# A geosynchronous satellite orbits at a distance from earth's center of about 6.6 earth radii and...

## Question:

A geosynchronous satellite orbits at a distance from earth's center of about 6.6 earth radii and takes 24 h to go around once. What distance in meters does the satellite travel in one day? What is its orbital velocity in m/s?

## Orbital speed:

The orbital speed of an object, revolving in a circular orbit, is defined as the ratio of the distance traveled(circumference) by this object in one revolution and the time taken in one revolution. Mathematically, it is expressed as follows:

{eq}v = \dfrac{ C}{ T } {/eq}

Where:

• {eq}C {/eq} is the distance traveled by the object in one revolution in its circular orbit.
• {eq}T {/eq} is the time taken by the object, revolving in the circular orbit, in one revolution.

## Answer and Explanation:

• The radius of the geosynchronous satellite's orbit is {eq}r = 6.6 R \,\,\, (\because R \text{ is the radius of the earth}) {/eq}
• Time taken by the geosynchronous satellite to make one revolution is {eq}T = 1.0 \, \rm day = 24 \, \rm h = 86400 \, \rm s {/eq}
• Radius of the earth is {eq}R = 6.37 \times 10^{6} \, \rm m {/eq}

The distance traveled by the satellite in one day will be equal to the circumference of the orbit, in which geosynchronous satellite revolves.

Therefore,

{eq}\begin{align} C &= 2 \pi r \\ &= 2 \pi ( 6.6 R ) \\ & = 2 \pi ( 6.6 ( 6.37 \times 10^{6} \, \rm m) ) \\ & = \color{blue}{\boxed {2.64 \times 10^{8} \, \rm m}} \end{align} {/eq}

Now,

The magnitude of the orbital velocity or speed of the satellite can be expressed as follows:

{eq}v = \dfrac{ C}{ T } {/eq}

After plugging in the values, we have:

{eq}v = \dfrac{2.64 \times 10^{8} \, \rm m }{ 86400 \, \rm s } {/eq}

Simplifying it further, we will get:

{eq}\color{blue}{\boxed{ v = 3.06 \times 10^{3} \, \rm m/s }} {/eq}