# Centripetal acceleration on Earth

1. May 24, 2010

### daysrunaway

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!

2. May 24, 2010

Try using this formula- 2$$\pi$$/$$\omega$$

$$\omega$$= 360/24*3600

Let me know if you got the answer.

3. May 24, 2010

### daysrunaway

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.

4. May 24, 2010

Is the mass of the satellite given?

5. May 24, 2010

### daysrunaway

No, it isn't.

6. May 24, 2010

### jyothsna pb

try using the equation GMm/r^2=mv^2/r
m-mass of satellite
M-mass of earth
v^2/r is the centripetal acceleration

7. May 24, 2010

### jyothsna pb

gravitational frce of earth is utilised for centripetal force

8. May 24, 2010

You are right. But what about the velocity?

9. May 24, 2010

### 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

10. May 24, 2010

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

11. May 24, 2010

### jyothsna pb

12. May 24, 2010

### jyothsna pb

u r welcome

13. May 24, 2010

And reply to my visitor msg pls

14. May 24, 2010

### daysrunaway

I didn't, though, and that's the source of my confusion. I know I have the right value for v but when I plug it in to v^2/r, I get 3.37 x 10^-2 m/s^2. This answer is not equal to the answer I found for the actual acceleration, which is 0.006 m/s^2. My question is why is my answer different from the real value?

15. May 24, 2010

### jyothsna pb

what is the value of v?

16. May 24, 2010

### daysrunaway

465.1 m/s

17. May 24, 2010

### jyothsna pb

there is some error in the velocity value

18. May 24, 2010

### jyothsna pb

we get acceleration value almost equal to g

19. May 24, 2010

### jyothsna pb

velocity in dis orbit must b approximately equal to 7.9*10^3m/s

20. May 24, 2010

### daysrunaway

I don't understand why that is, especially since Wikipedia says it is 451 m/s
en.wikipedia.org/wiki/Earth