# How does gravity behave outside the geoid?

I was watching a program about the geoid recently and the relative "height" of the Himalayan plateau gave me an idea. It makes sense for the geoid to be "highest" (is that the correct term?) there as there is more mass underneath you than in other places, however I remembered from physics that your distance from the center of the earth effects gravity as well.

So if you were on the peak of Mount Everest what would have a stronger impact on your weight as compared to say sea level at a comparable latitude (to control for the spin of the earth), there being more mass underneath you, or you being further from the center of the earth? In other words, would you feel slightly lighter or slightly heavier?

I tried working it out myself, but I haven't done any calculus in over fours years, so I couldn't figure it out.

Well, let us estimate. The Everest mountain under you is of order $h\sim 10 \textrm{km}$ thick. And lets say it has the same density as the Earth on average.

Everest creates the gravity force, which is guaranteed to be smaller than that created by a spherical shell of thickness $h$ and radius $R_{Earth}$ (because the additional material under you would only add up to gravity).

Now, the gravity force(acceleration) from the Earth is $g=\dfrac{G M_{Earth}}{R^2}=\dfrac{4\pi G\rho R^3}{3R^2} = \dfrac{4}{3}\pi G\rho R$. The gravity from the shell is $\delta g = \dfrac{G m_{shell}}{R^2}=\dfrac{4\pi G \rho R^2 h}{R^2}=4\pi G \rho h$.

The relative importance of the forces is then $\dfrac{\delta g}{g}=\dfrac{3h}{R}\sim\dfrac{30}{6000}=0.005$.

You may also check that the mountain will produce approximately two times smaller gravity force than a spherical shell we considered, which yields two times smaller answer.

Generally, gravity is something which is caused by all pieces of objects you consider, rather than only by the nearest ones.

My calculations show these two effects are both of order $\delta R/R$ and hence approximately equal. If Mount Everest is a pole, you would be lighter at the top. If it were a complete shell, you would be heavier, as mass increases as the cube and the force decreases as the inverse square, so the increase wins. You can calculate exactly if you assume that Mt Everest is a sphere, like a pea on a basketball. in this case, the decrease will be very nearly twice the increase so you would be slightly lighter. But if Mt Everests effective gravity is nearly twice as strong as the sphere, the two effects will cancel and you will weigh the same as if there were no mountain. So the result depends critically on the size, shape and density of Mt Everest.

If R is the radius of the earth and if r is the radius of the sphere substituting for Mt Everest, then the base case is $R^3/R^2$ , the base plus decrease is $R^3/(R+2r)^2$, and the increase is $r^3/r^2$. All times the same constant, which is $4 \pi G \rho /3$ as in the previous answer.

Put everything on a common denominator and expand to see the conclusion.