# A ball thrown in earth

A tunnel is dug through the earth and a point mass m falls from rest in it. I do know other methods but i want to do it using calculus , i tried it , but the answer i get is not matching . and also briefly tell how to accurately solve 2nd degree differential equations, i have take i account the shrinking of gravity while going down !
$$\This \note \is \now \closed!$$  Doubt resolved all thanks to Mark Heninngs sir !

Note by Brilliant Member
1 year, 7 months ago

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Assuming that the tunnel is radial...

At a distance $$x$$ from the centre of the earth, the gravitational attraction is proportional to $$x$$. Thus we obtain a differential equation $\frac{d^2x}{dt^2} \; = \; -k^2x$ lumping all the constants together into $$k$$. The simplest way to solve this DE is to look for solutions of the form $$x = e^{imt}$$. This function will be a solution provided that $$m^2 = k^2$$, so we have solutions $$e^{\pm ikt}$$. Since we can choose the initial speed and position of the particle, the general solution is $x \; = \; Ae^{ikt} + Be^{-ikt}$ Changing the shape of the arbitrary constants, this solution becomes $x \; = \; C \cos kt + D \sin kt$ The particle performs SHM.

You can integrate the DE in the way you tried (except that you missed out the minus sign in the DE), obtaining $\left(\frac{dx}{dt}\right)^2 + k^2x^2 \; = \; c$ If we start with the particle at rest at the earth's surface then, on the way down: $\frac{dx}{dt} \; = \; -k\sqrt{R^2 - x^2}$ We can separate variables and integrate to obtain $$x = R\cos kt$$. The downside of this method, compared with the previous one, is that you need to keep on solving different DEs' depending on the direction of motion of the particle (and the consequent choice of sign for the square root).

- 1 year, 7 months ago

Comment deleted Nov 25, 2016

The integral of $$\frac{1}{\sqrt{R^2-x^2}}$$ is $$\sin^{-1}\frac{x}{R}$$, giving $$\sin^{-1}\frac{x}{R} = t\sqrt{k} + c$$.

The first integral of the DE, with $$\dot{x}^2 + x^2$$ constant, is a statement of conservation of energy.

- 1 year, 7 months ago

but sir i have changed it to cos inverse yet the answer is coming wrong !

- 1 year, 7 months ago

i think the answer is give wrong sir !

- 1 year, 7 months ago

yes sir the answer given is wrong the answer i got was right !

- 1 year, 7 months ago

sir i knew that i was missing a negative sign (as x is decreasing) but since i had to square root every term after integrating once , i thought the minus is just creating difficulties so i dropped it for a second :( and this time i will do my method very clearly so that you can check it . also, i know no words how to thank you for the knowledge you impart me ! $$Thank\ you\ very\ much\ Sir$$

- 1 year, 7 months ago