An orbiting satellite stays over a certain spot on the equator of (rotating) Earth. What is the altitude of the orbit (called ageosynchronous orbit)?

${\mathrm{T}}_{\mathrm{satellite}}={\mathrm{T}}_{\mathrm{earth}}=86400\text{s\hspace{0.17em}\hspace{0.17em}}\left(\text{asthestatelliteisGeostationarysatellite}\right)$

$\mathrm{G}=6.67\times {10}^{-11}\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}$

${\text{M}}_{\text{earth}}=5.98\times {10}^{24}\text{\hspace{0.17em}kg}$

${\text{R}}_{\text{earth}}\text{=}6.37\times {10}^6\text{m}$

As the satellite is revolving around earth, using Kepler’s 3^rd law, first find the orbital radius of the satellite from the center of the earth.

To find the height of the orbit, subtract the radius of the earth from the orbital radius to get the required answer. According to Kepler’s third law, the squares of the orbital periods of the planets are directly proportional to the cubes of the semi major axes of their orbits.

Formula is as follow:

${\mathrm{T}}^2=\left(\frac{4{\mathrm{\pi }}^2}{\mathrm{GM}}\right){\mathrm{a}}^3$

where, T is time, G is gravitational constant, M is mass and a is altitude of the orbit.

Let, the height of satellite from the center of Earth be r. Using Kepler’s 3rd law,

${\mathrm{T}}^2=\left(\frac{4{\mathrm{\pi }}^2}{\mathrm{GM}}\right){\mathrm{a}}^3$

${\mathrm{a}}^3={\mathrm{T}}^2\times \left(\frac{\mathrm{GM}}{4{\mathrm{\pi }}^2}\right)$

$\mathrm{a}=\sqrt[3]{\left({\mathrm{T}}^2\times \left(\frac{\mathrm{GM}}{4{\mathrm{\pi }}^2}\right)\right)}$

$\mathrm{a}=\sqrt[3]{{\left(86400\mathrm{s}\right)}^2\times \left(\frac{6.67\times {10}^{-11}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2\times 5.98\times {10}^{24}\mathrm{kg}}{4{\mathrm{\pi }}^2}\right)}$

$\text{a}=4.225\times {10}^7\text{\hspace{0.17em}m}$

So, the height of the orbit from the surface of the earth will be,

$\mathrm{h}=\mathrm{r}-{\mathrm{R}}_{\mathrm{earth}}$.

$\begin{array}{c}\mathrm{h}=(4.225\times {10}^7\mathrm{m})-(6.37\times {10}^6\mathrm{m})\\ =3.59\times {10}^7\mathrm{m}\end{array}$

Hence, thealtitude of the orbit will be${\text{3.59×10}}^{\text{7}}\text{\hspace{0.17em}m}$.

Therefore, the altitude of the geostationary satellite can be found using the property of the geostationary satellite having the same time period or revolution as the time period of rotation of the earth.