# Does This Defy Classic Physics?

1. Aug 31, 2005

### redfish

Maybe i am looking at things wrong, but it seems that classic physics cannot tell me how much gravitational energy an object has at "rest", when it is at the ground potential- at zero.

Example. I have a table. I place a 100 kg object on the table. What is the gravitational energy of the object exerting on the table? The ground potential is the table top, not the ground. Classic mechanics does not seem to provide me an answer. PE = mgh does not work because i get zero. This is not right because it is obvious downward energy is being exerted on the table by the 100 kg weight. PE = mg does not work because if the object is raised above the table any distance less than a meter (such as 0.5 m), it will have a lower energy than the energy calculated at the ground potential- this is nonsensical.

This is bizarre. How come classic physics cannot calculate this? And how do i calculate? Do i use relativity or something? Or am i looking at the problem wrong?

I am stumped. Help, I need an expert! :surprised :surprised

2. Aug 31, 2005

### Andrew Mason

Gravitational potential energy is always relative to another mass/position. There is no such thing as absolute gravitational energy. The gravitational potential energy at an infinite distance from another mass is defined as 0. The gravitational potential energy of a mass at a distance less than infinity from another mass (say the earth) is always less than 0.

$$PE = -\frac{GMm}{R}$$

AM

3. Aug 31, 2005

### Staff: Mentor

Just to add to Andrew's remarks.
Two points:
(1) Gravitational force is what is exerted on the table, not gravitational energy.
(2) The "zero point" for calculating gravitational potential is completely arbitrary; whether you choose the ground or the table top, your force calculations will be the same.

For locations close to the earth's surface, $\Delta {PE} = mgh$ works just fine. You can measure the height from the ground or from the tabletop. All that matters is changes in the gravitational PE.

Raising an object 0.5m always increases the gravitational PE of the system, no matter what you choose as your reference point.

4. Aug 31, 2005

### redfish

Alright, if PE of an object at infinite distance is ZERO, what is it when it rests at the ground potential??? You see, classic physics cannot calculate this. But the resting weight DOES exert downward energy, as common sense tells you.

I want to know an energy figure, versus weight or force figure, so that i can calculate the loading limits of objects. Example, if a wood table can support 200 kg before breaking, i can then figure out what weight dropped at a height will break the table.

If classic physics can only give me a force figure for the resting weight, is it possible to convert this into an equivalent energy figure? Or how would i go about doing this apple and orange comparison?

5. Aug 31, 2005

### inha

Energy is a scalar quantity. It has no direction.

6. Aug 31, 2005

### Staff: Mentor

The PE when on the ground is given by the formula that Andrew provided:
$PE = -\frac{GMm}{R}$
Nonsense. The calculation is trivial.

That's a difficult engineering problem. For a taste, look here: https://www.physicsforums.com/showthread.php?t=80076 or https://www.physicsforums.com/showthread.php?t=63508

Force is not the same as energy.

Last edited by a moderator: Apr 21, 2017
7. Aug 31, 2005

### Chi Meson

It's funny that you have decided what classical physics "can" and "can't" do when it is clear that you have not taken a physics class yet. It's not that "Physics cannot calculte it." Isn't it just that you can't calculate it (yet)?
I'm sure there are many people here who will be glad to help you. Are you ready to be receptive and drop the "physics can't explain it" line?

8. Aug 31, 2005

### Gokul43201

Staff Emeritus
If the PE is defined as zero for points at infinity, then the PE is not zero anywhere else. So, this ill-defined "ground potential" is not a zero potential. In other words, you have described a contradictory (unphysical) situation and are asking for a calculation of some property of this situation.

However, if the ground potential (wherever it exists) were truly a zero potential, then the energy of an object there would truly be zero.

Yes, classical physics can only calculate properties of physical systems, not unphysical ones.

1. You can not "exert" energy.
2. Energy can not be upwards or downwards.
3. Your common sense just made up new meanings for accepted scientific terminology that has well-defined meanings.

This shows a possible root of the misconception. Failure is of two kinds : continuous failure, and impact failure. Impact strengths of materials are in fact quoted in energy (per unit area) units rather than force (or stress) units.

You're looking for the wrong conversion.

For a block that is gently placed on the table, compare the weight of the block with the allowable load for the table to determine what's safe. For a weight dropped from a height, compared the initial PE of the block (with respect to the table-top) with the impact toughness of the table. Of course, the geometry of the block is relevant as well. Failure in the two situations is different and they are tackled differently.

9. Aug 31, 2005

### redfish

Calculation is trivial? I would not be paying homage to the Physics gods on this forum if calculation was so trivial.

I think you folks are missing something here. Each one of these formulas you provided REQUIRE a distance. But when you make the distance 0, which what it would be when an object rests at the ground potential- or when two objects butt against each other- your result will be 0. This is nonsensical.

If i can determine the energy that an object exerts at the ground potential, it should be elementary to determine load handling. Example, if a resting 100 kg object exerts xx joules of energy on a table top, then using simple PE = mgh one can determine how high to drop a smaller object to create the same energy. What is so difficult about that? Well, it seems that classic physics does not allow one to calculate the energy of an object at rest, at the ground potential.

It is absurd that science dictates that an object MUST be at some height in order to calculate its energy. Again, i could be wrong in my thinking. But so far no one has been able to show me how i can calculate the energy of an object at rest using classic physics, which i should be able to do. And i don't see why it would be infinitely more difficult than using a formula such as PE = mgh. Something is missing here.

I need some serious help. This problem is starting to make me question the validity Science!

10. Aug 31, 2005

### Staff: Mentor

One thing you asked for is how to calculate the PE of an object. That is trivial to calculate. But that's not what you really want; you are confusing energy with impact force.

What are you talking about? Again, you fail to distinguish force from PE. If you take the table top as your (arbitrary) reference for calculating changes in PE, then obviously you measure the PE on the table as zero. This makes perfect sense.

This is gibberish. Objects exert forces not energy.
Again, you are confused. Calculating the kinetic energy that a falling object has when it hits a surface is trivial; calculating the impact force is not. (You'll need some empirical data to solve for the latter.)

But the energy depends on the height! You must realize that falling from a height of 10 meters will have different consequences than falling from a height of 1 meter.
Something's missing all right. Now you are asking for the energy (kinetic? potential?) of an object at rest? That's trivial! (But that's not really what you want.)

Why don't you pose the exact question that you want answered--the exact problem you are trying to solve. (Rather than make embarrassing statements about the failure of classical physics, which you apparently have never studied.)

11. Aug 31, 2005

### Staff: Mentor

ECHO, ECHO, echo...

12. Aug 31, 2005

### whozum

Do you know what potential energy is? You answered your own question here.

I want to give you an example of Gokul's proposition which was a great example of a contradiction in your thinking:

Lets say you had a glass coffee table. Whats the most weight you would place on it? Lets say 50kg, it would probalby hold that, right? You can set it there for hours and the glass probably wont buckle. The glass can obviously hold 50kg. By your logic, you are saying that I can drop a 50kg weight on the table and it would be able to hold it. Obviously not true, even a 5kg object would probably crack the glass.

Whats missing is your understanding of physics. We arent here to teach you conepts, we are here to help you get on the right track. People here have given you everything you need to know to understand where you are wrng, but you are not accepting the solution. That is beyond our help, it is your problem now, and you are refusing to work on the fundamental flaw in your understanding of energy and forces.

13. Aug 31, 2005

### Gokul43201

Staff Emeritus
I'm locking this thread. It reeks of trolling. The OP has not heeded any of the corrections pointed out to him/her and insists on repeating them with each new post.