Distance of closest approach for two protons

[AStaunton]

Problem is:

given the distance of closest approach in a coulombic fields is:

[tex]d=\frac{1}{4\pi\epsilon_{o}}\frac{Z_{1}Z_{2}}{mv^{2}}[/tex]

calculate the distance for two protons with velocity v=3X10^5 m/s and mass m=1.7X10^-27 kg

Can someone please advise on whether I should plug in the mass of an individual proton into the equation (the mass value I just stated), or should I use what I think is called the "weighted mass" given by:

[tex]\mu_{m}=\frac{m_{1}m_{2}}{m_{1}+m_{2}}[/tex]

where in my problem, m1=m2=mass of proton.

Similarly, what should be the value for velocity v to plug into equation? the velocity given for any of the two protons is what I stated above...but my concern is, should I assume a head on collision, in which case (by classical mechanics at least!) the velocity will be twice as large..to sum up what is the acceptable value for v and m to put into equation and also please advise on why this is the case and what is the standard way to interpret such problems.

Any feedback appreciated.

 

[gneill]

Consider two protons approaching a head-on collision in the center of mass frame of reference (that is, they have equal speeds in opposite directions). Write the expressions for the potential energy (electric) and total kinetic energy of the system.

When the point of closest approach occurs (at distance d), what will the kinetic and potential energies be?

 

[AStaunton]

total energy of system:

[tex](\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{4\pi\epsilon_{0}}\frac{Z_{1}Z_{2}}{r})+(\frac{1}{2}m_{2}v_{2}^{2}+\frac{1}{4\pi\epsilon_{0}}\frac{Z_{1}Z_{2}}{r})=constant[/tex]

as m1=m2, can drop subscript and say m1=m2=m similarly with v and Z. leads to:

[tex](mv^{2}+\frac{1}{2\pi\epsilon_{0}}\frac{Z^{2}}{r})=constant[/tex]

at closest approach kin energy=0 implies:

[tex]\frac{1}{2\pi\epsilon_{0}}\frac{Z^{2}}{r}=constant[/tex]

at infinity, pot energy =0 implies:

[tex]mv^{2}=constant[/tex]

equating gives:

[tex]mv^{2}=\frac{1}{2\pi\epsilon_{0}}\frac{Z^{2}}{r}[/tex]

implies:

[tex]r=\frac{1}{2\pi\epsilon_{0}}\frac{Z^{2}}{mv^{2}}[/tex]

where r=d.

is this the expression to use? ie. greater than the expression given in the problem by a factor of two?
when should [tex]\mu_{m}[\tex] be used?

 

[gneill]

You've counted the electrical potential energy twice.

 

[AStaunton]

ah...I thought It had to be counted once for each particle...
if I count it once, then the expression becomes that given in the question...
And in that case, I assume the mass to plug in should indeed be mass of prot and the velocity should be the velocity of anyone of the protons...
Why is the pot energy not counted twice?

 

[gneill]

There is only one potential energy because it depends upon relationship (charge, location) between the particles.

If you look up the definition for the electrical potential energy, it is equal to the work required to bring the charges from infinity to their final locations. Bringing them one at a time, since there's initially no charge to work against, the "first one's free", so to speak; it takes no work to bring the first particle into place because it doesn't have any external field to move through.

Bringing the second particle in, though, it has to work against (or with, depending upon the signs of the charges) the field of the charge that's already in place. So it's the work required to bring the second particle into place that makes up the potential energy.

 

[AStaunton]

I think I have seen [tex]\mu_{m}[/tex] used for the mass in problems similar to this..when is this valid?

 

[gneill]

I think I have seen [tex]\mu_{m}[/tex] used for the mass in problems similar to this..when is this valid?

Perhaps in solving the equations of motion in a collision, using conservation of momentum.

 

[ehild]

The motion of two interacting particles (point -like bodies) can be broken up to the motion of their centre of mass and and their relative motion with respect to each other. If there is no external force the CM moves with constant velocity along a straight line, and both the position of one particle with respect to the other one and their relative velocity changes with time as if a single particle of effective mass would do under the effect of the interaction between the particles. In your problem , the relative velocity is twice of the absolute one, and the effective mass is half of the real one. So the KE of the "effective particle" is 0.5*(m/2) * (2v)^2 = mv^2, the same as the sum of the KE-s of the individual particles.

ehild

 

[AStaunton]

just checking this equation for two protons...in order to get to a distance of 10^-15 meters aparts, the velocity has to be equal to 2.7X10^14ms-^1...although I know quantum tunneling is the way that they get to such distances, I am still surprised by how high this number is...6 orders of magnitude higher than speed of light...
again...I am not sure if I used the equation correctly..

 

[ehild]

What do you mean? The speed is given as v=3X10^5 m/s .

ehild

 

[AStaunton]

just trying out the equation for something else:
instead of taking it that two protons are going at 3X10^5m/s, I wanted to see what speed they'd have to go at in order to get 1 femto meter apart, so plug in 10^-15 meters for d and solve for v...

 

[ehild]

For such close approach that the protons "touch" each other you got an impossible speed, which means that Classical Mechanics can not be applied.

ehild

 

[AStaunton]

I just have a quick follow up question to this problem...

The equation stated right at the very top is for two positively charged bodies. if however we look at a positive and negative charge, the form of the equation stays the same, but now v is not the velocity needed to come to a certain distance d, rather it is the escape velocity and now d is not the final distance (distance of closest approach) but the starting distance.

Although it is not too surprising that this works out like that considering the only change is that we've gone from repulsive to attractive force, I still don't fully understand how for example the signs all work the same, as now we're dealing with a negative and a positive charge, it feels like there should be an extra negative sign floating about in the equation, is the reason this isn't the case down to the fact that we can choose the 0 point of pot energy arbitrarily?