Condition to fall into 1/r^2 potential

Question

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1. What the problem is asking (in plain language)

A particle of mass (m) comes from very far away (effectively from infinity) with a speed (v_{\infty}) and a lateral offset (impact parameter) (b) with respect to the centre of an attractive potential

[ V(r)= -\frac{k}{r^{2}},\qquad k>0 . ]

Because the particle also carries angular momentum, the centrifugal “push’’ can prevent it from reaching the origin.
If the impact parameter is small enough, the attraction wins and the particle “falls’’ all the way to (r=0).

Question:
What is the largest (critical) impact parameter (b_{\max}) for which the particle still falls to the centre? How do we obtain it?

2. Step‑by‑step derivation

2.1. Conserved quantities

Energy (mechanical, because the potential is time‑independent)

[ E = \frac12 m \dot r^{2}+ \frac{L^{2}}{2mr^{2}} + V(r) . ]

Angular momentum about the origin (central force → conserved)

[ L = m r^{2}\dot\theta = \text{constant}. ]

For a particle that starts at infinity with speed (v_{\infty}) and impact parameter (b),

[ L = m v_{\infty} b . ]

The total mechanical energy far away is simply kinetic, because the potential vanishes as (r\to\infty):

[ E = \frac12 m v_{\infty}^{2}>0 . ]

2.2. The effective radial potential

Group the terms that depend only on (r) into an effective potential:

[ V_{\text{eff}}(r)=\frac{L^{2}}{2m r^{2}}+V(r) =\frac{L^{2}}{2m r^{2}}-\frac{k}{r^{2}} =\frac{1}{r^{2}}\Bigl(\frac{L^{2}}{2m}-k\Bigr). ]

Thus the radial equation of motion becomes

[ \frac12 m\dot r^{2}+V_{\text{eff}}(r)=E . ]

2.3. When can the particle reach (r=0)?

The particle can get arbitrarily close to the origin iff the effective potential never rises above the total energy.
Because (V_{\text{eff}}\propto 1/r^{2}), its sign is determined by the coefficient

[ C \equiv \frac{L^{2}}{2m}-k . ]

If (C>0) then (V_{\text{eff}}(r)>0) for all (r); the particle feels a repulsive centrifugal barrier and will have a finite closest approach (r_{\min}>0).
If (C\le 0) then (V_{\text{eff}}(r)\le 0) and becomes arbitrarily large in magnitude (negative) as (r\to0).
The radial kinetic term (\frac12 m\dot r^{2}=E-V_{\text{eff}}(r)) stays positive for every (r), so the particle can keep falling to (r=0).

Hence the capture condition is

[ \boxed{\; \frac{L^{2}}{2m}\le k\; } . ]

2.4. Translate the condition to the impact parameter

Insert (L=m v_{\infty} b):

[ \frac{(m v_{\infty} b)^{2}}{2m}\le k \;\Longrightarrow\; \frac{m v_{\infty}^{2} b^{2}}{2}\le k . ]

Solve for (b):

[ b^{2}\le \frac{2k}{m v_{\infty}^{2}} \qquad\Longrightarrow\qquad b\le b_{\max}= \sqrt{\frac{2k}{m v_{\infty}^{2}}}. ]

2.5. Special case of the statement “potential (-1/r^{2})”

If the problem’s constant is hidden (i.e. the potential is written as (V(r)=-1/r^{2})), then (k=1) (in whatever system of units is being used). The critical impact parameter becomes

[ \boxed{\,b_{\max}= \sqrt{\frac{2}{m v_{\infty}^{2}}}\,}. ]

If the constant (k) is kept explicitly, use the formula with (k).

3. Final answer

For a particle of mass (m) arriving from infinity with speed (v_{\infty}) in the attractive inverse‑square potential

[ V(r)=-\frac{k}{r^{2}},\qquad k>0, ]

the largest impact parameter that still leads to a “fall to the centre’’ is

[ \boxed{b_{\max}= \sqrt{\frac{2k}{m\,v_{\infty}^{2}}}} . ]

If the potential is written as (V(r)=-1/r^{2}) (i.e. (k=1)), this reduces to

[ b_{\max}= \sqrt{\frac{2}{m\,v_{\infty}^{2}}}. ]

Impact parameters larger than (b_{\max}) give a finite periapsis distance; the particle never reaches the origin.

4. Common mistakes (and how to avoid them)

Mistake	Why it’s wrong	How to fix it
Forgetting the centrifugal term and using only (V(r)) in the energy equation.	The angular momentum creates an effective repulsive (L^{2}/(2mr^{2})) that is essential for the capture condition.	Write the full effective potential (V_{\text{eff}} = L^{2}/(2mr^{2}) + V(r)) before analyzing the motion.
Using (L = m v_{\infty} b) incorrectly (e.g., missing a factor of (v_{\infty}) or using (b) instead of (b^{2})).	The relation follows from (\mathbf{L}= \mathbf{r}\times m\mathbf{v}); a missing factor changes the final expression for (b_{\max}).	Remember that at infinity the velocity is essentially perpendicular to the radius vector, so (L = m v_{\infty} b).
Mixing up signs (thinking the effective potential is always repulsive).	The attractive (-k/r^{2}) can outweigh the centrifugal term, making (V_{\text{eff}}) negative.	Keep the sign of each term clear: (+L^{2}/(2mr^{2})) (repulsive) and (-k/r^{2}) (attractive).
Assuming the particle must have zero total energy to fall.	The total energy is positive (kinetic at infinity). Capture occurs because the radial kinetic energy never vanishes, not because (E=0).	Use the inequality (L^{2}/(2m)\le k); the value of (E) (positive) does not appear in the critical condition.
Neglecting that the condition is ≤ not <.	The borderline case (L^{2}=2mk) still leads to a fall (the particle reaches the centre with constant radial speed).	State the capture condition as (L^{2}\le 2mk) or equivalently (b\le b_{\max}).

Keeping these points in mind will give the correct critical impact parameter for any inverse‑square attractive potential.

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