Ball Dropped in Tunnel Through Earth: What Happens Next?

[Bartholomew]

The geometrical approach is far more intuitive than the calculus approach. This could turn into a lengthy debate so I'll say right now that you are not going to change my mind. You keep your opinion and I'll keep mine. Unless that really bothers you, in wihch case let's use the general philosophy forum.

 

[jlsoto]

Yes, that's right!



Well, if the tunnel were at equator, the trip would take just 4 or 5 seconds more, due the centrifugal force.

In other words, basically the same 42 minutes Gokul had said.

yes, but what if the tunnel is neither at the equator nor from north to south pole? what if it was let's say from San Francsico to Los Angeles? or from the north pole to Los Angeles, would it still be the same time?

 

[Doc Al]

yes, but what if the tunnel is neither at the equator nor from north to south pole? what if it was let's say from San Francsico to Los Angeles? or from the north pole to Los Angeles, would it still be the same time?

Set up the problem and see if that matters.

 

[jlsoto]

i'm just a simple layman that got interested in the subject. i have no idea of formulas or anything, you would have to explain it to me in terms that i would understand, that's why i asked.

 

[K.J.Healey]

I only get about 14.9 minutes: Plug into Mathematica:
F = -G m M /r[t]^2;
U = Integrate[F, r[t]];
\[Rho] = ME/((4/3)*Pi*RE^3);
M = \[Rho]*(4/3)*Pi*r[t]^3;
T = (1/2) m (r'[t])^2;
L = T - U;
a = D[L, r[t]];
b = D[D[L, r'[t]], t];
d = Solve[a - b == 0, r''[t]][[1]][[1]][[2]]
f = DSolve[{r''[t] == d, r[0] == RE, r'[0] == 0}, r[t],
t][[1]][[1]][[2]];
Time = -Solve[f == 0, t][[1]][[1]][[2]]
ME = 5.9742*10^24 _kg;
RE1 = 6356.75 *1000 _m;
RE2 = 6378.135 *1000 _m;
G = 6.667428*10^-11 (_m^3) _kg^-1 _s^-2;
RE = RE1;
Sqrt[Simplify[Time^2]*(1 _min^2/(60 _s)^2)]
RE = RE2;
Sqrt[Simplify[Time^2]*(1 _min^2/(60 _s)^2)]OUTPUT:
14.8658 minutes
14.9408 minutes

For the longest and shortest radii.

 

[Doc Al]

I only get about 14.9 minutes: Plug into Mathematica:
F = -G m M /r[t]^2;
U = Integrate[F, r[t]];
\[Rho] = ME/((4/3)*Pi*RE^3);
M = \[Rho]*(4/3)*Pi*r[t]^3;
T = (1/2) m (r'[t])^2;
L = T - U;
a = D[L, r[t]];
b = D[D[L, r'[t]], t];
d = Solve[a - b == 0, r''[t]][[1]][[1]][[2]]
f = DSolve[{r''[t] == d, r[0] == RE, r'[0] == 0}, r[t],
t][[1]][[1]][[2]];
Time = -Solve[f == 0, t][[1]][[1]][[2]]
ME = 5.9742*10^24 _kg;
RE1 = 6356.75 *1000 _m;
RE2 = 6378.135 *1000 _m;
G = 6.667428*10^-11 (_m^3) _kg^-1 _s^-2;
RE = RE1;
Sqrt[Simplify[Time^2]*(1 _min^2/(60 _s)^2)]
RE = RE2;
Sqrt[Simplify[Time^2]*(1 _min^2/(60 _s)^2)]


OUTPUT:
14.8658 minutes
14.9408 minutes

For the longest and shortest radii.

Not quite sure what you did here. Making the usual assumptions (uniform density, spherical, non-rotating Earth & frictionless/airless tunnel), the transit time from one end to the other is about 42 minutes. It's easy to show that the time does not depend on whether the tunnel goes through the center of the Earth or not. (Sorry for the delayed response; been away for a few days.)

 

[K.J.Healey]

I just used the Force/Potential of gravity to get equations of motion.
I get :
R'' = -((2 G ME r(t))/RE^3)
Where G is G
ME is mass of Earth (total)
and RE is surface radius of the earth.

I Made my "mass" of the Earth in:
F=-G m M[r] /r^2 a function of the distance from the center, that is calculated using the density of the Earth (total mass/total vol) times the current radius.

so as the object falls it feels less and less gravity.
The acceleration I got was above, which is linear in "r"

Solving this equation (simple ODE) you get a sinusoid that has:
r''=-K*r
so its standard cos+sin, but at t=0, r'[0] == 0 and r[0] == Radius of Earth
initial conditions.

My "K" is : 3.07034*10^-6 s^-2

And I solve t differential in the DSolve line. Plugging in a bunch of constants I get:

R[t]== RE cos((Sqrt[2] Sqrt[G] Sqrt[ME] t)/RE^(3/2))

which simplifies to:
R[t]==(6.37814*10^6)*Cos[0.00175224 t]
in meters, and seconds.
First 0 point is at about 900 seconds ~15 minutes.

 

[Doc Al]

I just used the Force/Potential of gravity to get equations of motion.
I get :
R'' = -((2 G ME r(t))/RE^3)
Where G is G
ME is mass of Earth (total)
and RE is surface radius of the earth.

Where did the 2 come from?

 

[K.J.Healey]

Where did the 2 come from?

When you take the potential
U = G m M[r] / r
And you calculate the mass of everything inside a circle of radius r:
M = density * (4/3) Pi r^3
so the potential goes like r^2
When you apply Euler-Lagrange the 2 comes down in the d/dr

 

[K.J.Healey]

Heres how i did it in mathematica:
<br /> F= -G m M /r[t]{}^{\wedge}2<br />
<br /> -\frac{G m M}{r(t)^2}<br />

<br /> U=\text{Integrate}[F,r[t]]<br />
<br /> \frac{G m M}{r(t)}<br />

<br /> \rho = \text{ME}/((4/3)*\text{Pi}*\text{RE}{}^{\wedge}3)<br />
<br /> \frac{3 \text{ME}}{4 \pi \text{RE}^3}<br />

M=\rho *(4/3)*\text{Pi}*r[t]{}^{\wedge}3<br />
<br /> \frac{\text{ME} r(t)^3}{\text{RE}^3}<br />

T=(1/2)m (r&#039;[t]){}^{\wedge}2<br />
<br /> \frac{1}{2} m r&#039;(t)^2<br />

<br /> L=T-U<br />
<br /> \frac{1}{2} m r&#039;(t)^2-\frac{G m \text{ME} r(t)^2}{\text{RE}^3}<br />

\text{c1}=D[L,r[t]]<br />
<br /> -\frac{2 G m \text{ME} r(t)}{\text{RE}^3}<br />
\text{c2}=D[D[L,r&#039;[t]],t]<br />

<br /> m r&#039;&#039;(t)<br />
\text{accel}=\text{Solve}[\text{c1}-\text{c2}==0,r\text{&#039;&#039;}[t]][[1]][[1]][[2]]<br />
<br /> -\frac{2 G \text{ME} r(t)}{\text{RE}^3}<br />

\text{motion}=\text{DSolve}[\{r\text{&#039;&#039;}[t]==\text{accel},r[0]==\text{RE},r&#039;[0]==0\},r[t],t][[1]][[1]][[2]]<br />
<br /> \text{RE} \cos \left(\frac{\sqrt{2} \sqrt{G} \sqrt{\text{ME}} t}{\text{RE}^{3/2}}\right)<br />
\text{Time}=-\text{Solve}[\text{motion}==0,t][[1]][[1]][[2]]<br />

<br /> \text{Solve}\text{::}\text{ifun}: \text{Inverse functions are being used by $\backslash $!$\backslash $(TraditionalForm$\backslash \grave{ }$Solve$\backslash <br /> <br /> $), so some solutions may not be found; use Reduce for complete solution information. $\backslash $!$\backslash $($\backslash $*ButtonBox[$\texttt{&quot;}<br /> <br /> \rangle\rangle \texttt{&quot;}$, ButtonStyle-$&gt;\texttt{&quot;}$Link$\texttt{&quot;}$, ButtonFrame-$&gt;$None, ButtonData:$&gt;\texttt{&quot;}$paclet:ref/message/Solve/ifun$\texttt{&quot;}<br /> <br /> $, ButtonNote -$&gt;$ $\texttt{&quot;}$Solve::ifun$\texttt{&quot;}$]$\backslash $)}<br />

<br /> \frac{\pi \text{RE}^{3/2}}{2 \sqrt{2} \sqrt{G} \sqrt{\text{ME}}}<br />

\text{ME}=5.9742*10{}^{\wedge}24 \text{$\_$kg};<br />

<br /> \text{RE1} = 6356.75 *1000 \text{$\_$m};<br />

<br /> \text{RE2} = 6378.135 *1000 \text{$\_$m};<br />

<br /> G = 6.667428*10{}^{\wedge}-11 \text{$\_$m}{}^{\wedge}3 \text{$\_$kg}{}^{\wedge}-1 \text{$\_$s}{}^{\wedge}-2<br />

<br /> \text{RE}=\text{RE1};<br />

<br /> \text{Sqrt}[\text{Simplify}[\text{Time}{}^{\wedge}2]*(1\text{$\_$min}{}^{\wedge}2/(60 \text{$\_$s}){}^{\wedge}2)]<br />

<br /> \text{RE}=\text{RE2};<br />

<br /> \text{Sqrt}[\text{Simplify}[\text{Time}{}^{\wedge}2]*(1\text{$\_$min}{}^{\wedge}2/(60 \text{$\_$s}){}^{\wedge}2)]<br />

14.8658 \sqrt{\text{$\_$min}{}^2}<br />

14.9408 \sqrt{\text{$\_$min}{}^2}<br />

 

[K.J.Healey]

The non generated one is just for :
G = 6.667428*10^-11 _m^3 _kg^-1 _s^-2

 

[Doc Al]

Heres how i did it in mathematica:
<br /> F= -G m M /r[t]{}^{\wedge}2<br />
<br /> -\frac{G m M}{r(t)^2}<br />

<br /> U=\text{Integrate}[F,r[t]]<br />
<br /> \frac{G m M}{r(t)}<br />

<br /> \rho = \text{ME}/((4/3)*\text{Pi}*\text{RE}{}^{\wedge}3)<br />
<br /> \frac{3 \text{ME}}{4 \pi \text{RE}^3}<br />

M=\rho *(4/3)*\text{Pi}*r[t]{}^{\wedge}3<br />
<br /> \frac{\text{ME} r(t)^3}{\text{RE}^3}<br />

T=(1/2)m (r&#039;[t]){}^{\wedge}2<br />
<br /> \frac{1}{2} m r&#039;(t)^2<br />

<br /> L=T-U<br />
<br /> \frac{1}{2} m r&#039;(t)^2-\frac{G m \text{ME} r(t)^2}{\text{RE}^3}<br />

Since I don't use Mathematica I can't comment on your use of it, but your potential energy term is incorrect.

The force is:
F = -\frac{m M G}{R^3} r

So:
U = \frac{m M G}{2 R^3} r^2

 

[K.J.Healey]

Ah that's totally it, can't integrate F = GmM/r^2 to U when M is a function of R.