The Work and Potential Energy of Two Masses at a Distance R

In summary, the problem is to calculate the work needed to bring two masses, m and n, from an infinite distance apart to a distance of R from each other. The potential energy of mass n at a distance r from m is given by U(r) = -Gmn/r. The work done on the system (against gravity) equals the change in the system's energy, which is negative since the applied force opposes gravity. This means that an external force must be applied to bring the masses together. However, the amount of work needed to stop the masses is the same as the amount needed to bring them together slowly with no kinetic energy. This raises the question of whether kinetic energy should be considered in this scenario. Overall, it is
  • #1
e(ho0n3
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Hi everyone,

Here is the problem: Two masses m and n are initially infinitely far appart from each other. Calculate the amount of work to get them a distance R from each other.

Taking mass m as my reference point, the potential energy of mass n a distance r from m is U(r) = -Gmn/r. Now, if only conservative forces are involved -W = U(r') - U(r) where W is the work need to move a mass a distance r' - r as I understand it. So -W = U(R) - 0 => W = -U(R) = Gmn/R. However, the book where I got this problem from has as the solution -Gmn/R. That means either W = U(r') - U(r) or U(r) = Gmn/r, but these would be contradictions.

Confuzzled here,
e(ho0n3
 
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  • #2
work against gravity

You are calculating the work done by gravity, but that's not what the question asks. The work done on this system (against gravity) equals the change in the system's energy. Since the applied force opposes gravity as the masses are brought closer, the work done is negative.
 
  • #3
Doc Al said:
You are calculating the work done by gravity, but that's not what the question asks. The work done on this system (against gravity) equals the change in the system's energy. Since the applied force opposes gravity as the masses are brought closer, the work done is negative.

I seem to be a little confused about the work done by a force on a body and the work done on a body. When I think of work, I think
[tex]W = \int_{p}^{p'}{\vec{F}\cdot\vec{dr}}[/tex]​
I guess I went wrong by assuming that gravity will be doing the work of getting m and n a distance R from each other. But as I understand from your explanation, there must be another force which must be applied for this to happen and that this force is opposing gravity.

I'm still confused though. Gravity is an attractive force, meaning that mass m will attract mass n and vice versa. Why would I apply a force that opposes gravity and impede m and n from getting closer.

e(ho0n3
 
  • #4
e(ho0n3 said:
I seem to be a little confused about the work done by a force on a body and the work done on a body. When I think of work, I think
[tex]W = \int_{p}^{p'}{\vec{F}\cdot\vec{dr}}[/tex]​
Correct, of course.
I guess I went wrong by assuming that gravity will be doing the work of getting m and n a distance R from each other. But as I understand from your explanation, there must be another force which must be applied for this to happen and that this force is opposing gravity.
What if the question asked: What work is done in separating two masses from distance R to infinity? I assume you would not hesitate is stating that positive work is done by some external force, right? Same thing here, only backwards. Somehow the masses are brought together (not just let go): the KE is kept at zero.

I will admit that it's a tricky question. :smile:

I'm still confused though. Gravity is an attractive force, meaning that mass m will attract mass n and vice versa. Why would I apply a force that opposes gravity and impede m and n from getting closer.
Sure, it's a bit of a semantic game. You can just let them fall together. In which case no external work is done, and the energy is constant. Or you can lower them together slowly, via an exernal force against gravity: in this case you've done negative work--and lowered the total energy of the two mass system.

It's slippery stuff.
 
  • #5
Doc Al said:
You can just let them fall together. In which case no external work is done, and the energy is constant. Or you can lower them together slowly, via an exernal force against gravity: in this case you've done negative work--and lowered the total energy of the two mass system.
I'm wondering, and this is to support you, not argue with you:

If you just let them go, you will eventually have to stop them. To do that, you would have to supply a force, infinite if you stop them dead in there tracks (decelerating for v to 0 instantaneously). In other words, isn't it a nontrivial issue that both initial and final states are stationary and therefore KE is not a factor? Of course, even this is an assumption that must be made in order to solve the problem.
 
  • #6
turin said:
If you just let them go, you will eventually have to stop them. To do that, you would have to supply a force, infinite if you stop them dead in there tracks (decelerating for v to 0 instantaneously).
Good observation, as usual. But note that the amount of work needed to stop the masses is the same amount of work needed to bring them together slowly with no KE. (No need to add in the complication of infinite force--assume some non-zero time for deceleration.)

In other words, isn't it a nontrivial issue that both initial and final states are stationary and therefore KE is not a factor? Of course, even this is an assumption that must be made in order to solve the problem.
Right--to even begin to answer this question you need to make some assumptions about KE. That's why I agree that it's a tricky question.
 

What is work?

Work is the measure of force applied over a distance. It is typically represented by the equation W = Fd, where W is work, F is force, and d is distance.

What is potential energy?

Potential energy is the energy stored in an object based on its position or configuration. It is typically represented by the equation PE = mgh, where PE is potential energy, m is mass, g is the acceleration due to gravity, and h is the height of the object.

How are work and potential energy related?

Work and potential energy are directly related, as work is the transfer of energy from one form to another. When work is done on an object, its potential energy can change, and vice versa.

What are some examples of potential energy?

Some examples of potential energy include a stretched spring, a raised weight, a compressed gas, and a stretched rubber band. These objects all possess potential energy due to their position or configuration.

Can potential energy be negative?

Yes, potential energy can be negative. This typically occurs when an object is in a position lower than its reference point, such as a weight that has fallen below its starting position. In this case, the potential energy would be negative, indicating that work was done by the object as it moved downward.

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