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akshansh (5)

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The escape velocity for a planet is V. A tunnel is dug along its diameter and a particle is dropped into the tunnel. At the centre the speed of particle will be? Answer will be in terms of V.
    
ganesha1991 (1700)

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from the bottom of the earth

by conservation of energy
K.Ei + P.Ei = K.Ef + P.Ef
0 + mg(2r) = 1/2mv'2 + mg'r
1/2mv'2 =2 mgr             because g' =0  at the centre of the earth
v'2 = 4gr
v' =  2 gr

now escape velocity V = 2gr

therefore the velocity at the centre wud be 2 *  V


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akshansh (5)

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That was a good joke!
I have asked for velocity not acceleration.
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ganesha1991 (1700)

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sorry didn't notice the question properly

i'll try to answer it
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akshansh (5)

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Your answer sqrt(2)*V is wrong!!!
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ganesha1991 (1700)

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then whats the answer
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akshansh (5)

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V / sqrt(2)
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uday_zingtudor (931)

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Easy one yaar

Conservation of energy!!!

-GMm/R + 0 = -GMm(3(a)^2-2(R)^2)/2a^3 + I/2mu^2

-GMm/R + 0 = -GMm(3(R)^2-2(R)^2)/2R^3 + I/2mu^2


that implies -GMm/R = -3GMm/2R+1/2mu^2


GM/r = u^2


we know that v^2 = 2GM/R


u=v/rt2

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don't get scared !!!
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