SHM and hole through the center of the Earth

[astenroo]

Homework Statement

View the Earth as a homogenous sphere with the density $$\rho$$. Assume a hole has been succesfully been drilled through the center of the Earth, and a body with mass m is dropped down the hole. Give an expression of the force as function of of the distance r from the Earth's centre. Show that this force gives rise to a harmonic oscillation, and give the period for the motion. Assume m = 1 kg. What is the velocity and kinetic energy at passing the centre?

I've seen this has been tried earlier, but I would like to give a shot at it myself. I'm currently taking a course in maths (methods for physicists) and the current topic is differential equations.
Ok, here goes

Homework Equations

F=-kx (this is what I need to show)

The Attempt at a Solution

So, this is where I think I should start out.

F=Gm $$\frac{dM}{dr^{2}}$$

or

m$$\frac{d^{2}r}{dr^{2}}$$=Gm$$\frac{dM}{dr^{2}}$$

So, am I on the right track or am making this more complicated than it is? Hints would be very nice at this stage. I have a hunch I'll end up with complex numbers...

 

[Doc Al]

How can you express the force on the mass m as a function of its distance from the center of the earth? (No complex numbers needed.) Hint: If the distance is r, how much mass is contained in the sphere of radius r?

 

[dacruick]

I wrote something but Doc Al has a better post:)

 

[astenroo]

How can you express the force on the mass m as a function of its distance from the center of the earth? (No complex numbers needed.) Hint: If the distance is r, how much mass is contained in the sphere of radius r?

The mass inside a sphere of radius r is V$$\rho$$, where V = $$\frac{4\pi r^3}{3}$$

If I use Newton's gravity law, then the force on m at distance r would be F=-GMm/r^2. But now the radius changes, which implies the force also changes... I'll just throw an idea

F = ma = m d2r/dt2

m d2r/dt2 = -Gm$$\frac{4\pi r^3}{3 dr^2}$$

 

[Doc Al]

You're on the right track. But you left out ρ in your final equation and added a 'd'.

The key point, which was proven by Newton, is that only the fraction of the total mass below the point 'r' contributes to the force on the mass m.

 

[astenroo]

You're on the right track. But you left out ρ in your final equation and added a 'd'.

The key point, which was proven by Newton, is that only the fraction of the total mass below the point 'r' contributes to the force on the mass m.

Yes indeed I did :)

So, the corrected formula should be

m $$\frac{d^{2}r}{dt^{2}}$$= -Gm$$\frac{4\pi r^{3}\rho}{3r^{2}}$$

At this point the m's cancel each other out and some of the r's and I end up with the following
$$\frac{d^{2}r}{dt^{2}}$$= -G$$\frac{4\pi \rho r}{3}$$ and

Then on the right hand side I basically end up with -kr... but, since r changes, am I allowed to make the assumption
$$\frac{d^{2}r}{dt^{2}}$$= -G$$\frac{4\pi \rho}{3}$$dr ?

Or should I be satisfied that

$$\frac{d^{2}r}{dt^{2}}$$ + G$$\frac{4\pi \rho}{3}$$r = 0, where G$$\frac{4\pi \rho}{3}$$ = k? Which according to the litterature characterizes all types of simple harmonic oscillations?

Hmm..I wonder if should've left the m's in the equation...

 

[Doc Al]

So, the corrected formula should be

m $$\frac{d^{2}r}{dt^{2}}$$= -Gm$$\frac{4\pi r^{3}\rho}{3r^{2}}$$

At this point the m's cancel each other out and some of the r's and I end up with the following
$$\frac{d^{2}r}{dt^{2}}$$= -G$$\frac{4\pi \rho r}{3}$$ and

All good.

Then on the right hand side I basically end up with -kr... but, since r changes, am I allowed to make the assumption
$$\frac{d^{2}r}{dt^{2}}$$= -G$$\frac{4\pi \rho}{3}$$dr ?

No, you can't just replace r with dr.

Or should I be satisfied that

$$\frac{d^{2}r}{dt^{2}}$$ + G$$\frac{4\pi \rho}{3}$$r = 0, where G$$\frac{4\pi \rho}{3}$$ = k? Which according to the litterature characterizes all types of simple harmonic oscillations?

Hmm..I wonder if should've left the m's in the equation...

Yes, you should have left the m's into make the equation exactly parallel to F = -kx.

$$m\frac{d^{2}r}{dt^{2}} = -Gm\frac{4\pi \rho}{3}r = -kr$$

 

[astenroo]

All good.No, you can't just replace r with dr.


Yes, you should have left the m's into make the equation exactly parallel to F = -kx.

$$m\frac{d^{2}r}{dt^{2}} = -Gm\frac{4\pi \rho}{3}r = -kr$$

$$m\frac{d^{2}r}{dt^{2}} + Gm\frac{4\pi \rho}{3}r$$= 0 should then be the required equation. Ok, so far so good.

Then the period is, don't know if I need to derive it from the diff, $$T = 2 \pi \sqrt{\frac{m}{k}}$$. From where it then is just to insert values and I should be done with periodic motion.

from the equation r(t)=A sin ($$\omega$$t + $$\varphi$$ I should be able to derive the kinetic energy and the velocity for the mass as it passes the centre.

 

[astenroo]

Aaargh... What is the major malfuntion with LaTex? The period should be 2 pi sqrt (m/k)...