Is it possible, two packages traversing through the earth at different lattidues

[eon714]

Ok now, here is the scenario; If you Imagine the Earth (for argument sake a perfect sphere with uniform density), and you were to drill two holes along different latitudes till you got to the other side, let's say one at the equator (with radius R) and the other really north let's say through the northern part of Canada (with radius r). Now since the Earth is shaped like a sphere the tunnel in the equator and the tunnel up north is going to be significantly different distances compared to each other. Clearly the tunnel cut through the northern part of the Earth is significantly shorter. The question posed is if two people were to drop a package into these tunnels simultaneously prove that it would take the same amount of time for the package to traverse their respective tunnels even though they are different distances. disregard wind resistance as well as core heat.

Now the approach I took for this question is as follows: I used Newton's Universial gravitation with a combination of Gauss's law (at the least the principle of Gauss's law) and got two gravitational forces for both holes as follows: F = ∫ g • dA =-4(π)GM

F = g• 4(π)R^2 = -4(π)dGM

we come to gravity as we know it for the entire sphere of Earth where M is the mass inside the gaussian surface :
Me is mass of Earth ;

g = -GMe/R^2

Now for the other gravity (g') felt in the other tunnel we apply the same principle with a gaussian surface but this time we must consider density and volume of the Earth since r<R.
Min = mass inside
Me = mass of Earth

F = ∫ g' • dA =-4(π)(Min)

Min = ρV' = {[(4/3)(π)(r^3)]/ [(4/3)(π)(R^3)]}Me

F = ∫ g' • dA =-4πG(r^3/R^3)Me

g'•4πr^2 = -4πG(r^3/R^3)Me

g' = -GrMe/R^3

Now to prove that the time it takes for a the packages to make it through their respective tunnels is the same simply use x(t) = 1/2at^2, where a is g or g'

for equator tunnel:

x(t) = 1/2at^2 = 1/2gt^2
2R= 1/2(GM/R^2 )t^2

t = (4R^3/GMe)^1/2

and for the tunnel (shorter in length) cut out up north:

2r = (1/2)(g')(t^2)= (1/2)(-GrMe/R^3)(t^2)

t' = (4R^3/GMe)^1/2

Interestingly enough t = t'

I was wondering if this was a valid approach to solving something like this. I understand gravitation is dependant on masses between two bodies as well as distance. So my asumption was to create a relationship between the two lengths with repsect to gravity. The integral accounts for the change in angle as the package passes through the tunnel. Of course we all know that theoretically the package would undergo Simple harmonic motion in a case like this.

Does anyone have other ways of going about this type of scenario, I'm curious to see what people have to say.

 

[MisterX]

The gravitational acceleration would not be constant so x(t) = (1/2)at^2 is not correct. For the longer hole through the equator (with perfect sphere of uniform density), the gravitational acceleration would be zero at the center.

For the shorter hole, the gravitational acceleration vector would not point up/down the hole. Instead it would point to the center of mass. Eventually the package would crash into the side of the hole.

There may be other mistakes here as well. Force is not the surface integral of acceleration, for example.

 

[eon714]

although you are correct about acceleration not being constant as the package traverses into the earth, you can formulate gravity using gaussian surfaces or at least many sources tell me so. for example i found this linked pdf file i went off ...

www.pgccphy.net/ref/gravity.pdf

 

[Superstring]

For a hole running through the diameter, using Gauss' Law I got:

$$g = -k^2r$$

Where:

$$k^2 = \frac{4}{3} \pi G \rho$$

With ρ being the average density of the earth.


since g=a:

$$\ddot{r}+k^2r=0$$

Which is the equation of a simple harmonic oscillator.


This means that:

$$r=r_0~cos(kt+\phi)$$


Solving for the period:

$$T=\frac{2\pi}{k}=\sqrt{\frac{3\pi}{G\rho }}$$


The time it takes to fall completely through the Earth is half the period, or:

$$t=\frac{\pi}{k}=\sqrt{\frac{3\pi}{4G\rho }}$$

 

[nasu]

For a tunnel running along a chord the component of g along the tunnel is
$$g_t=k^2 x$$
where x is the distance from the middle point of the tunnel (which is the equilibrium position) and k is the same constant as above.
So the period is the same as for the tunnel along the diameter.