# Simple rotational period problem

1. Jun 11, 2014

### Bob_Dole

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

Determine the length of day on a planet.

Imagine a planet with an acceleration at the equator of $a_{equator} = 10 \,\, m/s^2$ (ignoring rotation).

Object dropped at equator falls with an acceleration of:

$a_{obj} = 9.7 \,\, m/s^2$,

r = $6.2 \times 10^6 \,\, m$.

2. Relevant equations

I treated $a_{obj}$ as the centripital acceleration, and $a_{equator}$ as the total acceleration. I then solved for the tangential component $a_T$ using Pythagorean Theorem.

I then used the following formula,

$T = \sqrt{\frac{4 \pi^2 r}{a_T}}$.

3. The attempt at a solution

I am not getting the same result as the text, and am unsure as to where my thought process is going awry.

I get a value of 2.8 hr, whereas the value in the book is 7.9 hr.

Last edited: Jun 11, 2014
2. Jun 11, 2014

### dauto

Can you post the full text of the question? Seems like you gave us an abbreviated version.

3. Jun 11, 2014

### Bandersnatch

Hi Bob, welcome to PF!

Consider, if there were no gravity at all, what acceleration would the object need to have in order to move in circles?

4. Jun 11, 2014

### Bob_Dole

Centripetal acceleration of course, which would have some velocity perpendicular to this acceleration.

5. Jun 11, 2014

### Bandersnatch

Right, so, if the object, as seen from the surface, falls under the influence of gravity, does it mean the gravity supplies more, less or exactly the required amount of acceleration?

6. Jun 11, 2014

### Bob_Dole

The fact that the object "falls" toward the surface, would indicate that gravity is supplying more acceleration (greater than the fictitious centrifugal force).

Thus, I need to find out by how much. This would be $10 \,\, m/s^2 - 9.3 \,\, m/s^2 = 0.3 \,\, m/s^2$.

I am now embarrassed. Thanks for your help, and thank you for getting me to think!

7. Jun 11, 2014

### Bandersnatch

Just one more bit to iron out:
This is completely unnecessary, and indeed, meaningless. There is no tangential component to centripetal acceleration! It's always radial. It is velocity that is tangential.

So, just plug the centripetal acceleration you've found into the period equation you provided earlier, and you should be done.

8. Jun 11, 2014

### Bob_Dole

Yep, surely I was not using my noggin! Thanks again for your help. I got the correct answer just by subtracting the two.