- #1

magnolia1

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## 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?

## Homework Equations

a= v

^{2}/ r

T = (2 Pi r)/ V

## The Attempt at a Solution

Converting km to m: 790km*(10

^{3}m / 1km) = 7.90*10

^{5}m

Converting min to seconds: 100.5min*(60seconds/1min) = 6.03*10

^{3}seconds

Now that I have the conversions, I will solve for the unknown variables in the above formulas:

I think the radius is 7.90*10

^{5}m and period is 6.03*10

^{3}seconds

(a) Speed of satellite:

v=(2 pi (7.90*10

^{5}m)) / (6.03*10

^{3}s)

v=8.27*10

^{2}m/s

Correct answer: 7460 m/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 :)