac{mv^{2}}{R} = \frac{GMm}{R^{2}}$

where

v =  velocity of the satellite

R = radius of the earth = 6350 km = 6350000 m

G = gravitational constant = $6.674\times 10^{- 11} m^{3}/ks-s^{2}$

M = mass of earth = $5.972\times 10^{24} kg$

Therefore, the above eqn can be written as:

$v = \sqrt{\frac{GM}{R}}$

Now, for relativistic effects:

$\frac{v}{c} = \sqrt{\frac{GM}{Rc^{2}}} = 26.41\times 10^{- 6}$

Now,

r = R + 250

$\frac{v_{surface}}{c} = {\frac{1}{c}\frac{2\pi R}{24} = 1.54\times 10^{-6}$

Ratio of rate of satellite clock to surface clock:

$\frac{\sqrt{1 - \frac{v^{2}}{c^{2}}}}{\sqrt{1 - \frac{v_{surface}^{2}}{c^{2}}}} = 3.43\times 10^{-10}$

Clock on the satellite is slower than the one present on the earth:

$3.43\times 10^{-10}\times 24\times 3600 = 29.376 s$