Length of one day on another planet

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Homework Help Overview

The problem involves determining the length of one day on a hypothetical planet with specific gravitational and rotational characteristics. The planet has a radius of 6.2 x 10^6 m and an acceleration due to gravity of 9.70 m/s² at the equator, while the expected gravitational acceleration is 10 m/s².

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to use the formula for centripetal force and equate it to gravitational force, questioning the relevance of the given acceleration values. Some participants discuss the implications of gravitational force on objects at rest and in circular motion.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of gravitational force and its role in centripetal acceleration. There is no explicit consensus on the correct approach, and some participants express confusion regarding the necessity of certain calculations.

Contextual Notes

Participants note that the problem involves assumptions about the planet's rotation and the effects of gravity on objects at rest versus in motion. There is mention of a discrepancy between the original poster's calculations and a textbook answer, indicating potential misunderstandings or misapplications of the concepts involved.

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
 
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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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