Calculating Velocity of a Rock Dropped into a Hole Drilled Through the Earth

[pierce15]

Homework Statement

Inside the earth, the force of gravity is proportional to the distance from the center. If a hole is drilled through the Earth from pole to pole and a rock is dropped in the hole, with what velocity will it reach the center?

The Attempt at a Solution

I think that the proper way to set it up is

\frac{ d^2 x}{dt^2} = C (R - x),

where R is the radius of the earth. Then, since the acceleration is $g$ when the rock is at R, aka when x = 0, C must equal g / R. Thus, the equation is

\frac{d^2 x}{dt^2} = \frac{ g}{R} (R-x)

Is this the correct way to proceed?

\int \frac{d^2 x}{g/R (R - x)} = \int dt^2

 

[Mark44]

Homework Statement

Inside the earth, the force of gravity is proportional to the distance from the center. If a hole is drilled through the Earth from pole to pole and a rock is dropped in the hole, with what velocity will it reach the center?

The Attempt at a Solution

I think that the proper way to set it up is

\frac{ d^2 x}{dt^2} = C (R - x),

I would be inclined to let x be the distance from the center of the earth, with x = 0 being at the center and x = R being at the Earth's surface.

For a differential equation I would start with m x'' = Cx, which is a 2nd-order DE with constant coefficients. Presumably you have already worked some problems like this.

where R is the radius of the earth. Then, since the acceleration is $g$ when the rock is at R, aka when x = 0, C must equal g / R. Thus, the equation is

\frac{d^2 x}{dt^2} = \frac{ g}{R} (R-x)

Is this the correct way to proceed?

\int \frac{d^2 x}{g/R (R - x)} = \int dt^2

No. Your integral is not correct.

 

[pierce15]

Ok, so I have

\frac{ d^2 x}{dt^2} = \frac{g}{R} x

Can I do this?

\frac{ R}{g} \int \frac{ d^2 x}{x} = \int dt^2

Does that even make sense?

By the way, I haven't learned anything about second order d.e.'s yet

 

[Mark44]

Ok, so I have

\frac{ d^2 x}{dt^2} = \frac{g}{R} x

Can I do this?

\frac{ R}{g} \int \frac{ d^2 x}{x} = \int dt^2

Does that even make sense?

No, you can't do that, and it doesn't make sense.

By the way, I haven't learned anything about second order d.e.'s yet

Maybe I'm looking at this the wrong way, but I don't see any way around writing the equation as a 2nd order DE. I can't think of any trick to get it into the form of a 1st order DE.

 

[pierce15]

Well, I don't need to solve it, I only need to find an expression for $dx / dt$ since the problem is asking for the velocity at x = 0. Any ideas?

 

[Mark44]

The DE could be written this way:
m dv/dt = kx, x(0) = R

or dv/dt = (k/m)x
If you integrate the left side, you get v, but the right side would have to be left as an integral, since you're integrating a function of t (i.e., x(t)) with respect to t.

 

[ehild]

Ok, so I have

\frac{ d^2 x}{dt^2} = \frac{g}{R} x

You have a sign error. x is the distance from the centre of Earth, which decreases if the rock is falling down. The correct equation is \frac{ d^2 x}{dt^2} = -\frac{g}{R} x

The velocity v is the first time derivative of x, so d²x/dt²=dv/dt. You can consider v as function of x: v(x(t)). Whith the Chain Rule, dv/dt=dv/dx dx/dt = dv/dx v= 0.5 d(v²/dx). Your equation can be written as

\frac{ d(v^2)}{dx} = -\frac{g}{R} x.

that is a first order equation. Can you solve it for v²(x)?


ehild

 

[mfb]

Do you have to use a differential equation?
It is easier with energy conservation - ehild's last equation goes in the same direction.

 

[pierce15]

Nice catch, ehild. (By the way, I was defining up as positive, so the g i was using had a self-contained negative, but you're right, using using -g works better here)

\frac{dx^2}{dt^2} = v \frac{ dv}{dx} = - \frac{g}{R} x

\int v \, dv = -\frac{ g}{R} \int x \, dx

v^2 /2 = - \frac{g}{R} x^2 /2 + C

Since v= 0 when x = R:

0 = - gR + 2C \implies C = gR /2

Plugging in x = 0:

v^2 = gR \implies v = \sqrt{gR}

Interesting, even though I'm pretty sure hollowing out the Earth would take away its gravitational properties.

 

[D H]

Plugging in x = 0:

v^2 = gR \implies v = \sqrt{gR}

Correct.

This is an immediate consequence of conservation of energy -- assuming "the force of gravity is proportional to the distance from the center", that is. This isn't the case. That assumption requires a constant density throughout. The Earth's interior has anything but a constant density. The Earth's core is mostly iron and is highly compressed. Gravitational acceleration at the core/mantle boundary (about halfway down) is greater than acceleration at the surface of the Earth! That the Earth does not have a constant density has been known since 1774. Even though it's rather unrealistic, this constant density assumption makes for lots of interesting and fairly simple math/physics problems.

Interesting, even though I'm pretty sure hollowing out the Earth would take away its gravitational properties.

We're not hollowing out the Earth here. We're just making a hole through the Earth, presumably one with an insignificantly small diameter.

 

[pierce15]

We're not hollowing out the Earth here. We're just making a hole through the Earth, presumably one with an insignificantly small diameter.

Good point, but I think that's also impossible

 

[D H]

Of course it's impossible. That's not the point of the exercise.