Question: Gravitational Potential Energy in a Ball-Earth-Moon system?

AI Thread Summary
In the ball-Earth system, the potential energy of the ball decreases as it rises and then increases again when it falls, converting to kinetic energy during the descent. The total energy remains constant, represented by a flat line on a graph, while potential energy peaks and kinetic energy fluctuates. In the ball-Earth-Moon system, the changes in potential energy are minimal due to the Moon's lower gravitational influence compared to Earth. The overall energy dynamics are similar, but the Earth's mass and proximity result in a significantly larger impact on potential energy. Understanding these concepts clarifies the relationship between gravitational forces and energy transformations in these systems.
centauri
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Hello, here's a questions I was wondering if any of you could solve. I don't have the exact numbers, but the scenario is this: a guy standing on the Earth throws a ball upwards and catches it a few seconds later.

How does would affect the potential energy and/or mechanical energy of the ball-Earth system? (Specifically, how might the graph of energy vs. time look like?)

Then, how would this affect the potential energy and/or mechanical energy of the ball-Earth-Moon system?

Any responses would be much appreciated.

Thanks.
 
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Potential energy of the ball will be utilized when it will fall .The energy we gave to ball was stored in form of potential energy converted into kinetic energy at the time of falling.
 
iqra toheed said:
Potential energy of the ball will be utilized when it will fall .The energy we gave to ball was stored in form of potential energy converted into kinetic energy at the time of falling.

So the potential energy of the ball will decrease in the ball-Earth system...but how about the Ball-Earth-Moon system?
 
Please...anyone, this is a concept I'm really struggling with.
 
centauri said:
How does would affect the potential energy and/or mechanical energy of the ball-Earth system? (Specifically, how might the graph of energy vs. time look like?)

I'm not 100% sure if the following is correct, but from the work I've done in my physics 210 class I believe it works as follows.

Ignoring the energy that put the ball into motion in the first place, the graph of the total energy is a flat line, the potential energy rises to a peak before falling back down, and the kinetic energy starts high,drops to zero, then rises again as the ball accelerates downward. Once the person catches the ball the K.E. is zero and the potential energy is back to its original value.

centauri said:
Then, how would this affect the potential energy and/or mechanical energy of the ball-Earth-Moon system?

It would look very similar to the above since the change in potential energy of the ball in the Moon's gravity is much smaller than the change in potential energy due to Earth's gravity. The Earth is both much more massive and much closer than the Moon is, so the change in potential energy as the ball goes up and comes down is much larger for the ball-Earth than for the ball-Moon.
 
Thank you so much! That seems much clearer to me now.
 
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