# Gravitational Potential Energy of a bookbag

1. Jan 14, 2009

### Charlie G

This has troubled me ever since we went over gravity in science. I didnt ask the teacher becuase i thought the answer might be obvious and I might look like an idiot in front of the class lol.

But anyway, I was wondering where the energy comes from that an object has when it is in a gravitational field. Where does the energy come from that my bookbag has when I lift it from the floor. I thought its the same amount of energy I transferred to the bookbag when I lifted it, but what then about the case of a meteor or some other extra-terrestrial object?

2. Jan 14, 2009

### matt_crouch

basically when you lift the book bag up you are said to be doing "work" you are moving a force over a distance so to do this energy is required

(and yes this energy is the same energy that an object will have at that specific height)

so your book bag say it has a mass of 1 kg and is raised 2 meters off the ground at two meters the book bag will have an energy of 19.62 joules this is the same amount of energy required to raise it to that height (assuming your body is completely efficient which it isnt)

this energy comes from your muscles and your muscles get the energy from food. a meteor gains this potential energy as it enters a gravitational field, the strength of this field changes over distance.

as the meteor gets closer to the center of mass (center of earth in this example) the force pulling the meteor increases. as the object falls its potential energy gets converted into kinetic energy.. =]

3. Jan 14, 2009

### Charlie G

Oh ok thx for the help, Matt, that question was bugging me for a while.

4. Jan 14, 2009

### Magical

Keep raising the mass to the point in space between the Sun and Earth where the gravitational fields are in balance and there is no force in either direction.

Now it took work to raise the mass to this point, however it does not possess any potential energy at this point, as far as I can determine.

Where did the 'Energy' required to do the 'Work' go ?

It seems to have disappeared somewhere.

5. Jan 14, 2009

### Staff: Mentor

Just because there's no force does not mean there's no potential energy--it just means that the potential is a local minimum at that point.
Nope, it's there.

6. Jan 14, 2009

### rcgldr

The forces would be at a minimum, but not the potentials, they remain the same. Note that the mechanical work done is corresponds to the net change in the total GPEs (combined earth and sun GPE in this case). In the area where the forces are small, the change in the total GPE versus position is small.

For an analogy, imagine an object in a tunnel at the center of the moon. Even though the net force on an object at the center of the moon is zero, the GPE there is larger (more negative) than at the surface of the moon, and the escape velocity at the center of the moon would be greater than the escape velocity at the surface of the moon because of the difference in GPE.

Last edited: Jan 14, 2009
7. Jan 14, 2009

### Staff: Mentor

I meant to say that the potential is a local maximum, not minimum. (The main thing is that F = - dU/dx.)

8. Jan 14, 2009

### rcgldr

The confusing thing is GPE is negative or zero. Is it really a local maximum though? What if the object is moved in a direction that the forces of gravity between earth and sun perpendicular to that path remain equal? Seems to me that the work done by moving the object could be negative or positive depending on the direction the object is moved. In my example of the net force being zero at the center of the moon, the forces are zero when the GPE is at a minimum (most negative), so there doesn't seem to be a relationship between force and total GPE, just a relationship between force and change in GPE versus distance in particular direction.

As pointed out, the main thing is that F = -dU/dx. Work = change in GPE = force x distance, so the ratio of change in -GPE / distance (in a particular direction) = force.

Last edited: Jan 14, 2009
9. Jan 15, 2009

### Staff: Mentor

Good point. The initial point I was making is that the force being zero does not imply that GPE is zero or any other value (as you certainly know). If you just consider motion along the line joining earth and sun, the GPE as a function of position will be a maximum at the point where the net force is zero. It's not always a maximum (or minimum), just in the case under discussion. If you move the object only along a line perpendicular to the line between them, the potential will be a local minimum where the force is zero. What's always true is that the gradient of the potential is zero wherever the force is zero.

10. Jan 15, 2009

### rcgldr

Makes sense now. A local maximum at that point if the direction is on the line between the centers of the earth and sun, a local minimum at that point if the direction is perpendicular. It's direction sensitive.

11. Jan 15, 2009

### skeptic2

So for an object free falling from infinity to a black hole, even though it's velocity may become relativistic, it doesn't gain mass?

12. Jan 15, 2009

### matt_crouch

surely it would gain mass? as the velocity increases as does its energy so due to mass energy conservation it would increase?

13. Jan 15, 2009

### rcgldr

14. Jan 15, 2009

### skeptic2

1. By falling it is only converting the potential energy it already has to kinetic energy.
2. Even though it is being accelerated by gravity, to its own inertial reference frame it is at rest.
3. If it is gaining mass (energy) where is that mass coming from? Certainly not from the black hole.
4. If it gains mass as it falls, might it at some point have more mass than the black hole it is falling towards?

15. Jan 17, 2009

### stevebd1

A couple of equation relating to energy in Schwarzschild geometry-

'According to the law of constant E/m, an object has constant energy during its entire trip, whether it is far from the black hole or close to it. But when at rest far from the black hole, this stone has energy identical to its mass, or E/m=1. From the equation below, therefore, for every radius r on the inward plunge we have-

$$\frac{E}{m}=\left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}=1$$

The total energy of an object held at rest on a shell of radius ro is given by the expression-

$$\frac{E}{m}=\left(1-\frac{2M}{r_o}\right)^{1/2}$$

The shell observer stationed at ro uses special relativity to calculate the local energy of the passing object-

$$\frac{E_{shell}}{m}=(1-v_{shell}^2)^{-1/2}=\left(1-\frac{2M}{r}\right)^{-1/2}$$

So it appears the in-falling object has E=1, the stationary observer in the shell has E<1 (which means he needs power to remain still) who observes the in-falling object to have E>1 (I always think of remaining stationary in curved spacetime as being synonymous with traveling at relativistic speeds in flat space).

Source-
http://exploringblackholes.com/

Steve