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a hole is dug along the centre of the earth along the diameter, and a body of mass m is dropped in the tunnel, calculate the velocity of body at the centre of the earth , given that gravitaional potential at the centre of the earth is -3GM/2R, here M is mass of earth .
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7 Answers
1conservation of energy
- GMm/R = - 3GMm/2R + 1/2 mv2
1/2 mv2 = - GMm/R + 3GMm/2R
mv2 = GMm/R
v = √(GM/R)
1@ Ankit
Here, dont we take the mass of earth as Meff=(Mtotal/Vtotal)Veff
I mean the effective mass responsible for attraction reduces as you go deeper[7][7][7]
106yes but gravitational potential at a point = U/m ... So, U=m*V.. that is what ankit has done..
Initial KE=0 And PE = -GMm/R = for Meff
Finally KE = mv2/2 and PE = m*V = m*-3GM/2R.
Equating both... -GMm/R = mv2/2 - 3GMm/2R
==> v2/2 = GM/2R
==> v = √GM/R
1ashish yaar .. u just missed the deletion of 2 from the final answer :)
66It is easy to show that the body will perform SHM with amplitude equal to the radius R of the Earth and time period equal to that of a satellite orbiting very close to the Earth. From Kepler's law the time period is
T=\sqrt{\dfrac{4\pi^2}{GM}\,R^3}
According the angular frequency:
\omega = \dfrac{2\pi}{T}=\sqrt{\dfrac{GM}{R^3}}
Hence, velocity at the center
v = R\omega = \sqrt{\dfrac{GM}{R}}
