Question

A particle of mass m is released from a height R
(radius of earth) from the surface of earth. When it
reaches the earth's surface, it enters a tunnel
leading to its centre. Its speed at the centre is
[g = acceleration due to gravity on surface of earth]
(1)√gr
(2)√2gr
(3)√3gR
(4)√5gr​

Answer 1

Explanation:

Dear Student,

In this case, we will apply the conservation of energy.

Total energy of the particle at height R above the surface of the Earth is,

Ei=Ki+Ui⇒Ei=Ui=−GMm(R+R) [∵u=0]⇒Ei=−GMm2R .....(i)

Let the particle passes through the centre with velocity v.

At the centre of Earth, total energy of the particle is

Ef=Kf+Uf⇒Ef=12mv2−3GMm2R .....(ii)

Using the conservation energy

Ef=Ei12mv2−3GMm2R =−GMm2R⇒v2=2GMR⇒v=2GMR−−−−√