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!\This \note \is \now \closed! Doubt resolved all thanks to Mark Heninngs sir !

Note by A Former Brilliant Member
2 years, 11 months ago

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Assuming that the tunnel is radial...

At a distance xx from the centre of the earth, the gravitational attraction is proportional to xx. Thus we obtain a differential equation d2xdt2  =  k2x \frac{d^2x}{dt^2} \; = \; -k^2x lumping all the constants together into kk. The simplest way to solve this DE is to look for solutions of the form x=eimtx = e^{imt}. This function will be a solution provided that m2=k2m^2 = k^2, so we have solutions e±ikte^{\pm ikt}. Since we can choose the initial speed and position of the particle, the general solution is x  =  Aeikt+Beikt x \; = \; Ae^{ikt} + Be^{-ikt} Changing the shape of the arbitrary constants, this solution becomes x  =  Ccoskt+Dsinkt 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 (dxdt)2+k2x2  =  c \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: dxdt  =  kR2x2 \frac{dx}{dt} \; = \; -k\sqrt{R^2 - x^2} We can separate variables and integrate to obtain x=Rcosktx = 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).

Mark Hennings - 2 years, 11 months ago

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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 SirThank\ you\ very\ much\ Sir

A Former Brilliant Member - 2 years, 11 months ago

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