Question

a) What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 141 km above Earth's surface? (b) What is the period of revolution?

Answer 1

IN an orbit

Fc = Fg

m*v^2/r = G*M*m/r^

v = sqrt(GM/r)

M = Mass of earth = 5.98*10^24

G = gravitational constant = 6.67*10^-11 Nm^2/kg^2

r = orbital radis = (R+h)

R = radius of earth = 6.4*10^6 m

h = altitude = 141000 m

V = sqrt(6.67*10^-11*5.98*10^24/(6.4*10^6+141000))

v = 7809 m/s   <<<<======ANSWER

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time period T = 2*pi*r/V

T = 2*pi*r^(3/2)/sqrt(GM)

T = 2*pi*(6.4*10^6+141000)^(3/2)/sqrt(6.67*10^-11*5.98*10^24)

T = 5263 s <<<<<=======ANSWER