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Question

A GPS satellite orbits at an altitude of 2.06×104 km above the surface of Earth. (Recall that RE = 6.37 × 106 m.)

What is the speed of the satellite?

Answer 1

To find the the velocity of the GPS satellite, you need to just remember the equation for orbital speed:

 

v= √((Gm_e) /R)

where G= gravitational constant

m_e= mass of earth

R= radius of earth + altitude of satellite

 

G= 6.67 x10^-11 Nm²/kg²

m_e= 5.97 x10²⁴ kg

R= 6.37 x 10⁶ m + 2.06 x 10⁷= 2.697 x 10⁷= 2.70 x 10⁷ m

 

v= √((6.67 x10^-11 Nm²/kg² * 5.97 x10²⁴ kg)/ 2.70 x 10⁷ m)

v= 3840 m/s= 3.84 x 10³ m

Answer 2

Another way to work this in case you are not familiar with the equation v = √(GM/r) :

 

If you know that the period T of a GPS orbit is about 11 hours and 58 minutes, you can estimate the orbital speed from

 

v = 2πr/T

 

r = R_e + h = (6.37 × 10⁶ m) + (2.06 x 10⁷ m)] = 2.697 x 10⁷ m

 

T = [11(3600) + 58(60)] sec = 43080 sec

 

v = 2π(2.697 x 10⁷ m)/(43080 sec) ≅ 3930 m/s