Time of approach of two oppositely charged particle

In summary: Please help. The kinetic energy of the system is equal to the kinetic energy of the charge plus the kinetic energy of all the particles relative to the charge. In summary, the two point charges will collide in seconds if they are at a distance of 1 meter. The charge at the left is negative so the collision will happen to the right. The velocity of the particles will increase as the distance between the charges decreases.
  • #36
There is a hint given below with the question.
It is ----"Do it without integrating using Kepler's Law."
But Kepler's law are applicable to the planet moving in elliptical orbits.
Can it be applied here?
 
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  • #37
Yes, it can be. The electric field is also a central force, like gravity.
The two-body problem can be reduced by a single-body problem with the reduced mass method. The problem is equivalent if we consider a virtual body of reduced mass (μ=1/(1/m1 + 1/m2) about the centre of the force, where the distance from the centre is equal to the distance between the bodies in the original problem.
The straight line can be considered as a very elongated ellipse. At the limit, the focal point is at the opposite end of the distorted ellipse.
Kepler's third law states that the time period depends only on the average distance from the central body, on the semi-mayor axis.
You can determine what would be the period of a circular orbit with the radius as the semi-mayor axis.

ehild
 

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  • #38
An old thread for objects accelerating towards each other due to gravity, showing Kepler method (which still used integration) and method based on integrating acceleration to get velocity (rate of closure), then integrating velocity to get position.

https://www.physicsforums.com/showthread.php?t=635188
 
  • #39
Just before the collision their acceleration is infinite.Are their velocities also infinite just before the collision?
 
  • #40
If the charges on the particles would have been different then there would not be much problem I think.
But what would happen if their masses were different.I think in this, they will not meet at the center of path.I think they will meet at the CM of the system.But the motion of the particles will not be symmetrical in this case.How this will be solved?
 
  • #41
Satvik Pandey said:
Just before the collision their acceleration is infinite.Are their velocities also infinite just before the collision?

Yes. The total energy is finite, the potential energy is negative infinite at collision so the E must be positive infinite.

ehild
 
  • #42
You calculated with the distance of the particles previously. You can work with that distance again.

Chosse a system of coordinates with the CM as origin. Let be x1 and x2 the coordinates of the particles. Then m1x1+m2x2=0 x1=-x2m2/(m1)

Choose a new variable, r=x2-x1. Then x1=-rm2/(m1+m2) and x2=rm1/(m1+m2)
The velocities are
v1=-dr/dt m2/(m1+m2)
v2= dr/dt m1/(m1+m2)

The kinetic energy is KE=0.5 μ (dr/dt)2 where μ=m1m2/(m1+m2). μ is called the reduced mass of the particles.

The potential energy is PE=-q1q2/r. Initially the distance is R and the particles are in rest.

Write up the equation for conservation of energy: 0.5 μ (dr/dt)2 - q1q2/r=-q1q2/R and solve with the same method you have applied for the OP.

ehild
 
  • #43
ehild said:
The problem is equivalent if we consider a virtual body of reduced mass (μ=1/(1/m1 + 1/m2) about the centre of the force, where the distance from the centre is equal to the distance between the bodies in the original problem.

There are three terms involved Center of Mass , Center of force and Central body (M+m) . Isn't the Central body also the Center of force ? What are the roles of Center of Mass and Central body in the reduced mass approach ?

ehild said:
The straight line can be considered as a very elongated ellipse. At the limit, the focal point is at the opposite end of the distorted ellipse.

Could you explain it a bit more ? I am not very clear with this idea of masses orbiting in straight line .

Thanks
 
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  • #44
ehild said:
Yes. The total energy is finite, the potential energy is negative infinite at collision so the E must be positive infinite.

ehild
As the particles starts coming towards each other their potential begins to increase(E[itex]_{p}[/itex]=kq^2/r so if r decreases E[itex]_{p}[/itex] increases) and their kinetic energy also increases (because acceleration increases hence velocity also increases).Doesn't it violates law the of conservation of energy?
 
  • #45
ehild said:
Yes, it can be. The electric field is also a central force, like gravity.
The two-body problem can be reduced by a single-body problem with the reduced mass method. The problem is equivalent if we consider a virtual body of reduced mass (μ=1/(1/m1 + 1/m2) about the centre of the force where the distance from the centre is equal to the distance between the bodies in the original problem
ehild

What is centre of force?

ehild said:
where the distance from the centre is equal to the distance between the bodies in the original problem.
ehild
I don't understand what you trying to say from this line.
 
  • #46
Tanya Sharma said:
There are three terms involved Center of Mass , Center of force and Central body (M+m) . Isn't the Central body also the Center of force ? What are the roles of Center of Mass and Central body in the reduced mass approach ?

In case when the force of interaction is the Coulomb force, there is no "central mass".

Tanya Sharma said:
Could you explain it a bit more ? I am not very clear with this idea of masses orbiting in straight line .

Thanks

A straight line is just a very - very elongated ellipse. At the limit, it becomes a section of a straight line. The virtual body is at one end of the section initially, and the focus is on the other end.

ehild
 
  • #47
Satvik Pandey said:
As the particles starts coming towards each other their potential begins to increase(E[itex]_{p}[/itex]=kq^2/r so if r decreases E[itex]_{p}[/itex] increases) and their kinetic energy also increases (because acceleration increases hence velocity also increases).Doesn't it violates law the of conservation of energy?

The potential energy is negative. Ep = - kq2/r . It decreases if r decreases.

ehild
 
  • #48
Satvik Pandey said:
What is centre of force?


I don't understand what you trying to say from this line.

Both the Coulomb force and gravity are central forces. I used the term "centre of force" in the frames of the reduced mass method. It relates the original problem of a single particle orbiting around a centre. The force points towards that centre and its magnitude is given, kq2/r2 this time. The distance from the centre of that virtual particle is the distance between the real particles.

Browse "two-body problem" and "reduced mass".

ehild
 
  • #49
I searched and found that central forces depends on the distance between the particles and they are directed along the line joining them.
I found reduce mass of two particles system is m1*m2/m1+m2 and this is an imaginary particle.But what is the position of this imaginary particle and how this will help me in solving problem which I have posted in post#40.
 
  • #50
Satvik Pandey said:
But what is the position of this imaginary particle and how this will help me in solving problem which I have posted in post#40.

The reduced mass would be at the same distance as between the two charged particles in the original problem.Initially it is at a distance of 1 m .

ehild has given you pretty much everything in post#42 to solve the problem.

Write up the equation for conservation of energy: 0.5 μ (dr/dt)2 - q1q2/r=-q1q2/R and solve with the same method you have applied for the OP.

Here 'r' is the instantaneous separation .It will be solved on similar lines to what you have done earlier.
 
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  • #51
Tanya Sharma said:
The reduced mass would be at the same distance as between the two charged particles in the original problem.Initially it is at a distance of 1 m .

What I have I understood till here is here----
The two-body problem can be reduced by a single-body problem with the reduced mass method.In this we assume that there is a 'reduce mass'(m1m1/m1+m1) instead of two body.Am I right till here.
You said that the reduced mass would be at the same distance as between the two charged particles in the original problem.But when we measure distance we need to specify an origin from where it is measured.This 'distance' in this case is measured from which origin.
It would be nice if you could explain it with a figure.
View attachment 71218
In this figure consider the mass of q1 as m1 and q2 as m2.Let m1 <m2. Could you please represent the location of reduced mass.This a new concept for me therefore I am facing some problem in understanding it.
 
  • #52
Satvik Pandey said:
What I have I understood till here is here----
The two-body problem can be reduced by a single-body problem with the reduced mass method.In this we assume that there is a 'reduce mass'(m1m1/m1+m1) instead of two body.Am I right till here.
You said that the reduced mass would be at the same distance as between the two charged particles in the original problem.But when we measure distance we need to specify an origin from where it is measured.This 'distance' in this case is measured from which origin.

The origin can be anywhere .For example consider charge q1 at (1,0) and charge q2 is at (2,0) or charge q1 at (4,0) and charge q2 is at (5,0) . The thing that matters is their relative separation.

Note carefully that 'r' in the above equation represents the relative distance between the charges i.e x2-x1.Initially r= 1m but decreases gradually . If you think in terms of hypothetical reduced mass μ ,it is at a distance 'r' from the center of force (some hypothetical source exerting force kq2/r2),you may consider the center of force to be at the origin and the reduced mass to be at (1,0) initially .It is as if some mass μ is under the influence of an attractive force kq2/r2 just as Earth is under the influence of gravitational force by sun given by GMm/r2 .

Read the below post made by ehild in an earlier thread .

Here is the derivation how a two-body problem can be reduced to one body-problem. I write it in one dimension,but the derivation is the same for 3D.

There are two point masses, m1, m2. Their coordinates are x1 and x2. m1 exerts force f12 on m2 and m2 exerts force f21 on m1. Assume the forces act along the line connecting the masses.

f12=-f21=F

[itex]m_1\ddot x_1=-F[/itex].....(1)
[itex]m_2\ddot x_2=F[/itex].....(2)

Add the equations:
[itex]m_1\ddot x_1+m_2\ddot x_2=0[/itex]....(3)

The coordinate of the centre of mass is X
[itex]X=\frac{m_1 x_1+m_2 x_2}{m_1 +m_2}[/itex]
so you can write equation (3) as [itex](m_1+m_2)\ddot X=0[/itex]

The centre of mass moves with constant velocity. You can choose the frame of reference fixed to the CM: Then X=0,

[itex]m_1x_1+m_2x_2=0[/itex].

Divide equation (1) with m1, equation(2) with m2 and subtract them.

[itex]\ddot x_1=-\frac{F}{m_1}[/itex]
[itex]\ddot x_2=\frac{F}{m_2}[/itex]
[itex]\ddot x_2-\ddot x_1=\frac{F}{m_2}+\frac{F}{m_1}[/itex]

Denote x2 -x1 = x
and
[itex]\frac{1}{μ}=\frac{1}{m_1}+\frac{1}{m_2}[/itex]

You get the equation for the relative coordinate x=x2-x1 in terms of the reduced mass μ and the force of interaction F

[tex]μ \ddot x= F[/tex], the same, as in case of a single body of mass μ.
 
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  • #53
Tanya Sharma said:
The origin can be anywhere .For example consider charge q1 at (1,0) and charge q2 is at (2,0) or charge q1 at (4,0) and charge q2 is at (5,0) . The thing that matters is their relative separation.

Note carefully that 'r' in the above equation represents the relative distance between the charges i.e x2-x1.Initially r= 1m but decreases gradually . If you think in terms of hypothetical reduced mass μ ,it is at a distance 'r' from the center of force (some hypothetical source exerting force kq2/r2),you may consider the center of force to be at the origin and the reduced mass to be at (1,0) initially .It is as if some mass μ is under the influence of an attractive force kq2/r2 just as Earth is under the influence of gravitational force by sun given by GMm/r2 .

Read the below post made by ehild in an earlier thread .
Thank you for the explanation
.When two body problems are reduced to one body problem, in this,is center of force also hypothetical?
 
  • #54
ehild said:
You calculated with the distance of the particles previously. You can work with that distance again.

Chosse a system of coordinates with the CM as origin. Let be x1 and x2 the coordinates of the particles. Then m1x1+m2x2=0 x1=-x2m2/(m1)

Choose a new variable, r=x2-x1. Then x1=-rm2/(m1+m2) and x2=rm1/(m1+m2)
The velocities are
v1=-dr/dt m2/(m1+m2)...(1)
v2= dr/dt m1/(m1+m2)

The kinetic energy is KE=0.5 μ (dr/dt)2 where μ=m1m2/(m1+m2). μ is called the reduced mass of the particles.

The potential energy is PE=-q1q2/r. Initially the distance is R and the particles are in rest.

Write up the equation for conservation of energy: 0.5 μ (dr/dt)2 - q1q2/r=-q1q2/R ...(2)and solve with the same method you have applied for the OP.

ehild
From eq(1) I put the value of dr/dt in eq (2)
0.5m1*m2/m1+m2 * v1^2(m1+m2)^2/m2^2=q1q2(1/R+1/r)
If I write V1 as dx/dt and transpose other element in LHS to RHS.And then after taking sq.root on both side if I integrate LHS within limits x1 to 0 and RHS within 0 to T will I get the answer.
 
  • #55
I do not see what you want to do.ehild
 
  • #56
ehild said:
I do not see what you want to do.


ehild

I just want to find time period in which m1 and m2 will converge at a point.
 
  • #57
Instead of x1 and x2, use the reduced mass method and solve the problem for r(t), the distance between the particles.

From conservation of energy:
0.5 μ (dr/dt)2 -k q1q2/r=-kq1q2/R. r is the distance between the particles, R is the initial distance, R=1 m and μ is the reduced mass.

[tex]\frac{dr}{dt} = -\sqrt {\frac{2q_1q_2 k}{ μ }\left(\frac{1}{r} -\frac{1}{R}\right)}[/tex]

Integrate as before, r goes from R to 0.

ehild
 
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  • #58
ehild said:
Instead of x1 and x2, use the reduced mass method and solve the problem for r(t), the distance between the particles.

From conservation of energy:
0.5 μ (dr/dt)2 -k q1q2/r=-kq1q2/R. r is the distance between the particles, R is the initial distance, R=1 m and μ is the reduced mass.

[tex]\frac{dr}{dt} = -\sqrt {\frac{2q_1q_2 k}{ μ }\left(\frac{1}{r} -\frac{1}{R}\right)}[/tex]

Integrate as before, r goes from R to 0.

ehild

Thank You
 
  • #59
ehild said:
Instead of x1 and x2, use the reduced mass method and solve the problem for r(t), the distance between the particles.

From conservation of energy:
0.5 μ (dr/dt)2 -k q1q2/r=-kq1q2/R. r is the distance between the particles, R is the initial distance, R=1 m and μ is the reduced mass.

[tex]\frac{dr}{dt} = -\sqrt {\frac{2q_1q_2 k}{ μ }\left(\frac{1}{r} -\frac{1}{R}\right)}[/tex]

Integrate as before, r goes from R to 0.

ehild

Thank you ehild.This my hopefully last query in this question.In reducing two body problem into single body problem reduced mass is located at a distance from the center of force which is equal to distance of separation between two bodies.But what is the magnitude and direction of force which acts on the reduced mass.Is magnitude of force acting on reduced mass is equal to the force acting on any individual particle.And if the direction of force on particle is towards the CM of system then the direction of force on reduced mass is towards the center of force.Am I right.
 
  • #60
Satvik Pandey said:
Thank you ehild.This my hopefully last query in this question.In reducing two body problem into single body problem reduced mass is located at a distance from the center of force which is equal to distance of separation between two bodies.
Yes.

Satvik Pandey said:
But what is the magnitude and direction of force which acts on the reduced mass.Is magnitude of force acting on reduced mass is equal to the force acting on any individual particle.
Yes. Note that the force is interaction between the real particles. The force the imaginary particle "feels" is equal to that force of interaction, but as if it was exerted by a centre of force in the origin.
Satvik Pandey said:
And if the direction of force on particle is towards the CM of system then the direction of force on reduced mass is towards the center of force.Am I right.

If the force of interaction acts along the line connecting the real particles, then it is also towards to or away from the CM of the two-body system.
The imaginary particle also feels central force, acting along the line connecting the particle and the centre (origin)

ehild
 
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  • #61
The trick of applying Kepler's third law is in assuming that motion is elliptic, but the ellipse is "stretched" very severely. Taking this to the limit, it is quite obvious that the time to collision is half the period in elliptic motion. What remains is a reformulation of the law for charged particles.
 
  • #63
ehild said:
Yes.


Yes. Note that the force is interaction between the real particles. The force the imaginary particle "feels" is equal to that force of interaction, but as if it was exerted by a centre of force in the origin.


If the force of interaction acts along the line connecting the real particles, then it is also towards to or away from the CM of the two-body system.
The imaginary particle also feels central force, acting along the line connecting the particle and the centre (origin)

ehild

Thank You ehild. You helped me a lot.
 
  • #64
You are welcome:smile:

ehild
 

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