# Acceleration due to gravity

## 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

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).

berkeman
Mentor
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...

That's a reasonable point to add in (w $$\times$$ r)