Elementary Dynamics (1 Viewer)

[deadpatch]

Suppose a tunnel is constructed through the centre of the Earth along its diameter. A particle is released at a distance of 2R from the Earth's centre, where R is the radius of the Earth. While above the Earth's surface, the particle experiences an acceleration which is proportional to the inverse square of its distance from the Earth's centre. While inside the tunnel, the accln is direction proportional to the distance from the Earth's centre. Assume that air resistance can be negligible. Let g be the accln at the earth's ruface.

If T₁ is the time elapsed to first reach the Earth's surface, and T₂ is the time for the particle to travel from the earth's surface to the earth's centre, find the time taken for the particle to go back to the starting point.

Solutions say its is 4(T₁+T₂)

Can anyone explain?

 

[Trebla]

Basically it falls from 2R above the centre straight through the centre to the other side of the Earth until it is 2R relative to the centre on the other side before it repeats the same motion but in the opposite direction. (note that the acceleration equation under the surface of Earth describes simple harmonic motion)