$g=GM/r^2$; $M=$mass of Earth. So in center $r=0$, so $g$ should be infinity

Question

From the equation of Newtonian gravitational force, we know that $$F=\frac{Gm_1m_2}{r^2}.$$ If one of the two objects is Earth, then from $F=GMm/r^2$ and we know $g=GM/r^2$. So 'g' should be infinity, when the object is on the center.

Possible duplicates: http://physics.stackexchange.com/q/2481/2451 , http://physics.stackexchange.com/q/18446/2451 and links therein. — Qmechanic, Sep 09 '15 at 19:41
What would you plug in for the mass at the center in your equation? — Karan Singh, Sep 09 '15 at 19:42
You should learn to use MathJax for equations. See http://meta.math.stackexchange.com/q/5020/ — Bill N, Sep 09 '15 at 22:26

score 2 · Answer 1 · answered Sep 09 '15 at 22:22

This is a Gauss's Law problem using mass. The only mass important in the $g=GM/r^2$ equation is the mass inside the radius $r$. The mass outside this radius is irrelevant in the first order.

At $r<R_E$, $$M=\rho (4/3)\pi r^3$$ where $\rho$ can be approximated (poorly) by

$$\rho=\frac{3M_E}{4\pi R_E^3}.$$

$$M=M_E\frac{r^3}{R_E^3}$$ so $g$ below the Earth will behave approximately like $$g=GM_E\frac{r}{R_E^3}$$ which goes to zero as $r\rightarrow 0$.

score 1 · Answer 2 · answered Sep 09 '15 at 19:41

1

That equation holds for a point mass (or a radially-symmetric mass when you are outside the shells). The earth is not a point mass, so you can't get to r=0 from the whole thing.

The equation fails as soon as you drill below the surface.

answered Sep 09 '15 at 19:41

BowlOfRed

40,008

score 1 · Answer 3 · answered Sep 09 '15 at 19:48

At the center of the earth you would experince no gravtational force. (Since the force composants from the surrounding matter will cancel each other.)

Reference https://www.physicsforums.com/threads/how-strong-is-gravity-in-the-center-of-the-earth.203955/

score 1 · Answer 4 · answered Sep 09 '15 at 19:50

As a thought experiment, imagine you are sinking through the Earth, from the surface of the towards the center.

At any point as you sink, you can divide the mass of the Earth into two parts: the sphere below you and the spherical shell above you.

The net gravitational attraction at any position within a spherical shell is zero, as Newton proved.

The remaining spherical mass, that beneath you, if you assume constant density, is proportional to the volume of the sphere. Spherical volume varies as the cube of the distance from the center of the Earth. Gravitational attraction is proportional to that mass, and inversely proportional to the square of the distance; r³/r² leaves you with just r; so gravitational attraction at the surface is linearly proportional to the distance from the center. As you sink to zero radius, the mass beneath you also goes to zero, and gravitational attraction falls linear along the way.

$g=GM/r^2$; $M=$mass of Earth. So in center $r=0$, so $g$ should be infinity

4 Answers4