- #1

zenterix

- 530

- 74

- Homework Statement
- Imagine a shaft going all the way through the Earth from pole to pole along its rotation axis. Assuming the Earth to be a homogeneous ball and neglecting air drag, find

(a) the equation of motion of a body falling down into the shaft

(b) time it takes the body to reach the other end of the shaft

- Relevant Equations
- I started with the equation

$$F_g=\frac{Gm_1m_2}{y^2}$$

Then, by the 2nd Law

$$F_g=\frac{Gm_1m_2}{y(t)^2}=m_1y''(t)\tag{1}$$

$$y''(t)=\frac{Gm_2}{y(t)^2}\tag{2}$$

I don't know how to solve (2) but it doesn't seem to be correct.

I then thought about the following picture

Consider the Earth as made up of infinitesimally small pieces of mass ##dm##.

Given a point on the shaft with height ##y## ( the red point on the ##y##-axis above), we can try to calculate the gravitational force at that point.

Collectively, the pieces of Earth between the red curve and the green part of the circle above it contribute zero to the gravitational force because of cancellation of the forces due to symmetry.

This green curve at the top is parametrized as

$$\vec{c}(\theta)=R\hat{r}(\theta)=R(\sin{\theta}\hat{i}+\cos{\theta}\hat{j})$$

The red curve is given by

$$\vec{p}(\theta) = c_x(\theta)\hat{i}+[y-(c_y(\theta)-y)]\hat{j}$$

$$=R\sin{\theta}\hat{i}+(2y-R\cos{\theta})\hat{j}$$

For a point ##(x_{dm},y_{dm})## with mass ##dm##, we can compute the light green vector in the following picture

It is ##(x_{dm},y_{dm})-(0,y)## with magnitude

$$\lVert (x_{dm},y_{dm})-(0,y)\rVert=\sqrt{x_{dm}^2+(y_{dm}-y)^2}$$

$$=\sqrt{R^2\sin^2{\theta}+(y_{dm}-y)^2}$$

The gravitational force at the point on the shaft is then

$$dF_g=\frac{Gm_1dm}{(x_{dm}^2+(y_{dm}-y)^2)^{3/2}} (x_{dm}\hat{i}+ (y_{dm}-y)\hat{j})$$

$$=\frac{Gm_1dm}{(R^2\sin^2{\theta}+(y_{dm}-y)^2)^{3/2}}(R\sin{\theta}\hat{i}+(y_{dm}-y)\hat{j})$$

Let me explain what I would like to do here.

We have a given point on the shaft with a given height ##y##.

Note that for this ##y##, the corresponding ##\theta## is

$$\cos{\theta}=\frac{y}{R}$$

$$\theta=\cos^{-1}{\frac{y}{R}}$$

Given a ##\theta## between ##0## and ##\cos^{-1}{\frac{y}{R}}##, I'd like to integrate the gravitational force ##dF_g## for ##y_{dm}## between the red curve and the bottom of the circle.

That gives the contribution from the purple line in the picture above on the right side of the circle.

By symmetry, the left side of the circle cancels the ##x## component and doubles the ##y## component of the integral.

Thus, given a ##\theta## the integral is

$$\int_{-R\cos{\theta}}^{2y-R\cos{\theta}} \frac{2Gm_1dm}{(R^2\sin^2{\theta}+(y_{dm}-y)^2)^{3/2}}((y_{dm}-y)\hat{j})$$

I'm not sure at this point what to do with the ##dm## but it should be expressed as a ##dy_{dm}##.

I have lots more calculations but I want to stop here so that this initial reasoning can be critiqued.

Does this approach make any sense?

Note that the full calculation would include:

1) Varying ##\theta## from ##0## to ##\cos^{-1}{\frac{y}{R}}##, varying an angle ##\phi## from ##0## to ##\pi## and

2) Doing a similar calculation for the light green lines in the picture above.

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