Length of one day on another planet

AI Thread Summary
To determine the length of one day on a planet with a radius of 6.2 x 10^6 m and an acceleration of 9.70 m/s^2 at the equator, the centripetal force equation Fc=4π²mr/T² is used, with gravitational force considered as the centripetal force. The calculation led to a result of approximately 1.3 hours for one revolution, which contradicts the textbook answer of 7.9 hours. The confusion arises from the need to account for the difference between the planet's acceleration and the gravitational acceleration experienced by an object at the equator. Clarification is sought on the importance of this adjustment and how it relates to centripetal acceleration. Understanding this concept is crucial for accurately determining the planet's rotational period.
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
 
Physics news on Phys.org
Fgravity does more than keep a person in a cicular orbit. It also keeps his feet on the ground ... with a given force mg.
 
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
 
So ?
 
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 ?
 
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?
 
If the planet wouldn't rotate, what would the scale indicate ?
 
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
 
Is that understandable ?
 
  • #10
BvU said:
Is that understandable ?
I don't understand why you're supposed to do that, but that's what my teacher said
 

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