Gravity and Rotation?

Question

Suppose the space shuttle is in orbit 430 km from the Earth's surface, and circles the Earth about once every 93.2 minutes. Find the centripetal acceleration of the space shuttle in its orbit. Express your answer in terms of g, the gravitational acceleration at the Earth's surface.

Answer 1

Centripetal force on the space shuttle is
F= ms* v^2 /(h+Re)

centripetal acceleration = v^2 /(h+Re) -----(1)

orbital speed v = (h+Re) w = (h+Re) 2pi/T

w=2 pi/T and T = period of revolution in orbit of
(1) becomes

centripetal acceleration = 4* pi^2* (h+Re) / T^2

orbit radius
(h+Re) = R + 430 km = 6400+430 = 6830 km = 6830000 m
T = 93.2*60 secs

centripetal acceleration =
= 4* (3.14)^2* (683*10^4) / (93.2*60)^2
fc = 8.614 m/s^2

g = 9.81 m/s^2

fc = g * 0.878
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Note the fc can be made to directly involve "g"

ms * fc = ms * G Me/(R+h)^2 = [G Me/Re^2] [ 1/(1+h/Re)^2]

fc = g [1/(1+h/Re)^2] = 4* pi^2* (h+Re) / T^2

this can give fc without using T

fc = g [1/(1+h/Re)^2] = g [1/(1+0.671875)^2 ]
= g [1/(1.671875)^2] = g [0.937]^2 = g * 0.878