- #1
jactre
- 4
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An Earth satellite moves in a circular orbit 589 km above Earth's surface with a period of 96.26 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?
I know I need the equations for 1) period and 2) centripetal acceleration and 3) r.
1) T= (2pi x r)/v (seconds)
2) a= v^2/r (m/s^2)
3) 589 km
I converted T=92.26 min into seconds and got 5775.6s and converted 589km to meters. I then plugged these numbers into equation 1 to solve for velocity (v= (2pi x r)/T). I used this velocity as well as 589,000m for r to find acceleration using equation 2.
The answers I calculated for both velocity and acceleration are wrong, and I think my problem lies in my value for r.
Did I interpret this incorrectly? Any input on what r should be (or if I made another mistake) is greatly appreciated! Thank you!
I know I need the equations for 1) period and 2) centripetal acceleration and 3) r.
1) T= (2pi x r)/v (seconds)
2) a= v^2/r (m/s^2)
3) 589 km
I converted T=92.26 min into seconds and got 5775.6s and converted 589km to meters. I then plugged these numbers into equation 1 to solve for velocity (v= (2pi x r)/T). I used this velocity as well as 589,000m for r to find acceleration using equation 2.
The answers I calculated for both velocity and acceleration are wrong, and I think my problem lies in my value for r.
Did I interpret this incorrectly? Any input on what r should be (or if I made another mistake) is greatly appreciated! Thank you!
