Length of one day on another planet

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Balsam
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Homework Statement


Imagine another planet with an acceleration of 10m/s^2 at its equator when ignoring the rotation of the planet. The radius of the planet is 6.2 x 10^6m. An object dropped at the equator yields an acceleration of 9.70m/s^2. Determine the length of one day on this planet.
r=6.2x10^6m
g=9.70m/s^2
a=10m/s^2--> I don't know why we were given the first acceleration, I don't think it's equal to ac.

Homework Equations


I used Fc=4pi^2mr/T^2 and isolated for T.

The Attempt at a Solution


I used the above equation, plugging in mg for Fc because I think the force of gravity is the centripetal force. The 'm' variables canceled out since I just divided them out. Then, I plugged all of my other given values in and solved for T. Since T is the time it takes in seconds to complete one revolution, I divided my result by 3600 to get the time it takes in hours to complete one revolution. My final answer was about 1.3hours. However, the book's answer is 7.9hours. I don't know where I went wrong
 
on Phys.org
BvU said:
Fgravity does more than keep a person in a cicular orbit. It also keeps his feet on the ground ... with a given force mg.
But the formula I used just asked for Fc
 
BvU said:
If a person with a mass of 80 kg stands on a scale, the scale will show 80 x 9.7 kg, so the scale will push up with 776 N. Where does that force come from ?
It's the force of gravity, but how is that important in centripetal acceleration?
 
BvU said:
If the planet wouldn't rotate, what would the scale indicate ?
Apparently you have to subtract the acceleration of the dropped object from the acceleration of the planet
 
BvU said:
Is that understandable ?
I don't understand why you're supposed to do that, but that's what my teacher said