Question

According to an observer at Earth's equator, by how much would his clock and one on a satellite in geosynchronous orbit differ in one day? (Geosynchronous orbit means an orbit period of one day-always in the same place in the sky)

Short answer

His clock and one on a satellite in geosynchronous orbit differ in one day by $4.006\times {10}^{-5}\mathrm{s}$.

Step-by-step solution

Significance of the geosynchronous orbit

The geosynchronous orbit is mainly described as an orbit which mainly matches the rotation of the Earth. The rotation of this orbit is about 24 hours.

Determination of the time

The equation of the velocity of that orbit is expressed as:

$\mathrm{v}={\mathrm{v}}_{\mathrm{Satellite}}-{\mathrm{v}}_{\mathrm{Earth}}$

Here, v is the velocity of that orbit,$v_{Earth}$ is the velocity of the Earth and$v_{Satellite}$ is the velocity of the satellite.

The above equation can also be expressed as:

$\mathrm{v}=\sqrt[3]{\frac{2\mathrm{\pi GM}}{\mathrm{t}}}-\frac{2{\mathrm{\pi r}}_{\mathrm{Earth}}}{\mathrm{t}}$

Here, G is the gravitational constant, M is the mass of the Earth,${\mathrm{r}}_{\mathrm{Earth}}$ is the radius of the Earth and t is the time taken for one complete rotation of the Earth.

Substitute$6.67\times {10}^{-11}{\mathrm{m}}^3/\mathrm{kg}.{\mathrm{s}}^2$ for G,$5.98\times {10}^{24}\mathrm{kg}$ for M,role="math" localid="1658320978791" $24\times 60\times 60\mathrm{s}$ for t and$6371\times {10}^3\mathrm{m}$ for${\mathrm{r}}_{\mathrm{Earth}}$ in the above equation.

$\begin{array}{rcl}\mathrm{v}& =& \sqrt[3]{\frac{2\left(3.41\right)\left(6.67\times {10}^{-11}{\mathrm{m}}^3/\mathrm{kg}.{\mathrm{s}}^2\right)\left(5.98\times {10}^{24}\mathrm{kg}\right)}{24\times 60\times 60\mathrm{s}}}-\frac{2\left(3.14\right)\left(6371\times {10}^3\mathrm{m}\right)}{24\times 60\times 60\mathrm{s}}\\ & =& \sqrt[3]{\frac{2.5\times {10}^{15}{\mathrm{m}}^3/{\mathrm{s}}^2}{8.64\times {10}^4\mathrm{s}}}-\frac{\left(4.0009\times {10}^7\mathrm{m}\right)}{8.64\times {10}^4\mathrm{s}}\\ & =& 3068.78\mathrm{m}/\mathrm{s}-463.06\mathrm{m}/\mathrm{s}\\ & =& 2605.72\mathrm{m}/\mathrm{s}\end{array}$

The equation of the time period of Earth if the velocity exceeds the speed of light is expressed as:

${\mathrm{t}}_1=\left(1+\frac{{\mathrm{v}}^2}{2{\mathrm{c}}^2}\right)\mathrm{t}$

Here, t₁ is the time period of Earth in the first case and c is the light’s speed.

Substitute 2605.72 m/s for v and$3\times {10}^8\mathrm{m}/\mathrm{s}$ for c in the above equation.

$\begin{array}{rcl}{\mathrm{t}}_1& =& \left(1+\frac{{\left(2605.72\mathrm{m}/\mathrm{s}\right)}^2}{2{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2}\right)\mathrm{t}\\ & =& \left(1+\frac{6.7\times {10}^6{\mathrm{m}}^2/{\mathrm{s}}^2}{2\left(9\times {10}^{16}{\mathrm{m}}^2/{\mathrm{s}}^2\right)}\right)\mathrm{t}\\ & =& \left(1+\frac{6.7\times {10}^6{\mathrm{m}}^2/{\mathrm{s}}^2}{18\times {10}^{16}{\mathrm{m}}^2/{\mathrm{s}}^2}\right)\mathrm{t}\\ & =& \left(4.7\times {10}^{-11}\right)\mathrm{t}\end{array}$

From the general relativity, the equation of the time period of Earth can be expressed as:

${\mathrm{t}}_2=\mathrm{t}\left(1-\frac{\mathrm{GM}}{{\mathrm{c}}^2}\left(\frac{1}{{\mathrm{r}}_{\mathrm{Erath}}}-\frac{2\mathrm{\pi }}{\mathrm{vt}}\right)\right)$

Here, t₂ is the time period of the Earth in the second case.

Substitute the values in the above equation.

$\begin{array}{rcl}{\mathrm{t}}_2& =& \mathrm{t}\left(1-\frac{\left(6.67\times {10}^{-11}{\mathrm{m}}^3/\mathrm{kg}.{\mathrm{s}}^2\right)\left(5.98\times {10}^{24}\mathrm{kg}\right)}{{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2}\left(\frac{1}{6371\times {10}^3\mathrm{m}}-\frac{2\left(3.14\right)}{\left(2605.72\mathrm{m}/\mathrm{s}\right)\mathrm{t}}\right)\right)\\ & =& \mathrm{t}\left(1-\frac{\left(3.98\times {10}^{14}{\mathrm{m}}^3/{\mathrm{s}}^2\right)}{9\times {10}^{16}{\mathrm{m}}^2/{\mathrm{s}}^2}\left(1.56\times {10}^{-7}{\mathrm{m}}^{-1}-\frac{\left(6.28\right)}{\left(2605.72\mathrm{m}/\mathrm{s}\right)\mathrm{t}}\right)\right)\\ & =& \mathrm{t}\left(-3.4\times {10}^{-3}\mathrm{m}\left(1.56\times {10}^{-7}{\mathrm{m}}^{-1}-\frac{2.41\times {10}^{-3}\mathrm{s}/\mathrm{m}}{\mathrm{t}}\right)\right)\\ & =& -5.3\times {10}^{-10}\mathrm{t}+8.1\times {10}^{-6}\mathrm{s}\end{array}$

The equation of the difference between the time interval is expressed as:

${\mathrm{t}}_3={\mathrm{t}}_2-{\mathrm{t}}_1$

Here, t₃ is the difference between the time interval.

Substitute$24\times 60\times 60\mathrm{s}$ for t in the above equation.

$\begin{array}{rcl}{\mathrm{t}}_3& =& \left(4.7\times {10}^{-11}\right)\left(24\times 60\times 60\mathrm{s}\right)-\left(-5.3\times {10}^{-10}\left(24\times 60\times 60\mathrm{s}\right)+8.1\times {10}^{-6}\mathrm{s}\right)\\ & =& 4.06\times {10}^{-6}\mathrm{s}-\left(-4.5\times {10}^{-5}\mathrm{s}+8.1\times {10}^{-6}\mathrm{s}\right)\\ & =& 4.06\times {10}^{-6}\mathrm{s}+3.6\times {10}^{-5}\mathrm{s}\\ & =& 4.006\times {10}^{-5}\mathrm{s}\end{array}$

Thus, his clock and one on a satellite in geosynchronous orbit differ in one day by $4.006\times {10}^{-5}\mathrm{s}$.