A space shuttle is orbiting around Earth, at distance r= 6.77 x 10 6 m away from

Question

A space shuttle is orbiting around Earth, at distance r= 6.77 x 106 m away from the center of Earth.

The space shuttle is orbiting once a day above the equator of Earth (Synchronous Satellite), answer the questions below.

a. Find the speed of the shuttle.

b. Compute the centripetal acceleration of the shuttle, Ac

c. Estimate the effect of the centripetal from (b) by calculate the percentage of Ac compared to the gravitational acceleration, Ag.

d. After considering the centripetal acceleration from (b), what is the "free fall" acceleration in the shuttle?

e. Estimate the measure of the astronaut's weight in the shuttle assuming her mass is 60 kg.

Explanation / Answer

a)

Gravitational force = Centripetal force
GMm/r2= mv2/r
G is the universal gravitational constant, G = 6.674 x 10-11 m3 kg-1 s-2
M is the mass of earth, M = 5.972 x 1024 kg
v = sqrt[GM/r]
= sqrt[(6.674 x 10-11) x (5.972 x 1024) / (6.77 x 106)]
= 7.673 x 103 m/s

b)
Centripetal acceleration, Ac = v2/r
= (7.673 x 103)2 / (6.77 x 106)
= 8.696 m/s2.

c)
Gravitational acceleration, Ag = GM/r2.
= [(6.674 x 10-11) x (5.972 x 1024) / (6.77 x 106)2]
= 8.696 m/s2.
This is 100% as that of Ac.

d)
Since there is no net acceleration, free fall acceleration = 0

e)
Weight = 0 since there is no acceleration

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