Time of approach of two oppositely charged particle

[Satvik Pandey]

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.

 

[Tanya Sharma]

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

m_1\ddot x_1=-F.....(1)
m_2\ddot x_2=F.....(2)

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

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

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

m_1x_1+m_2x_2=0.

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

\ddot x_1=-\frac{F}{m_1}
\ddot x_2=\frac{F}{m_2}
\ddot x_2-\ddot x_1=\frac{F}{m_2}+\frac{F}{m_1}

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

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

μ \ddot x= F, the same, as in case of a single body of mass μ.

 

[Satvik Pandey]

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?

 

[Satvik Pandey]

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.

 

[ehild]

I do not see what you want to do.ehild

 

[Satvik Pandey]

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.

 

[ehild]

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.

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

Integrate as before, r goes from R to 0.

ehild

 

[Satvik Pandey]

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.

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

Integrate as before, r goes from R to 0.

ehild

Thank You

 

[Satvik Pandey]

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.

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

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.

 

[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.

Yes.

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.

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

 

[voko]

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.

 

[rcgldr]

In case this was missed, the same basic integral is solved in the older thread about two objects colliding due to gravity:

https://www.physicsforums.com/showthread.php?t=635188

 

[Satvik Pandey]

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.

 

[ehild]

You are welcome

ehild