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 :-)

 

[rwisz]

Possible response:

Hello, thank you for presenting your solution attempt for this difficult planetary motion problem. Your approach of solving for the radius of Planet X and the mass of the astronaut is correct. However, your calculation for the day length may not be accurate due to a few potential factors.

Firstly, the loss of weight at the equator may not necessarily be equal to the centripetal force. It could also be affected by the planet's rotation rate and its gravitational field. Therefore, it would be best to approach this problem using the concept of rotational dynamics and conservation of angular momentum.

Secondly, the distance from the north pole to the equator is given in kilometers, but your calculations use meters. This could lead to a significant difference in the final answer.

Lastly, it is important to consider any other variables that may affect the day length on Planet X, such as the planet's mass and density. It would be helpful to double-check your calculations and also consider these additional factors in order to arrive at a more accurate answer. Keep up the good work!