Acceleration due to gravity

  • #1
208
7

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?
 

Answers and Replies

  • #2
52
0
yes indeed - something missing in the problem formulation?
 
  • #3
208
7
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.
 
  • #4
208
7
I'll just give the good old g=-(GM/r^2).
 
  • #5
berkeman
Mentor
57,708
7,740
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 centripital 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...
 
  • #6
208
7
That's a reasonable point to add in (w [tex]\times[/tex] r)
 

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