# Kinetic energy and potential energy relationship

1. Nov 11, 2015

### gracy

I know,kinetic energy=$\frac{1}{2}$$mv^2$
But I saw one other equation it is
kinetic energy=$\frac{1}{2}$$|P.E|$
Then I started looking for in which specific condition is it true?Because I know it is not always applicable .But I could not find that particular condition when it is true.So my question is kinetic energy=$\frac{1}{2}$$|P.E|$ when and how?Please give me hints.
Thanks!

2. Nov 11, 2015

### Chandra Prayaga

You will have to be more complete in your description. What is your system, is it a particle, a point charge, an electric dipole, etc. Then you should specify what the quantities are in your expression. What is P, what is E, are they vectors, scalars, etc.

3. Nov 11, 2015

### UncertaintyAjay

Where did you see the equation

4. Nov 11, 2015

### nasu

See "Virial theorem" for a general answer and the conditions necessary for this to apply.

5. Nov 11, 2015

### haruspex

Seems that "P.E" is supposed to be "P.E.", i.e. potential energy.
For an orbiting body mass m in the gravitational field of a body mass M, P.E. is taken to be 0 at infinity, so the P.E. at radius r is negative:
$P.E.=-\frac{GMm}r$.
The speed, v, must be such that the centripetal force keeps it in orbit:
$\frac {mv^2}r = \frac{GMm}{r^2}$
The K.E. is $\frac 12 mv^2$.
Combining these gives K.E. = -P.E./2.

6. Nov 11, 2015

### gracy

But I am leaning electrostatic field and potential nowadays.I saw this equation there only.But yes,it does talk about gravitational force .The condition described there is that "the centripetal force is provided by the electrostatic force of attraction".

7. Nov 11, 2015

### gracy

So no other cases in which this
K.E. = |P.E./2|.
is true.

8. Nov 11, 2015

### haruspex

No doubt other cases could be constructed, but that is the one that comes to mind.
It would extend at least to orbits driven by any inverse square law of attraction, so would include a negative charge orbiting a positive one.
Note also that it only applies to bodies in circular orbits. Otherwise the two energies are varying with a constant sum, which would clearly break this relationship.

9. Nov 11, 2015

### Chandra Prayaga

haruspex above is quite right. Also, the same result holds for electrostatics too, which is where gracy says she has seen it. The same result that holds for the gravitational force will also hold for the electrostatic force, for example, in the classical model of a hydrogen atom, where the electron orbits around the proton. Here too, the potential energy is - Ke2 /r, where K = 1/4πε0 and the kinetic energy will similarly work out to be K.E = - P.E/2. The algebra in both cases, gravitational and electrostatic, is identical.

10. Nov 11, 2015

### gracy

and what about inverse square law of repulsion?

11. Nov 11, 2015

### haruspex

How would that lead to a stable orbit?

12. Nov 11, 2015

### gracy

So ,basically there should be a body in circular orbit around the other body (so two bodies)driven by any inverse square law of attraction
These are the two conditions for this rule to follow
K.E. = |P.E./2|.
One more thing I would like to ask what if there are more than two bodies involved?

13. Nov 11, 2015

### haruspex

How will you define P.E. of a given body in this case? It will depend on the positions of the other two.

14. Nov 12, 2015

### gracy

Did you answer the question yourself?Or you were only giving me a hint.

15. Nov 12, 2015

### haruspex

I'm saying I don't see how you could say anything equivalent to that equation when there are three bodies.
The 'three body problem' is a classic in mechanics. There is no general analytic solution.

16. Nov 12, 2015

### gracy

Ok.Actually I don't have in depth knowledge in this.My teacher says you don't need to go in much detail.So,you are saying that equation is not applicable in case of more than 2 bodies.

17. Nov 12, 2015

### haruspex

Yes.

18. Nov 12, 2015

### nasu

In the case of a system of many bodies a more general relationship exists, however.
If the system is stable and bound by potential forces (like gravity or electrostatic), the average over time of the total KE and PE satisfy the same relationship.
$KE_{ave}=\frac{1}{2} PE_{ave}$
Even for potentials with other dependence of distance (not 1/r but 1/r^n) there is a similar relationship, but the fraction is some other number and not 1/2.
If you are interested in more details you can see
https://en.wikipedia.org/wiki/Virial_theorem
or other sites about virial theorem.

You don't need for high school level problems, though.
But it may help to see a little of the bigger picture.