# A GPS satellite orbits at an altitude of 20,200 km above the surface of the earth. What is the...

## Question:

A GPS satellite orbits at an altitude of 20,200 km above the surface of the earth. What is the speed of the satellite?

## Speed of Satellite:

When a satellite in put at any altitude above the earth surface, gravitational force helps it to stay at its position as it provides the necessary centripetal force.

{eq}\dfrac {mv^2}{R} = \dfrac {Gmm_e}{R^2} {/eq}

The velocity of the satellite can be given by the following formula:

{eq}v = \sqrt {\dfrac {Gm_e}{R}} {/eq}

where,

'G' is the gravitational constant = {eq}6.67 \times 10^{-11} \ \dfrac {Nm^2}{{Kg}^2} {/eq}

{eq}'m_e' {/eq} is the mass of earth ={eq}5.97 \times 10^{24} \ Kg {/eq}

'R' is the radius of orbit = r + h

where,

'r' is the radius of earth = {eq}6378 \ km. {/eq}

h = altitude of the satellite above the earth surface.

Given:

Altitude of satellite, {eq}h = 20,200 \ km = 20,200\times 10^3 \ m {/eq}

{eq}\\ {/eq}

{eq}R = r + h \\ \Rightarrow R = 6378 + 20,200 \ km \\ \Rightarrow R = 26578 \ km \\ \Rightarrow R = 26578 \times 10^3 \ m \\ {/eq}

{eq}\\ {/eq}

The velocity of the satellite is given by:

{eq}v = \sqrt {\dfrac {Gm_e}{R}} \\ \Rightarrow v = \sqrt {\dfrac {(6.67 \times 10^{-11}) (5.97 \times 10^{24})}{ 26578 \times 10^3}} \\ \Rightarrow v = \sqrt {\dfrac { 3.98 \times 10^14 }{ 26578 \times 10^3}} \\ \Rightarrow v = 3870.69 \ \dfrac {m}{s} \\ \Rightarrow v = 3.87 \ \dfrac {km}{s} \\ {/eq}

Therefore, the speed of satellite orbiting at an altitude of 20,200 km above the earth surface will be {eq}3.87 \ \dfrac {km}{s}. {/eq}