Question

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

Answer 1

In order to solve this question we need to understand the relativistic effect. So when a particle moves with speed close to light like muons then relativistic effect comes into account. To relativistic effect, length of particle moving would seem to be contracted from a frame which is at rest in comparison to moving frame and the time would be dilated for particle moving with velocity close to light.

Complete step by step answer:
Time dilation can be experienced in real life by the twin paradox in which one of twins stays on earth and one moves in a spaceship close to the speed of light then after some time on earth if the traveler twin returned to earth then he would be younger than the twin who stayed on earth.

Mass of earth is given as, $M = 6 \times {10^{24}}kg$.
Radius of earth is, $R = 6400 \times {10^3}m$.
And the value of gravitation constant is, $G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$.
And let the mass of satellite be, $m$.Using non-relativistic effect to calculate satellite speed.Let the satellite speed be “v”.Then the gravitational attraction between satellite and earth would be given as,
${F_g} = \dfrac{{GMm}}{{{R^2}}}$
Also the centripetal force be given as,
${F_c} = \dfrac{{m{v^2}}}{R}$..............(Since the satellite is orbiting close to earth)

So for the satellite to be at equilibrium, we equate both the force and get, ${F_g} = {F_c}$.
Putting values we get, $\dfrac{{GMm}}{{{R^2}}} = \dfrac{{m{v^2}}}{R}$.
So the velocity of the satellite is, $v = \sqrt {\dfrac{{GM}}{R}}$.
Putting values we get,
$v = \sqrt {\dfrac{{(6.67 \times {{10}^{ - 11}}N{m^2}k{g^{ - 2}}) \times (6 \times {{10}^{24}}kg)}}{{(6400 \times {{10}^3}m)}}}$
$\Rightarrow v = 7910\,m{\sec ^{ - 1}}$
Since the light speed is, $c = 3 \times {10^8}m{\sec ^{ - 1}}$
So, $(\dfrac{v}{c}) = \dfrac{{7910}}{{3 \times {{10}^8}}}$
$(\dfrac{v}{c}) = 2.637 \times {10^{ - 5}}$

So we use relativistic constant as,
$\gamma = \dfrac{1}{{\sqrt {1 - {{(\dfrac{v}{c})}^2}} }}$
Putting values we get,
$\gamma = \dfrac{1}{{\sqrt {1 - {{(2.637 \times {{10}^{ - 5}})}^2}} }}$
$\Rightarrow \gamma = {(1 - {(2.637 \times {10^{ - 5}})^2})^{ - (\dfrac{1}{2})}}$
$\Rightarrow \gamma = {(1 - (6.95 \times {10^{ - 10}}))^{ - (\dfrac{1}{2})}}$
Since $v < < c$ so by Direct binomial approximation we can write it as,
$\gamma = 1 - \dfrac{1}{2}(6.95 \times {10^{ - 10}})$
$\Rightarrow \gamma = 1 - (3.48 \times {10^{ - 10}})$
So the time taken by the satellite to complete one revolution is $T = \dfrac{{2\pi R}}{v}$.
Putting values we get,
$T = \dfrac{{2\pi (6400 \times {{10}^3}m)}}{{7910m{{\sec }^{ - 1}}}}$
$\Rightarrow T = 5080\sec$

This time period is improper time as it is measured by an observer on earth.Let the proper time measure be $t$ as observed by the observer at the satellite itself. Since we know the time dilation formula is given as. $t = \gamma T$
Putting values we get, $t = (1 - (3.48 \times {10^{ - 10}})) \times (5080)$
$\dfrac{t}{{5080}} = 1 - (3.48 \times {10^{ - 10}})$
$\Rightarrow \dfrac{t}{{5080}} - 1 = - (3.48 \times {10^{ - 10}})$
$\Rightarrow \dfrac{{(t - 5080)}}{{5080}} = - (3.48 \times {10^{ - 10}})$
$\Rightarrow (t - 5080) = - (3.48 \times {10^{ - 10}}) \times 5080$
$\Rightarrow (t - 5080) = - 1.77 \times {10^{ - 6}}\sec$
$\therefore (t - 5080) = - 1.77\mu \sec$

So satellite clocks fall behind by $1.77\mu \sec$ in one revolution.

Note: It should be remembered that negative signs indicated that time falls behind the earth clock. If you observe closely then the time by which it falls behind is very less, so there is almost no difference in satellite clock and earth’s clock and this is due to the fact that velocity of satellite is much less in comparison to velocity of light.