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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Mar 2008 15:27:28 IST
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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.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Mar 2008 15:35:57 IST
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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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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Mar 2008 15:38:54 IST
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That was a good joke! I have asked for velocity not acceleration.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Mar 2008 15:43:24 IST
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sorry didn't notice the question properly
i'll try to answer it
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Mar 2008 08:24:11 IST
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Your answer sqrt(2)*V is wrong!!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Mar 2008 09:25:41 IST
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then whats the answer
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Mar 2008 07:00:38 IST
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V / sqrt(2)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Mar 2008 08:00:45 IST
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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
~Cheerio!!!
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Talk less work more!! {To be simplistic and 2 gain respect}
Eat less work more!!! {To "build" ur body}
Work less Do more!!! {2 make ur life big}
don't get scared !!!
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