Work is done or not?

1. Jun 24, 2013

I had posted this question before,but did not get any reasonable answer.
A Ball is floating in the space.There are nothing nearby(Or Nothing even far away)It is stationary(Not moving at all)-Meaning it does not have kinetic energy,Or gravitational potential energy.Suddenly,A planet appears by supernatural powers,It exerts force on the ball and the ball falls.(It gains kinetic energy)From where?It did not had any energy before.

2. Jun 24, 2013

Staff: Mentor

When the planet appears from nothing, so does the system's potential energy.

3. Jun 24, 2013

Staff: Mentor

What you have proposed is a completely unrealistic situation, and as such, there is no realistic answer. Objects do not magically appear in reality.

4. Jun 24, 2013

If so,Assume that the a large asteroid is getting closer to it

5. Jun 24, 2013

Staff: Mentor

I'm sorry, I don't quite understand you. Assume a large asteroid also exists in this universe and is approaching the ball?

6. Jun 24, 2013

The ball is alone.Not influenced by gravity or any other force.There is nothing nearby or far away.An asteroid 10000000000 light years away Starts its journey towards the ball.As the asteroid gets closer to the ball,the ball gets kinetic energy.(Because of gravity of the asteroid)From where?

7. Jun 24, 2013

bp_psy

From the potential energy. $U=-G \frac{ m_1 m_2}{r}+C$.

8. Jun 24, 2013

What's that?Energy coming from no where?

9. Jun 24, 2013

bp_psy

No energy coming from the beginning of the universe.

10. Jun 24, 2013

Bandersnatch

In the case of an asteroid closing in from far away, the gravitational potential energy exists at all times, and is the greatest when the asteroid is the farthest(i.e., at infinity). You can't remove the ball from the source of gravity so far away that its gravity dissapears. The farther you go the greater(that is, less negative) its potential energy is, because you need to do more work to put it there in the first place, and the field can do more work on the ball(impart it with more kinetic energy) as it falls.

As far as the appearance of the field goes, I'll repeat Alan Guth's argument from his "Inflationary Universe" to show that the creation of gravitational field releases energy.

Shell theorem shows that from a vantage point outside the shell, the gravitational field looks the same as if all of the mass of the shell were concentrated in its centre. The same theorem shows that inside the shell the gravitational field is exactly zero.

Now, let this shell contract under its own gravity until it reaches radius R2.
The contraction allows for energy to be extracted. To help with visualisation of this process, Guth uses the imagery of electric generators attached via ropes to the shell, so that as it contracts it pulls on the ropes and spins the generators.

After the contraction, the gravitational field as seen from above the original radius(R1) is exactly the same - still looking just as if all the mass were concentrated in the shell's centre.
The gravitational field inside the smaller shell is still zero.
However, in the volume contained between the original(R1) and the resultant(R2) radii of the shell, there is now a gravitational field where before the contraction it wasn't there.

So, then, after the contraction, two things have changed:
1.Energy was extracted from the system.
2.A new gravitational field appeared.

The conclusion is that the appearance of the gravitational field is connected with the release of energy.

11. Jun 24, 2013

Potential energy=mgh
We assume that g is nearly 9.81m/s2 but as you go away from the source,the gravity force weakens.So the g is small and potential energy also small.
Then how can potential energy get so big that you can't even imagine?(As h is 10000000000 ly s How greater will the potential energy be?)How can you say that potential energy is greatest at infinity then?

12. Jun 24, 2013

Bandersnatch

Think of it in terms of work. You need to do work to lift something. You need to do more work to lift it higher. You need to do most work to lift it highest. Highest possible is infinity. Therefore you need to do the most work to lift an object to infinity.

13. Jun 24, 2013

A.T.

No. g is not the potential, just the gradient of it.

14. Jun 24, 2013

I want to know it interms of gravitational potential energy.

15. Jun 24, 2013

For example,I need to do less work on the moon,Because the moon exerts less force.And also the potential energy a ball has on a certain height is smaller compared to earth.Why wouldn't this happen going far away?

16. Jun 24, 2013

Bandersnatch

First thing to understand about the potential energy(U) is that it doesn't make much sense to talk about absolute values of it. You can only talk about U differences with respect to some base point in the gravitational field.
It's the $ΔU$ that is meaningful.

The reason lies in the equation provided by bp_psy in post #7, which is the general equation for gravitational potential energy(mgh can be extracted from it, under certain assumptions, but I'll get to that in a minute). The C in it is the integration constant. It appears when one integrates an equation, and which cannot be disregarded, despite the fact that it's value can be anything you want.

This means that were you try to calculate U at some distance R from the centre of the field, you end up with some value x+C, where only x is calculated and C is whatever you name it. If U can be anything, then how can you say it's something concrete?

However, when you take the difference in U between two points $ΔU=U-U_0$, you end up with the C's dissapearing from the equation:

$ΔU=-G \frac{ M m}{R_1}+C - (-G \frac{ M m}{R_0}+C)=-G \frac{ M m}{R_1}+G \frac{ M m}{R_0}+C-C=-G \frac{ M m}{R_1}+G \frac{ M m}{R_0}$

Now you have a concrete result with no ambiguity that the free variable C causes.

To talk about potential energy at some point R, we must therefore always define from which point R0 is the energy going to be counted.

When you talk about U on the Moon being lower than U on Earth in post #15, you are not using the same base reference, and you're not talking about the same gravitational field.
You should stick to e.g., the Earth's surface, and the Earth's gravity, and compare how much work do you need to do to lift a ball a few metres above ground with the work required to lift the ball from the ground on Earth to the Moon. You can see how the second task requires much more work.
You can do the same with respect to the surface of the Moon and the Moon's gravitational field. Now the ball on Earth has got much higher potential energy w/r to the Moon's gravitational field.
Or you can use the surface of the Moon and the Earth's gravitational field or vice versa.

Sometimes, it's convenient to use certain points as base reference to simplify the calculations. We chose the free variable to have such a value, so as to make $U_0=0$ for some convenient value of R

For example, when we use the surface of the Earth as the base point, and under the assumption that we do not lift/drop anyting very high(h<<R; where R is the radius of the Earth), we get:

$ΔU=U-U_0=U$
$U=-G \frac{ M m}{R+h}+G \frac{ M m}{R}=-G M m(\frac{1}{R+h}-\frac{1}{R})=-G M m(\frac{R}{R(R+h)}-\frac{R+h}{R(R+h)})=-G M m\frac{R-R-h}{R(R+h)}$

With the assumption h<<R, R+h can be treated as just R to a good degree of accuracy.

$U=-G M m\frac{-h}{R^2}$

knowing that the acceleration in the gravitational field $g=G\frac{M}{R^2}$ we get:

$U=mgh$
This is the potential energy at height h above the base point of Eath's surface, where potential energy is 0.

This approximation obviously doesn't work any more if the height is significant as compared to the radius of the Earth.

Another convenient approach is to use as the base point the potential energy at infinity from the source:

$R_0=∞$:
$U=-G \frac{ M m}{R_1}+G \frac{ M m}{R_0}$

and the secon term goes to 0 as R0 goes to infinity. All we're left with is:

$U=-G \frac{ M m}{R_1}$

You can see from the last equation, that the farther you get from the source(higher values of R1), the less negative is U. The greatest value it can have, is zero, when R1 also reaches infinity.

17. Jun 24, 2013

Staff: Mentor

That equation is not accurate for large distances, where g is changing. It assumes a constant g for simplification.

Modifying that equation to include a changing g requires calculus.