NASA launches a satellite into orbit at a height above the surface of the Earth

Question

NASA launches a satellite into orbit at a height above the surface of the Earth equal to the Earth's mean radius. The mass of the satellite is 660 kg. (Assume the Earth's mass is 5.97 x10^24 kg and its radius is 6.38 x 10^6 m.)

(a) How long does it take the satellite to go around the Earth once? Consider Kepler's Third Law. ___ h

(b) What is the orbital speed of the satellite? What distance does the satellite cover in one revolution? ___ m/s

(c) How much gravitational force does the satellite experience? Consider Newton's Universal Law of Gravitation. ____N

Explanation / Answer

part a )

T^2 = 4pi^2*(2r)^3/G(Me+Ms)

r = 6.38 x 10^6 m

Me = mass of earth = 5.97 x 10^24 kg

T = 14351.78 s = 3.987 hrs

part b )

G*Me.ms/r^2 = ms*v^2/r

v = sqrt(GMe/2r)

v = 5586 m/s

revolution

T = 2pi*(2Re)/v

T = 3.98 hr

part c )

F = GMems/(2r)^2

F = 1614.15 N

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