Centripetal Acceleration and Satellite Question

[chihockey7]

1. The astronaut orbiting the Earth in Figure P4.30 is preparing to dock with a Westar VI satellite. The satellite is in a circular orbit 600 km above the Earth's surface, where the free-fall acceleration is 8.21 m/s2. The radius of the Earth is 6400 km.
a) Determine the speed of the satellite. m/s
b) Determine the time required to complete one orbit around the Earth. min


2. Homework Equations [/b]
centripetal acceleration= v squared/ radius
distance*speed= time


3. The Attempt at a Solution [/b]
To solve for the speed I used 8.21 as the centripetal acceleration and made that equal to v squared/7000 and then solved for v.

8.21=v squared/7000
v= 239 m/s

I then took 14000*3.14 for the circumference of the orbit and multiplied that by 239 m/s for the total time to orbit and divided that by 60 to put the answer in minutes. I got 2919.94 minutes.

However, both answers are wrong. Can I not use the free fall acceleration for centripetal acceleration?

 

[cepheid]

Welcome to PF,

I then took 14000*3.14 for the circumference of the orbit and multiplied that by 239 m/s for the total time to orbit

Are you trying to claim that distance*speed = time? You might want to rethink that one...

 

[cepheid]

However, both answers are wrong. Can I not use the free fall acceleration for centripetal acceleration?

Yes, you can, and you should, because gravity is what is providing the centripetal force that keeps the satellite moving in a circle. You might have a computation error. Can you post your calculations for part a?

 

[chihockey7]

for part a I used the equation centripetal acceleration= v^2/ radius

8.21= v^2/(6400+600)
57470=v^2
v= 239.73

what is wrong with this calculation?

Concerning finding the time. I completely forgot that t=distance/speed.

 

[chihockey7]

I just realized my problem. I did not do the simple unit conversions from km to m. Everything worked out fine. Thanks for all of your help!