# Radial acceleration around a planet

1. Oct 17, 2008

### anteaters

1. The problem statement, all variables and given/known data

Consider a planet of radius 7.31 x 10^6 m with a rotation period of 25.4 hours. Compute the radial acceleration of a point on the surface of the planet at the equator owing to its rotation about its axis.

2. Relevant equations

T = [2(pi)r]/v
a_r = -[(v^2)/r]

3. The attempt at a solution

so i know one period (T) is 25.4 hours, and r = 7.31 x 10^6 m. so i use that and find velocity, and tried to plug that into the radial acceleration equation, but i got the wrong answer. what am i doing wrong? is it something to do with units of time?

2. Oct 17, 2008

### anteaters

oh, and my answer was 2.88 x 10^8 m/s2

3. Oct 17, 2008

### D H

Staff Emeritus
Did you convert the hours to seconds? If not, you are spot-on regarding time.

That means you did something else wrong in addition to the time units error. Please show your work.

4. Oct 17, 2008

### anteaters

25.4 hours = 91440 seconds = 2(pi)(7.31 x 10^6)/v
so v = 502.3 m/s
a_r = [(502.3 m/s)^2]/(7.31 x 10^6) = 0.0345 m/s^2

would that be the right answer? it seems kind of small.

5. Oct 17, 2008

### D H

Staff Emeritus
Looks fine to me.

6. Oct 17, 2008

### anteaters

thanks a lot D H.