# Centripetal acceleration of the satellite?

## Homework Statement

An Earth satellite moves in a circular orbit 790 km above the Earth's surface. The period of the motion is 100.5 min.
(a) What is the speed of the satellite?
(b) What is the magnitude of the centripetal acceleration of the satellite?

a= v2 / r

T = (2 Pi r)/ V

## The Attempt at a Solution

Converting km to m: 790km*(103m / 1km) = 7.90*105m
Converting min to seconds: 100.5min*(60seconds/1min) = 6.03*103seconds

Now that I have the conversions, I will solve for the unknown variables in the above formulas:
I think the radius is 7.90*105m and period is 6.03*103seconds

(a) Speed of satellite:
v=(2 pi (7.90*105m)) / (6.03*103s)
v=8.27*102m/s

because the speed is incorrect, the error follows through to part (b) when substituting in speed.
How is it that it's so far off?
Thanks :)

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ehild
Homework Helper
r should be the radius of the orbit, centred at the centre of Earth. The height of the satellite above the surface of Earth is 790 km, how far is it from the centre of Earth?

ehild

• 1 person
I had to look it up, but 6371 km is from the surface of Earth to its centre. So I will try now with the new radius of (6371km+790km) * (10^3m/1km) = 7.16*10^6metres.
ANS: 7460.63 = 7461m/s

ehild
Homework Helper
I had to look it up, but 6371 km is from the surface of Earth to its centre. So I will try now with the new radius of (6371km+790km) * (10^3m/1km) = 7.16*10^6metres.
ANS: 7460.63 = 7461m/s

You are welcome. 