Centripetal acceleration of a geostationary satellite

[MissJewels]

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
a_c = (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:
a_c = (4∏²35 800 000) / 86400²
a_c = 0,189

The answer SHOULD be 0.223 m/s²

Help!

 

[D H]

I have highlighted your error:

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

r[/size] = 35800 km

Do you see the problem?

 

[MissJewels]

I have highlighted your error:



Do you see the problem?

ahaha... isn't altitude the radius?

 

[D H]

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.

 

[MissJewels]

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

 

[MissJewels]

Nvm, I got it, thanks again !

 

[D H]

Very good, and you're welcome.

 

[mnicorici]

how do you calculate the distance from the center of the Earth for a geosynchronous satellite orbit?