How close can two protons get if....

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PAK108
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Mentor note: Thread got moved to the homework section

How close can two protons get if one is at rest and the other has a kinetic energy equal to the average energy at T =107 K?

I know that the kinetic energy of the moving proton is 3/2kT, but what is the kinetic energy of the proton
at rest? This question is from my book in astronomy. The answer are supposed to be r= r = 1.1 × 10−12m

Any help appreciated
 
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PAK108 said:
but what is the kinetic energy of the proton
at rest?
Zero. What else?

At the moment of closest approach, both protons will move and have non-zero kinetic energy.

Is this homework?
 
PAK108 said:
I know that the kinetic energy of the moving proton is 3/2kT, but what is the kinetic energy of the proton
at rest?
In your opinion what should be kinetic energy of a body which is not in motion?
the other protons kinetic energy is given so by virtue of its energy it can approach the other proton -in spite of repulsion from it so you must calculate the net work done by it in getting closer...
 
mfb said:
Zero. What else?

At the moment of closest approach, both protons will move and have non-zero kinetic energy.

Is this homework?

Aha, I think I understand now, that means that all the kinetic energy becomes potential energy when the electron
are at the closest approach...thank you :)

Yes, this is homework (I realized that this should be posted at the homework section after posting. I am sorry for posting here)
 
PAK108 said:
Aha, I think I understand now, that means that all the kinetic energy becomes potential energy when the electron
are at the closest approach...thank you :)
No it does not.
For that to occur, both would have to stop at the same time, which violates momentum conservation.

I moved the thread to the homework section.
 
It's not clear from the problem statement that the "at rest" proton is meant to be able to move, or if it is to remain at rest (fixed in place). Is the problem statement complete and exactly as it was given?
 
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gneill said:
It's not clear from the problem statement that the "at rest" proton is meant to be able to move, or if it is to remain at rest (fixed in place). Is the problem statement complete and exactly as it was given?

Yes this is the complete problem statement

I tried to solve the problem like this:

3/2kT + mc2=k(e2/r2), solving for r gave r=1.2*10-9, which is like 1000 order bigger than the correct answer :(
 
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PAK108 said:
mc2
what this energy term ( equivalent to mass) is doing there -is the problem talks about velocities comparable to velocity of light
 
mfb said:
That does not even have matching units, and I don't understand where you got that equation from.

The equation was just put together from the equation for "thermal energy and rest energy" = "potential energy"

the k's on the left hand side is not the same on the right hand side
LHS k=1,38065*10-23 J/K
RHS k=8,99*109 Nm2/C2
 
drvrm said:
what this energy term ( equivalent to mass) is doing there -is the problem talks about velocities comparable to velocity of light

No, the problem does not talk about velocities comparable to light...basically I was just trying something since the units matched...the chapter mentions the rest energy of a proton (E=mc2) so I just gave it a shot
 
PAK108 said:
The equation was just put together from the equation for "thermal energy and rest energy" = "potential energy"
Why should that be true? Do you annihilate a particle to extract its rest energy? Even if that would be true, why one and not both?

PAK108 said:
the k's on the left hand side is not the same on the right hand side
That is clear and no problem. The potential energy still does not follow a 1/r2 law.
 
PAK108 said:
o, the problem does not talk about velocities comparable to light...basically I was just trying something since the units matched...
well you can not do these things as physics theory does not mixes Newtonian and sTR -it will be a mess.
 
mfb said:
The potential energy still does not follow a 1/r2 law.
Ah of course, it is supposed to be 1/r, that means that the calculations are completely off