Posted by Sierra on .
Suppose the space shuttle is in orbit 390 km from the Earth's surface, and circles the Earth about once every 92.4 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.
*note my teacher has not taught me anything to do with w = 2pi/T as some help websites have used this route to find the answer, obviously there's another way that my teacher has taught us how to solve this kind of problem and i already know the answer is 0.89g i just don't know how to get there. Thanks.

Physics Please Help ASAP 
drwls,
Use the fact that the radius of the Earth is R = 6378 km.
At an altitude of 390 km, the shuttle is R = 6378+390 = 6768 km from the center of the earth. Centripetal a equals gravity g', which is proportional to 1/R^2. Therefore at 6768 km,
g' = g*(6368/6758)^2 = 0.8879 g 
Physics Please Help ASAP 
Sierra,
thank you so much! super helpful i didn't realize i was overlooking the radius of earth!

Physics Please Help ASAP 
drwls,
You're welcome. There are other ways to come up with the same answer using the universal G and the period of the satellite orbit, but this method is quicker.