## Centripetal acceleration on Earth

1. The problem statement, all variables and given/known data
An object orbits the earth at a constant speed in a circle of radius 6.38 x 106 m, very close to but not touching the earth's surface. What is its centripetal acceleration?

2. Relevant equations
a = v2/r = 4$$\pi$$2v/T2
v = 2$$\pi$$r/T

3. The attempt at a solution
I plugged in r = 6.38 x 106 m and T = 24.0 h = (24.0 h x 3600 s / 1 h) = 86,400 s into the equation above and found a = 3.37 x 10-2 m/s2. However, I looked up the centripetal acceleration on the earth's surface and found out it is 0.006 m/s2. I can't understand why my answer is wrong. Can anyone point out the error in my logic?

Thanks!
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 Try using this formula- 2$$\pi$$/$$\omega$$ $$\omega$$= 360/24*3600 Let me know if you got the answer.
 Thanks for the suggestion RoughRoad but I don't really understand how I am supposed to use this equation. I found that my calculation for the angular velocity (2piR/T) is approximately equal to the wikipedia value for angular velocity, so I really am confused now because I can't see what I'm doing wrong in just squaring that value and dividing by 6.38 x 106.

## Centripetal acceleration on Earth

Is the mass of the satellite given?
 No, it isn't.
 try using the equation GMm/r^2=mv^2/r m-mass of satellite M-mass of earth r-radius of orbit v^2/r is the centripetal acceleration
 gravitational frce of earth is utilised for centripetal force

 Quote by jyothsna pb try using the equation GMm/r^2=mv^2/r m-mass of satellite M-mass of earth r-radius of orbit v^2/r is the centripetal acceleration
You are right. But what about the velocity?
 you are asked to find the centripetal acceleration you just have to find the value of v^2/r which gives the centripetal acceleration

 Quote by jyothsna pb you are asked to find the centripetal acceleration you just have to find the value of v^2/r which gives the centripetal acceleration

Oh yeah! How can I be so foolish. Thanks for helping!