# Gravity on flat object [duplicate]

I was wondering how gravity would behave on object of different shapes.

1. If the Earth was squeezed into a thin disk what would the gravitional acceleration be at the center of the flat surface? Would it be really low because the amount of matter beneath me would be small? If I stood on the edge would the gravitational acceleration be enormous because the amount of matter beneath me was huge?

2. If an object is lowered into the Mariana trench will the effect of gravity increase because it gets closer to the center of the Earth?

## marked as duplicate by ja72, Kyle Kanos, Brandon Enright, Qmechanic♦Jul 16 '14 at 18:08

• I'm assuming that by squeezing the Earth you mean deforming the Earth to the shape of a thin circular flat disc. Is that correct? – mpv Jul 16 '14 at 15:48
• Yes. Making it like a pancake with same density as now – stefan Jul 16 '14 at 16:44
• Your Q2 is unrelated to Q1 and should really be posted as a separate question. Actually I think it's an interesting one and the answer is not at all clear to me. – John Rennie Jul 16 '14 at 17:16
• @JohnRennie: Your link in your first comment is to this question. – Kyle Kanos Jul 16 '14 at 17:52
• The duplicate, and rmhleo's answer only give the field at the centre of the flat surface. The strength of the field when standing in the middle of the rim is still an open question. – John Rennie Jul 16 '14 at 18:44

In the Newtonian framework, you just need to solve the integral $$\int{ \frac{ G \rho_{(r)}}{|\vec{r} - \vec{r}_o|^2} \frac{\vec{r} - \vec{r}_o}{|\vec{r} - \vec{r}_o|}}dV$$ for the volume in question, where $\vec{r}$ is the distance from an arbitrary reference point to the element of matter where density is $\rho_{(r)}$, and $\vec{r}_{o}$ the distance from the same reference point to the position where you are finding the gravitational force value. And the answers would be:

1) Assuming you mean squeezing the sphere-like Earth into a disc-like shape, you would get: $$\int{ \frac{ G\sigma_{(r)}}{|\vec{r} - \vec{r}_o|^2}\frac{\vec{r} - \vec{r}_o}{|\vec{r} - \vec{r}_o|}}dS$$ where if you are near the center of this disc, you could assume is an infinite surface and disregard border effects. Also assuming constant $\sigma_{(r)}=\sigma_o$, choosing our reference point in the surface, and the point of interest at a height $h$ over the plane, the symmetry conditions with respect to $\theta$ yields: $$\int_{h}^{\infty}{ \frac{ \pi\sigma_oGh }{r^2} }dr = \pi\sigma_oG$$ A constant value! This is typical from fields depending on the inverse squared distance. Furthermore since $\sigma_o = \frac{M_T}{\pi R_T^2}$, where $M_T$ and $R_T^2$ are respectively the Earth's mass and radius (of the disc which I intentionally chose the same as sphere radius), we get

$$\pi\sigma_oG = G \frac{M_T}{R_T^2}$$

The force in the surface of Earth-disc (and unlike in the Earth-sphere, in every point over it) is the same as the value in the surface of the of the Earth-sphere!

2) I will leave to you the calculations, using the same first formula, but you will see that indeed in the bottom of the Marianna Trench you should feel a smaller attraction force. In fact if you open a hole through the Earth the intensity of the force would decrease linearly with the the distance to to the center. The reason for this is that inside mass shells the attraction from its different parts compensate among them, as you can see for yourself if you solve the integral to find the force in a point inside a mass shell. But from symmetry only, you see that in the center of a spherical mass shell the force should be zero, right?

Your question 1 is non-sensical. If the earth were sqeezed (I assume you really mean "compressed") to be much smaller, then where does your flat surface come from? One can imagine a small sphere, but there would be nothing "flat".

As for question 2, there are two competing factors at work. First, the effect of gravity could be lowered because some of the mass of the earth is now pulling away from the center of mass. This effect actually makes you weigh a little less at the bottom of the Mariana trench than you would in a ship above it on the surface.

Second, the density of the earth is non-uniform. If most of the mass of the earth were concentrated in a small sphere at its center, then you would weigh more in the trench because the additional gravitational force due to being closer to this mass is more than the gravitational force of the part of the earth that is now above you.

The center of the earth is believed to be iron, which certainly is more dense than seawater. My original gut feel was that you'd still weigh less at the bottom of the trench, but apparently the higher concentration of mass in the center of the earth is the dominating effect according to a comment by AlanSE.