Calculating Distance from Center of Earth to Point with 0.50g Acceleration

[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 can't seem to find a way to relate Keplers Laws to this problem. Are there other equations I could use?
Thanks in advance!

 

[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.

 

[franznietzsche]

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 can't seem to find a way to relate Keplers Laws to this problem. Are there other equations I could use?
Thanks in advance!

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.

 

[HallsofIvy]

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 can't seem to find a way to relate Keplers Laws to this problem. Are there other equations I could use?
Thanks in advance!

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 m_e is the mass of the earth, m₂ is the mass of the "test object". If r= R, the radius of the earth, then [itex]\frac{Gm_{2}m_{2}}{R^2}=m_2 g[/tex] so that $\frac{Gm_e}{R^2}= g$ as Leong said. <br /> <br /> HOWEVER, only the mass BELOW the level of the object affects the gravitational pull. With m_e 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 <b>below</b> radius r is $\rho\frac{4}{3}\pi r^3= m_e\frac{r^3}{R^3}= m_e\(\frac{r}{R}\)^3$.<br /> <br /> 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.<br /> <br /> We must have $\frac{m_er}{R^3}= \frac{g}{2}=/frac{m_e}{R^2}$ . Solve that for r (which is remarkably simple!).[/itex]