Speed of an object thrown down a hole to the center of the earth

[qpt]

I am supposed to calculate how fast an object is moving at the center of the Earth if thrown down some hole that was drilled through the earth. I am supposed to set up a dr/dv differential equation and solve it. I am supposed to use Gauss's law. I am to assume the Earth is uniformly dense.

Here is what I have so far, and you will see why I am puzzled:

int(g . da) = Menc / G

(g is the force, G is the universal gravitational constant).

So g = Fg/m = GMe/r^2, no problem there. I can take it out of the integral by symmetry:

GMe/r^2 int(da) = Menc / G

Now the da is what is getting me. A = 4 pi r^2, so da = 8pi*r dr, but then I have int (8pi*r dr) and then my dr's vanish. I am supposed to get this in terms of a DE with dr/dv so I can't have my dr's vanishing.

One thing I thought about doing, and I don't know if this is right or not:

int(da) = 4pi* (dr)^2

And Menc = int(k*dv) (v is volume, k is some constant),
Menc = 4/3 (Pi*(dr)^3).

When I collect the Menc and the da, all but one dr cancels, leaving me with.

g = (4/3)Pi*dr / (G)

Is this the right approach?

My other problem, assuming that what I have is true, is that I can rewrite g in terms of Force, which has an acceleration, which has a dv/dt term in it. But now I have 3 differentials; dv/dt and dr. dt is the odd guy out:

[some coefficient here] dv/dt = (4/3)Pi*dr / G.

I can think of no way to "chain rule" dt out. Is there something obvious I'm missing.

Thanks in advance.

 

[cookiemonster]

This is one of my favorite problems, actually. However, I never used the approach you are. I used a conservation of energy approach.

However, for Gauss's law,

$$\int_\textrm{surface}F \cdot dA = \frac{M_\textrm{earth}}{G}$$

you're probably going to want to convert the integral into spherical coordinates. When you do that, you must replace $dA$ with $r^2\textrm{sin}\theta dr\,d\theta\,d\phi$ where r is the radius, $\theta$ is the angle off the z-axis, and $\phi$ is the angle in the xy-plane.

Once you evaluate the integral, there should be no differentials left.

Just out of curiosity, why do you want to know the gravitational field? Do you intend on using it to calculate potential?

cookiemonster

 

[M98Ranger]

It looks like you are on the right track with your approach. I can see why you are puzzled with the dr's canceling out, but that is actually the correct result. The reason for this is that when we are dealing with Gauss's law, we are looking at the net gravitational force acting on an object. This force is dependent on the mass enclosed within a certain radius, not the actual radius itself. So, when we integrate over the entire volume of the Earth, we are essentially summing up all of the mass that is contributing to the gravitational force at the center. This is why the dr's cancel out, because we are not concerned with the actual distance from the center, but rather the mass enclosed within that distance.

To solve for the speed at the center of the Earth, we can use the equation you have derived:

g = (4/3)Pi*dr / G

We can then rewrite this in terms of dv/dt:

g = (4/3)Pi*r*dv/dt / G

We can then rearrange this to solve for dv/dt:

dv/dt = (3/4)G*r*g / Pi

Now, we can integrate both sides with respect to time to get the velocity at the center of the Earth:

v = (3/4)G*r*g*t / Pi + C

Where C is the constant of integration. We can determine this constant by setting v = 0 when t = 0, since the object is initially thrown with no initial velocity. This results in:

C = 0

So, our final equation for the speed at the center of the Earth is:

v = (3/4)G*r*g*t / Pi

Where r is the radius of the Earth and g is the gravitational acceleration at the surface of the Earth. This equation can be used to calculate the speed at any given time t after the object is thrown down the hole.

I hope this helps clarify your approach and solve the problem. Good luck!