Question

A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.97 x 10^4 s. What is the speed at which the satellite travels?

Answer 1

Answer:

v = 5030.4 m/s

Explanation:

let Fc be the centripetal force and Fg be the gravitational force of attraction on the satelite. let m be the mass of the satelite and M be the mass of earth.

At any point in the orbit, the satelite experiences forces , Fc and Fg such that:

Fc = Fg

m×v^2/r = G×M×m/(r^2)

m and r are the same throughout the equation,so that:

v^2 = G×M/(r)

  v =  \sqrt{G×M/(r)}

In a circular orbit, the displacement the satelite should cover is 2×π×r in a period :

T = 2×π×r /v

so  that:

v = 2×π×r /T

then

2×π×r /T = \sqrt{G×M/(r)}

           r =  \sqrt[3]{T^2×G×M/(4×π^2)}

           r =   \sqrt[3]{(1.97×10^4)^2×(6.67408×10^-11)×(5.98×10^24)/(4×π^2)}

           r =  15772060.56 m

then :

v = 2×π×r /T

  = 2×π×(15772060.56) /(1.97×10^4)

  = 5030.4 m/s

therefore, the satelite travels at a speed of 5030.4 m/s.