How do I find the gravitational acceleration at the earth’s surface for a shuttle?
Suppose the space shuttle is in orbit 470 km from the Earth’s surface, and circles the Earth about once every 94.0 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.
centrip accel = v^2/r
we need to find the speed and the radius
the speed is the distance traveled/time
the distance is the radius of the orbit, or 2 pi r; the time is the period P
so v=2 pi r/P and v^2=4 pi^2 r^2/P^2
then a=v^2/r = 4 pi^2 r/P^2
r=radius of earth + 470 km
= 6400 km + 470 km = 6870 km
= 6.87×10^6m
P=94 mins = 94 minsx60 sec/min = 5640s
a=4 pi^2 (6.87×10^6)/(5640)^2
= 8.5 m/s/s => 8.5/9.8 = 0.87 g