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?

Respuesta :

Answer:

[tex]T=66262.4s[/tex]

Explanation:

From the question we are told that:

Altitude [tex]A=2.90 *10^7[/tex]

Mass [tex]m=5.97 * 10^{24} kg[/tex]

Radius [tex]r=6.38 *10^6 m.[/tex]

Generally the equation for Satellite Speed is mathematically given by

[tex]V=(\frac{GM}{d} )^{0.5}[/tex]

[tex]V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}[/tex]

[tex]V=3354.83m/s[/tex]

Therefore

Period T is Given as

[tex]T=\frac{2 \pi *a}{V}[/tex]

[tex]T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}[/tex]

[tex]T=66262.4s[/tex]