# Earth and Gravity help

1. Nov 9, 2004

### envscigrl

Problem:
Calculate the distance from the center of the Earth to a point below the surface where the acceleration due to gravity is 0.50g, where g is the acceleration due to gravity at the Earth's surface.
I am really stuck on this. I tried using T^2 = 4pi^2 / GMe *r^3
but it didnt work it could be because M is supposed to be the mass of the sun but I decided it should be the mass of the earth! But it wasnt, it didnt work. I just cant seem to find a way to relate Keplers Laws to this problem. Are there other equations I could use?

2. Nov 9, 2004

### Leong

Universal gravitational law :
$$\frac{Gm_{1}m_{2}}{r^2}=m_{2}g$$
m1 is the mass of the eart from its center up to the point. g is the gravitational acceleration at that point. m2 is the mass of an object you place at that point.
to find m1, you have to know the density of earth.

3. Nov 9, 2004

### franznietzsche

You used kepler's law. It makes no mention of force. It would work for a body in orbit of the earth, T being the period of the orbit. But that is not the question, so that equation is completely irrelevant. use the equationg given by leong. consider:

$$g=G*\frac{M}{r^2}$$

so

$$0.5g = G*\frac{M}{r_{new}^2}$$

and solve for $$r_{new}$$.

Note: $$M$$ is the mass of the earth.

4. Nov 10, 2004

### HallsofIvy

Staff Emeritus
I don't know why you would use that formula- that's for time and radius of an orbit!

In general, gravitational force is $\frac{Gm_{e}m_{2}}{r^2}$ where me is the mass of the earth, m2 is the mass of the "test object". If r= R, the radius of the earth, then $\frac{Gm_{2}m_{2}}{R^2}=m_2 g[/tex] so that [itex]\frac{Gm_e}{R^2}= g$ as Leong said.

HOWEVER, only the mass BELOW the level of the object affects the gravitational pull. With me as the total mass of the earth and R the radius of the earth, and &rho; the (average) density of the earth, $m_e= \frac{4}{3}\piR^3 \rho$ so $\rho= \frac{3m_e}{4\piR^3}$. The mass of the earth below radius r is $\rho\frac{4}{3}\pi r^3= m_e\frac{r^3}{R^3}= m_e$$\frac{r}{R}$$^3$.

The gravitational pull on an object of mass m1 at distance r from the center of the earth is $\frac{m_em_1$$\frac{r}{R}$$^3}{r^2}= \frac{m_em_1r}{R^3}$ and we want that equal to (1/2)gm.

We must have $\frac{m_er}{R^3}= \frac{g}{2}=/frac{m_e}{R^2}$ . Solve that for r (which is remarkably simple!).