# Homework Help: Digging a hole through the earth

1. Feb 5, 2010

### aloshi

have examined how g varies with distance from the Earth's surface. but how to change g if, instead dig down to the center of the earth?
if we do Start in the center of the Earth (see picture) how will the value of g varies from there to 2r height above the Earth's surface. Suppose that Earth's density is constant throughout the Earth's volume.

http://www.pluggakuten.se/wiki/images/5/5a/Martin.jpg [Broken]

i do sow:
$$M = \rho \cd V = \rho \fr{4\pi r^3}{3}\\ F_1=m\cdot g\\ F_2=G\frac{mM}{r^2}\\ F_1=F_2\rightarrow \\ g=G\frac{M}{r^2}\rightarrow g=G\frac{\fr{\rho 4\pi r^3}{3}}{r^2 }\rightarrow \\ g=G\frac{\rho 4\pi r}{3}$$

but its WRONG,

Last edited by a moderator: May 4, 2017
2. Feb 5, 2010

### HallsofIvy

Re: gravitational

Why do you say it's wrong? It looks correct to me. The gravitational force due to any mass above you is canceled by a corresponding mass in the opposite direction so you only use the mass below you which is $(4/3)\pi r^3 \rho$. Since we can treat that mass as if it were all at the center of the earth the acceleration is $g= (4/3)\pi r^3 \rho G/ r^2= (4/3)\pi\rho G r$, exactly what you have.

Last edited by a moderator: May 4, 2017
3. Feb 5, 2010

### aloshi

Re: gravitational

but what is r? in this formula $g= (4/3)\pi r^3 \rho G/ r^2= (4/3)\pi\rho G r$
r must be distance between celestial bodies, but in the formula are the r=radius.

4. Feb 5, 2010

### aloshi

Re: gravitational

Now I'm mixing up all sorts of things in each other. What is meant by the change of g instead dig down to the center of the earth?
it means that we have different masses for the Earth?like;
http://www.pluggakuten.se/wiki/images/6/6b/3.JPG [Broken]

But what happens to gravity when earth curvature is half instead, see picture below;
http://www.pluggakuten.se/wiki/images/7/75/2.JPG [Broken]

does I thinking right? sorry, my english is bad

Last edited by a moderator: May 4, 2017