Register to reply 
Centripetal acceleration of an orbit to the earth... help! 
Share this thread: 
#1
Nov1411, 03:20 PM

P: 34

1. The problem statement, all variables and given/known data
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? 2. Relevant equations I believe I should use a_{c} = (4∏^{2}r) / T^{2} Converted: T= 24 hr = 86400s r = 35800 km = 35 800 000 m = 3.58 x 10^{7} m 3. The attempt at a solution I plugged in the values: a_{c} = (4∏^{2}35 800 000) / 86400^{2} a_{c} = 0,189 The answer SHOULD be 0.223 m/s^{2} Help! 


#2
Nov1411, 03:31 PM

Mentor
P: 15,066

I have highlighted your error:



#3
Nov1411, 03:54 PM

P: 34




#4
Nov1411, 04:00 PM

Mentor
P: 15,066

Centripetal acceleration of an orbit to the earth... help!
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
Nov1411, 04:13 PM

P: 34

THANKS 


#6
Nov1411, 04:20 PM

P: 34

Nvm, I got it, thanks again !



#7
Nov1411, 04:43 PM

Mentor
P: 15,066

Very good, and you're welcome.



#8
Nov1911, 10:51 AM

P: 1

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



Register to reply 
Related Discussions  
What causes the centripetal acceleration of the earth?  Classical Physics  4  
Centripetal acceleration on Earth  Introductory Physics Homework  26  
Centripetal acceleration of Earth around Sun  Introductory Physics Homework  3  
Centripetal Acceleration of Earth  Introductory Physics Homework  7  
Centripetal acceleration of Earth around Sun?  Introductory Physics Homework  6 