How Does Earth's Shape Affect Gravity at Different Locations?

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Homework Help Overview

The discussion revolves around the calculation of the percent difference in gravitational acceleration (g) at the poles and the equator, assuming the Earth is a sphere with a radius of 4000 miles. Participants are exploring the implications of this assumption on the problem's outcome.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants question whether the percent change in g should be considered zero under the spherical assumption, while others suggest that the problem may be missing important details. There is discussion about the implications of treating the Earth as an oblate spheroid and the potential effects of centripetal acceleration.

Discussion Status

The conversation is ongoing, with participants raising valid points about the assumptions in the problem statement. Some have suggested that additional factors, such as centripetal acceleration, might need to be considered, although the original wording of the problem seems to limit this exploration.

Contextual Notes

Participants note that previous questions have required more complex considerations, leading to uncertainty about whether additional factors should be included in this problem. The specific wording of the problem may influence how participants approach the calculation.

atomicpedals
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Homework Statement



Taking the Earth to be a sphere of 4000mi radius, compute the percent difference in g (the acceleration due to gravity) at the poles and the equator.


The Attempt at a Solution



If we assume the Earth is a sphere with a given radius, then shouldn't the percent change in g be zero?
 
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yes indeed - something missing in the problem formulation?
 
The problem as it appears in the text is "Taking the Earth to be a sphere of 4000 miles radius, compute the percent difference in g, the acceleration due to gravity, at the poles and equator."

So as stated I it's zero. However, as every-other question has required much much more I thought I was missing something!

Were the question to consider the Earth as it is (an oblatespheroid) there would be a small difference.
 
I'll just give the good old g=-(GM/r^2).
 
atomicpedals said:
The problem as it appears in the text is "Taking the Earth to be a sphere of 4000 miles radius, compute the percent difference in g, the acceleration due to gravity, at the poles and equator."

So as stated I it's zero. However, as every-other question has required much much more I thought I was missing something!

Were the question to consider the Earth as it is (an oblatespheroid) there would be a small difference.

Maybe factor in the difference in centripetal acceleration? The wording of the problem would seem to exclude that, but I don't see any other difference offhand. If they said "the apparent acceleration due to gravity", it would be easier to include that other term...
 
That's a reasonable point to add in (w [tex]\times[/tex] r)
 

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