Ball Dropped in Tunnel Through Earth: What Happens Next?

[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.