Difficult Planetary motion problem

[Hockeystar]

Homework Statement

Planet X rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs 950.0 N on the Earth weighs 917.0 N at the north pole of Planet X and only 860.0 N at its equator. The distance from the north pole to the equator is 1.883×10⁴ km , measured along the surface of Planet X. How long is the day on Planet X?

Homework Equations

A lot

The Attempt at a Solution

First we solve the radius of Planet:

circumfrence = 0.5pi(r)
r=11987550m

Next we solve m: 950N/9.8 m/s² = 96.9kg

Then the tricky tricky part. Should I assume the loss of weight is equal to the centripetal force? In that case I have

mg_{north pole} - mg_equator = m4pi²r²/T²
T= 9.82e⁷s

However my answer is incorrect. Is my theory sound? Did I make a calculation error?

 

[Delphi51]

It all looks good except I think you have an extra r in m4pi²r²/T², which makes a large difference in the answer.

 

[D H]

mg_{north pole} - mg_equator = m4pi²r²/T²

Look at your units. The left hand side has units of mass*acceleration or mass*length/time². The units on the right hand side are mass*length²/time². Once you get into the habit of checking units, it takes but a few seconds to double check that you have consistent units. When you don't have consistent units, as is the case here, you *know* you have made an error somewhere.

 

[Hockeystar]

Thanks for the help guys :-)