Question

A space shuttle is orbiting around Earth, at distance r = 6.77 x 10⁶ m away from the center of Earth. At this distance, gravitational acceleration is equal to g = 8.69 m/s². There is an astronaut aboard the space shuttle, she is 1.70 m tall. The difference in gravitational acceleration betweeen her feet and her head is Δg = -4.36 x 10^-6 m/s₂

Considering the above information, answer parts a, b, c, d, and e below. Include your explanation and show your calculation in detail.

a) Assuming the space shuttle is orbiting once a day above the equator of Earth, estimate the speed of the shuttle.

b) For this satellite (NOT Synchronous Satellite), compute the centripetal acceleration of the shuttle, a_c, using the speed obtained in a).

c) Estimate the effect of the centripetal acceleration from b) by calculate the percentage of a_c compared to the gravitational acceleration, a_g.

d) After considering the centripetal acceleration from b), what is the ‘free fall’ acceleration in the shuttle?

e) If an object is dropped from her head, how long does it take until it hits the floor?

Answer 1

a. assuming space craft orbits earth once per day

r = 6.77*10^6 m

speed = v

v = 2*pi*r/t = 2*pi*6.77*1066/24*60*60 = 492.3282 m/s

b. centripital acceleration = v^2/r = 0.03580312381 m/s/s

c. ac/ag = 0.03580312*100/8.69 = 0.412 %

d. hence free fall acceleration in shuttle = 8.69 - 0.035 = 8.655 m/s/s

e. time taken to reach floor = t

distance covered = s

we can consider constant acceleration as dg is veruy small in that distnace

hence

1.7 = 0.5*(8.655)*t^2

t = 0.626766711 s