Period of motion of an object dropped through the Earth

[kubaanglin]

Homework Statement

An object of mass $m$ moves in a smooth, straight tunnel dug between two points on the Earth’s surface. Show that the object moves with simple harmonic motion, $a = - ω^2 x$. Find the period of this motion. You can assume that the Earth’s density is uniform.

Homework Equations

$$F=G\frac{Mm}{r^2}$$
$$V_{sphere}=\frac{4}{3}πr^3$$
$$ρ = \frac{M}{V}$$

The Attempt at a Solution

I have been staring at this problem for the past 30 minutes. I would be very appreciative if someone could give me a hint on how to start this. I intuitively understand why the object would experience simple harmonic motion, I just can't figure out how to express it mathematically, especially since the hole does not have to go through the center of Earth.

 

[Simon Bridge]

For SHM, $F\propto r$ when r is the displacement from an equilibrium point.
Notice that the two points on the surface of the Earth do not have to be diametrically opposite each other.
However - you have $F\propto r^{-2}$ for the case that r>R. But that is not the case here.

How does M vary with r when r<R (R=radius of the earth)?

 

[kubaanglin]

$M=\frac{4}{3}πr^3ρ=(\frac{4}{3}πr^3)(\frac{M_{Earth}}{\frac{4}{3}πR^3})=\frac{r^3M_{Earth}}{R^3}$
$a=G\frac{M}{r^2}=G\frac{\frac{r^3M_{Earth}}{R^3}}{r^2}=G\frac{rM_{Earth}}{R^3}$
Is this correct so far?

 

[SammyS]

$M=\frac{4}{3}πr^3ρ=(\frac{4}{3}πr^3)(\frac{M_{Earth}}{\frac{4}{3}πR^3})=\frac{r^3M_{Earth}}{R^3}$
$a=G\frac{M}{r^2}=G\frac{\frac{r^3M_{Earth}}{R^3}}{r^2}=G\frac{rM_{Earth}}{R^3}$
Is this correct so far?

That is the magnitude of radial acceleration if the object were not constrained to move in the tunnel.

Multiply by mass, m, to find the magnitude of the gravitational force. What component of that force is in the direction of the tunnel?

 

[kubaanglin]

$F = G\frac{rmM_{Earth}}{R^3}\cosθ$ where $θ$ is the angle between $r$ and the direction of motion. $\cosθ = \frac{x?}{r}$.

 

[haruspex]

where θ is the angle between r and the direction of motion.

If you fall into a vertical hole, what is the angle between your direction of motion and the line joining you to the centre of the Earth?

 

[kubaanglin]

The angle would be $0°$ so then $\cos(0°)=1$. That leaves me with $F = G\frac{rmM_{Earth}}{R^3}$. How would that account for the component of the force if the tunnel was not through the center of the Earth?

 

[haruspex]

The angle would be $0°$ so then $\cos(0°)=1$. That leaves me with $F = G\frac{rmM_{Earth}}{R^3}$. How would that account for the component of the force if the tunnel was not through the center of the Earth?

I'm sorry - I misread the question.
So let x be the distance from the midpoint of the tunnel. Express theta as a function of x and r.

 

[kubaanglin]

So $\cos(θ)=\frac{x}{r}$ and then $F = G\frac{mMx}{R^3}$?

 

[haruspex]

So $\cos(θ)=\frac{x}{r}$ and then $F = G\frac{mMx}{R^3}$?

Yes.

 

[kubaanglin]

Okay, so then
$a=ω^2x=G\frac{Mx}{R^3}$
$ω=\sqrt{\frac{GM}{R^3}}=\frac{2πr}{T}$
$T=\frac{2πr}{\sqrt{\frac{GM}{R^3}}}$
Is this correct?

 

[haruspex]

Okay, so then
$a=ω^2x=G\frac{Mx}{R^3}$
$ω=\sqrt{\frac{GM}{R^3}}=\frac{2πr}{T}$
$T=\frac{2πr}{\sqrt{\frac{GM}{R^3}}}$
Is this correct?

What is r there? Check the dimensional consistency.

 

[kubaanglin]

Oh, $ω=\frac{2π}{T}\neq\frac{2πr}{T}$
Then this should be the answer:
$$T=\frac{2π}{\sqrt{\frac{GM}{R^3}}}$$
$$T=\frac{2π}{\sqrt{\frac{(6.67⋅10^{-11})(5.96⋅10^{24})}{(6.37⋅10^6)^3}}}=5066 s=84.44 min$$

 

[haruspex]

Oh, $ω=\frac{2π}{T}\neq\frac{2πr}{T}$
Then this should be the answer:
$$T=\frac{2π}{\sqrt{\frac{GM}{R^3}}}$$
$$T=\frac{2π}{\sqrt{\frac{(6.67⋅10^{-11})(5.96⋅10^{24})}{(6.37⋅10^6)^3}}}=5066 s=84.44 min$$

Looks right.

 

[kubaanglin]

Thank you so much for your help!