Question

A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take the mass of the Earth to be 5.97 x 10^24 kg and its radius 6.38 x10^6 m.

Required:
What is the orbital period of this GPS satellite?

Answer 1

`T=66262.4s`

Step-by-step explanation:

From the question we are told that:

Altitude
`A=2.90 *10^7`

Mass
`m=5.97 * 10^(24) kg`

Radius
`r=6.38 *10^6 m.`

Generally the equation for Satellite Speed is mathematically given by

`V=((GM)/(d) )^(0.5)`

`V=((6.67*10^(-11)*5.97 * 10^(24))/(6.38 *10^6+2.90 *10^7) )^(0.5)`

`V=3354.83m/s`

Therefore

Period T is Given as

`T=(2 \pi *a)/(V)`

`T=(2 \pi *(6.38 *10^6+2.90 *10^7)/(3354.83)`

`T=66262.4s`