Does a free falling ball drops on Earth violate this theory?

In summary, when considering a free falling ball and the Earth as an isolated system, the ball should continue to drop over time. The total momentum of the 'ball + earth' system remains unchanged, but the ball may not bounce back with the same velocity depending on the nature of the collision. The change in momentum of the Earth is not negligible and should be taken into account. Momentum and energy conservation will dictate the behavior of the ball and Earth system.
  • #1
rajivrout
2
0
Does a free falling ball drops on Earth violate this theory?

If we consider the ball and Earth in an isolated system then the ball should continue to drop over time.

If the impulse given is extremely negligible that it doesn't cause any significant change in the path of earth, then the whole momentum must be conserved by the ball and it should bounce back with the same velocity.

Which is the missing point?
 
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  • #2
The total momentum of the 'ball + earth' system remains unchanged. Whether the ball bounces back with the same speed depends on the nature of the collision, not on momentum conservation. Regardless of the speed of the bouncing ball, the total momentum of the system remains unchanged.
 
  • #3
The change of momentum of the Earth isn't exactly negligible. It is exactly the same change in momentum as for the ball. However, you are right that the change in momentum of the Earth corresponds to a negligible change in velocity of the Earth due to its large mass.
Usually a ball won't bounce back with the full velocity, since it will transfer momentum to particles in the Earth (which will scatter in all directions). This could create a tiny shock wave in the earth, but again due to the spread it is not detectable.
 
  • #4
@pf mentor
My point is m(e)*v(e)+m(b)*v(b) must be unchanged after colission. Since m(e)*v(e) is equal to zero and there is no (negligible) change in m(b), the final velocity before colission must be equal to the initial velocity after colission. If the initial velocity while bouncing is same as the final velocity at the time of drop then there can not be any change in displacement. Yes if the material is more elastic then it will help the object to restore the kinectic energy and it will bounce back better, but let's talk abt only momentum.
@gerenuk
Do you think the momentum transferred to the particles should be considered? Why can't we just consider these simply as two objects? Let's forget Earth consider a big stone.
 
  • #5
rajivrout said:
@gerenuk
Do you think the momentum transferred to the particles should be considered? Why can't we just consider these simply as two objects? Let's forget Earth consider a big stone.
I'm not sure what you want to point out. The Earth is a big object only?
Maybe I make another comment. Momentum and energy is always perfectly conserved if only you take all objects into account.

You are right if you mean that, provided the Earth does not absorb any energy and does not deform, then energy and momentum conservation will dictate that the ball will have the same magnitude of velocity when it reaches a point that it was at before.
 
  • #6
rajivrout said:
@pf mentor
My point is m(e)*v(e)+m(b)*v(b) must be unchanged after colission.
That's true.
Since m(e)*v(e) is equal to zero and there is no (negligible) change in m(b), the final velocity before colission must be equal to the initial velocity after colission.
Not true. Don't confuse having the same speed with having the same velocity. If the ball bounces back with the same speed but in the opposite direction, its momentum is not the same.
If the initial velocity while bouncing is same as the final velocity at the time of drop then there can not be any change in displacement. Yes if the material is more elastic then it will help the object to restore the kinectic energy and it will bounce back better, but let's talk abt only momentum.
Let's work out the details:
Initial velocity of ball: - v(b) (it's moving towards the earth)
Initial momentum of ball: -m(b)v(b)
Initial velocity of earth: 0 (assume it's zero)
Initial momentum of earth: 0

Final velocity of ball: + v(b) (assume it bounces back with same speed, so now it's moving away from earth)
Final momentum of ball: +m(b)v(b)

Change in momentum of ball = final - initial = +m(b)v(b) - -m(b)v(b) = +2m(b)v(b)

Thus momentum conservation allows us to conclude that the final momentum of the Earth is now -2m(b)v(b). The Earth is now moving away from the ball. The speed of the Earth might be tiny, but you cannot neglect its momentum.
 

1. Does a free falling ball on Earth violate the theory of gravity?

No, a free falling ball on Earth does not violate the theory of gravity. In fact, it is a perfect example of how gravity works. The ball is falling towards the center of the Earth due to the force of gravity.

2. Why does the ball fall towards the Earth instead of floating in space?

The ball falls towards the Earth because of the force of gravity. Gravity is a natural force that exists between all objects with mass. The larger the mass of an object, the stronger its gravitational pull.

3. How does the theory of gravity explain a free falling object?

The theory of gravity explains that the force of gravity between two objects depends on their masses and the distance between them. When an object is free falling, it is being pulled towards the center of the Earth by the force of gravity.

4. Does the mass of the ball affect its rate of free fall?

Yes, the mass of the ball does affect its rate of free fall. According to the theory of gravity, objects with larger masses have a stronger gravitational pull and therefore fall faster than objects with smaller masses.

5. Can a free falling ball ever reach a constant velocity?

Yes, a free falling ball can reach a constant velocity once it reaches terminal velocity. Terminal velocity is the maximum speed that an object can reach when falling through a medium, such as air. At this point, the force of air resistance is equal to the force of gravity, resulting in a constant velocity.

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