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.
