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

In summary, the conversation discussed the phenomenon of simple harmonic motion (SHM) when a ball is dropped through a hole in the Earth. The question raised was about the conditions for SHM to occur, specifically when the displacement is small and the force only depends on the linear term kx. However, the displacement in this scenario is very large, potentially comparable to the radius of the Earth. The questioner asked for clarification on the higher order terms that may affect the motion.
  • #1
Kolahal Bhattacharya
135
1
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?
 
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  • #2
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.
 
  • #3


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.
 

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

1. What is the displacement of a ball falling through Earth's radius?

The displacement of a ball falling through Earth's radius is the distance that the ball travels from its starting point to its ending point. In this case, the starting point is the surface of the Earth and the ending point is the center of the Earth.

2. How is the displacement of a ball affected by Earth's radius?

The displacement of a ball falling through Earth's radius is affected by the size of the Earth's radius. A larger radius would result in a longer displacement, while a smaller radius would result in a shorter displacement.

3. What is the formula for calculating the displacement of a ball falling through Earth's radius?

The formula for calculating the displacement of a ball falling through Earth's radius is d = (1/2)gt^2, where d is the displacement, g is the acceleration due to gravity, and t is the time it takes for the ball to fall through the Earth's radius.

4. How does the acceleration due to gravity affect the displacement of a ball falling through Earth's radius?

The acceleration due to gravity plays a crucial role in determining the displacement of a ball falling through Earth's radius. The larger the acceleration due to gravity, the faster the ball will fall and the longer the displacement will be.

5. Can the displacement of a ball falling through Earth's radius be influenced by external factors?

Yes, the displacement of a ball falling through Earth's radius can be influenced by external factors such as air resistance and the density of the Earth's layers. These factors can affect the acceleration due to gravity and therefore impact the displacement of the ball.

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