# A satellite with an orbital period of exactly 24.0 h is always positioned over the same spot on...

## Question:

A satellite with an orbital period of exactly 24.0 h is always positioned over the same spot on Earth. This is known as a geosynchronous orbit. Television, communication, and weather satellites use geosynchronous orbits.

At what distance would a satellite have to orbit Earth in order to have a geosynchronous orbit?

## Kepler's Third Law:

Kepler's Third Law is the last of the three laws of Kepler which explains the behavior of planetary motion. The third law relates the period of the orbit and the distance of a orbiting satellite to its host celestial body. It is formulated as the ratio of the square of the period over the cube of the distance equates to 1 or the square of the period is proportional to the cube of the period. Newton's Theory of Gravitation modified Kepler's Third Law and relates it to the host celestial body mass. It is mathematically formulated as:

{eq}T^{2}=\frac{4\pi^{2}}{GM}(r^{3}) {/eq}

Where:

{eq}T {/eq}=Period of the orbit

{eq}G {/eq}=Gravitational Constant

{eq}M {/eq}=Mass

{eq}r {/eq}=Distance

The distance from the earth's surface the satellite have to orbit to have a geosynchronous orbit is {eq}35.8\times10^{6} m {/eq}

Solution:

Converting the Period to seconds

{eq}24 hr \times \frac{60 mins}{1 hr} \times \frac{60 seconds}{1 min} {/eq}

={eq}8.64\times10^{4} s {/eq}

Using Newton's-Revised Kepler Third Law

{eq}T^{2}=\frac{4\pi^{2}}{GM}(r^{3}) {/eq}

Rearranging to get distance then

{eq}r=\sqrt\frac{GMT^{2}}{4\pi^{2}} {/eq}

={eq}\sqrt\frac{(6.67\times10^{-11} \frac{m^{3}}{kg \cdot s^{2}})(5.97\times10^{24} kg)(8.64\times10^{4} s)^{2}}{4\pi^{2}} {/eq}

={eq}\sqrt{7.53\times10^{22} m^{3}} {/eq}

={eq}4.22\times10^{7} m {/eq}

This is the distance from the Earth's center, to get the distance from the earth's surface

{eq}r_{orbit}=r_{total}-r_{earth} {/eq}

{eq}r_{orbit}=(4.22\times10^{7} m)-(6.4\times10^{6} m) {/eq}

={eq}3.58\times10^{6} m {/eq} 