# Centripetal acceleration of an orbit to the earth help!

## Homework Statement

A geostationary satellite goes around the Earth in 24hr. Thus, it appears motionless in the sky and is a valuable component for telecommunications, including digital television. If such a satellite is in orbit around the Earth at an altitude of 35 800 km above the Earth's surface, what is the module of its centripetal acceleration?

## Homework Equations

I believe I should use
ac = (4∏2r) / T2

Converted:
T= 24 hr = 86400s
r = 35800 km = 35 800 000 m = 3.58 x 107 m

## The Attempt at a Solution

I plugged in the values:
ac = (4∏235 800 000) / 864002
ac = 0,189

The answer SHOULD be 0.223 m/s2

Help!

D H
Staff Emeritus

If such a satellite is in orbit around the Earth at an altitude of 35 800 km above the Earth's surface ...

r = 35800 km

Do you see the problem?

Do you see the problem?

D H
Staff Emeritus
No. Altitude is the distance between the satellite and the closest point on the surface of the Earth. Radius is the distance to the center of the Earth.

No. Altitude is the distance between the satellite and the closest point on the surface of the Earth. Radius is the distance to the center of the Earth.

OOOOH so i add the radius with the altitude, and THATS the r value i use! Right?
THANKS

Nvm, I got it, thanks again !

D H
Staff Emeritus