Centripetal acceleration of an orbit to the earth help!

  • Thread starter MissJewels
  • Start date
  • #1
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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!
 

Answers and Replies

  • #2
D H
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I have highlighted your error:

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?
 
  • #3
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I have highlighted your error:



Do you see the problem?
ahaha... isnt altitude the radius?
 
  • #4
D H
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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.
 
  • #5
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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
 
  • #6
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Nvm, I got it, thanks again !
 
  • #7
D H
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Very good, and you're welcome.
 
  • #8
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how do you calculate the distance from the center of the Earth for a geosynchronous satellite orbit?
 

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