Exploring the Displacement of a Ball Falling Through Earth's Radius

[Kolahal Bhattacharya]

We all know that a ball when dropped into a straight hole through the earth, it begins shm.The question I want to raise here is we know that SHM happens when displacement is very small so that F(x) depends only on the linear term kx and not on the higher order terms:x62, x63 etc.Here the displacement is very big, may well be compared to The Earth's radius.where I am going wrong?

 

[Doc Al]

What higher order terms are you talking about? In the usual "ball dropped through a hole through the Earth" problem, we assume a uniform density and small hole, so that F(x) = kx throughout the motion.

 

[kyle8921]

Thank you for your question. It is true that SHM (simple harmonic motion) occurs when the displacement is small, as the restoring force is directly proportional to the displacement. However, in the scenario of a ball falling through the Earth's radius, we must take into account the changing gravitational force as the ball moves closer to the center of the Earth.

As the ball falls, the distance from the center of the Earth decreases, resulting in a stronger gravitational force and a larger displacement. This means that the restoring force is not solely dependent on the linear term kx, but also on the gravitational force, which is a non-linear term.

Therefore, SHM is not applicable in this situation, as the displacement is not small and the restoring force is not solely dependent on the linear term. Instead, we would need to use the equations of motion to accurately describe the ball's motion as it falls through the Earth's radius.

I hope this helps clarify any confusion. Please let me know if you have any further questions.