Assuming that the earth attracts points inside it with a force which varies as distance from its centre, show that if a straight frictionless airless tunnel be made from one point of the earth's surface to any point, a train would traverse the tunnel in slightly less that three quarter of an hour. Assume the earth to be a homogeneous sphere of radius $6400$ km.
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Hint. The gravitational force on mass $m$ at distance $x$ from the centre of earth, towards the center, is $$\frac{G(M(x/R)^3)m}{x^2}=\frac{GMmx}{R^3}$$ (the earth attracts a point inside it with a force proportional to its distance from the centre).
Hence the motion of the mass obeys the linear differential equation $$mx''=-\frac{GMmx}{R^3}$$ that is $$x''+\frac{GM}{R^3}\cdot x=0.$$ By solving it, we obtain that the solution is periodic with a period given by $$T=2\pi\sqrt{\frac{R^{3}}{{GM}}}=2\pi\sqrt{\frac{R}{g}}\approx 6.28\sqrt{\frac{6400\cdot 10^3}{9.81}}\approx 5072\, sec.$$ Note that $T$ does not depend on where the tunnel is located (its distance from the center of the earth).
Robert Z
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But 5072 sec is more than an hour. The question needs it to be slightly less than 3\4 of an hour. – user467745 Aug 20 '17 at 12:57
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1@user467745 A period is equal to TWO transversal trips... – Robert Z Aug 20 '17 at 13:09