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$$\int_0^R mg\frac{r}{2}dr=\frac{m}{2}v_{max}$$ and can't write $$\frac{m}{2}v_{max}=mgR$$ don't we usually equal potential energy to kinetic energy?

A stone is falling through a tunnel towards the center of the Earth. Just the ush =)

And it goes on suggesting that the time it arrives at the center is $$T_{1/2}=\frac{\pi}{\omega_0}$$ since the movement can be characterized as a harmonic oscillator.

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When you equate kinetic and potential energy via $\frac{1}{2}mv^2=mgh$, you are implicitly assuming that the gravitational field is constant, i.e. it has constant magnitude $mg$ and is directed 'downwards'. This is a fine approximation if you confine yourself to work near the surface of the earth all the time, but that is not the case here. In this case, the magnitude of the gravitational field varies significantly and so cannot be assumed to be constant. Its magnitude is given by $$ G\frac{mM}{r^2}, $$ where $G$ is Newton's gravitational constant, $M$ is the mass of the earth enclosed in a sphere centered at the center of the earth of radius $r$, $m$ is the mass of the stone, and $r$ is the radial distance from the center of the earth to the stone.

In this problem, $r$ will vary all the way from $0$ to the radius of the earth, and so it is clearly not an accurate approximation to assume that the gravitational force will be constant.

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    $\begingroup$ In that equation $M$ is the mass enclosed in a Gaussian surface about the center of the Earth. If I remember correctly, the field turns out to go like $GmMr/R^3$ where $R$ is the radius of the Earth. $\endgroup$ – user12029 Nov 11 '14 at 17:10
  • $\begingroup$ Why is it then that kinetic energy does not vary? $\endgroup$ – user42141 Nov 11 '14 at 17:36
  • $\begingroup$ @user42141 The kinetic energy does vary . . . $v$ depends on $r$. $\endgroup$ – Jonathan Gleason Nov 11 '14 at 21:19

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