Simplified Derivation of the Collision Probability of Two Objects in Independent Keplerian Orbits

Consider the motion of two bodies, body 1 and body 2, whose fixed Keplerian orbits are intersecting each other. Their motions are approximated to be linear near the orbit intersection where collision is possible, as illustrated in Figure 1. Each body's position can be written as

$$\begin{eqnarray}&&{\boldsymbol{\rho }}(t)={\boldsymbol{r}}+t{\boldsymbol{v}},\end{eqnarray} \tag{ 1 }$$

where ${\boldsymbol{r}}$ is an arbitrary position vector along the line at time t = 0 and ${\boldsymbol{v}}$ is the constant velocity in the neighborhood of the orbit intersection point. We can set up a position relation between the two bodies using the point of intersection:

$$\begin{eqnarray}&&{{\boldsymbol{r}}}_{1}+{t}_{1}{{\boldsymbol{v}}}_{1}={{\boldsymbol{r}}}_{2}+{t}_{2}{{\boldsymbol{v}}}_{2},\end{eqnarray} \tag{ 2 }$$

where subscripts denote the object number (Figure 1). In this configuration, each body passes the orbit intersection point at time t₁ and time t₂, respectively. The encounter velocity is

$$\begin{eqnarray}&&{\boldsymbol{U}}={{\boldsymbol{v}}}_{1}-{{\boldsymbol{v}}}_{2}.\end{eqnarray} \tag{ 3 }$$

By taking the cross product with the encounter velocity on both sides of Equation (2) and rearranging the terms, we get

$$\begin{eqnarray}&&({{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2})\times {\boldsymbol{U}}=({t}_{1}-{t}_{2})({{\boldsymbol{v}}}_{1}\times {{\boldsymbol{v}}}_{2}).\end{eqnarray} \tag{ 4 }$$

Any pair of bodies moving with non-parallel constant velocities has a unique time when the distance between the two bodies, $| {{\boldsymbol{r}}}_{1}+t{{\boldsymbol{v}}}_{1}-{{\boldsymbol{r}}}_{2}-t{{\boldsymbol{v}}}_{2}|$, becomes minimum. By calling this specific time t = 0, we get the minimum distance ${D}_{\min }=| {{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}|$.

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At the minimum distance, the encounter velocity vector is normal to the relative position vector. This is a trivial consequence of the fact that at the minimum mutual distance,

$$\begin{eqnarray*}&&\displaystyle \frac{d}{{dt}}| {{\boldsymbol{\rho }}}_{1}(t)-{{\boldsymbol{\rho }}}_{2}(t){| }^{2}=0,\end{eqnarray*}$$

and the left-hand side is proportional to $({{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2})\cdot {\boldsymbol{U}}$. As this concept is a core part in further derivations, we write it down as a mathematical theorem below for frequent reference.

Theorem 1. Two moving points have their local minimum distance when their relative position vector is normal to their encounter velocity vector.

This is true for any two moving points on regular curves, except when the relative distance of the two moving points vanishes and the direction of their relative position becomes meaningless. We mention that Theorem 1 encapsulates the same concept as in the definition of the so-called "b-plane" adopted by Greenberg et al. (1988) in the context of Öpik–Wetherill formulas. The b-plane is defined as the plane passing through the target body and being normal to the inbound asymptotic (unperturbed) relative velocity. The impact parameter b, which is the magnitude of the projection of the inbound asymptote on the b-plane, is related to the minimum encounter distance by a scaling factor that accounts for the gravitational interaction between the two bodies. In fact, the gravitational interaction between the bodies makes the two bodies approach closer than the distance b (this is the so-called "gravitational focusing" effect); we will include this effect when we discuss the collision radius later. We note that gravitational focusing also causes a change in the relative velocity of the two bodies, but this is usually insignificant and is neglected in the calculation of P in the Öpik–Wetherill approach. Throughout this paper, we use "encounter velocity" to refer to the relative velocity that is unperturbed by the mutual gravitational interaction of body 1 and body 2.

From Theorem 1 and Equation (4), we can find that the time interval between body 1 and body 2 passing the orbit intersection point, ${\rm{\Delta }}t=| {t}_{1}-{t}_{2}|$, is given by

$$\begin{eqnarray}&&{\rm{\Delta }}t=\displaystyle \frac{{D}_{\min }U}{| {{\boldsymbol{v}}}_{1}\times {{\boldsymbol{v}}}_{2}| },\end{eqnarray} \tag{ 5 }$$

where U is the scalar magnitude of ${\boldsymbol{U}}$.

For convenience we choose body 1 as the body passing the intersection point ahead of the other body (${t}_{1}\lt {t}_{2}$) and illustrate this in Figure 2. Note that both t₁ and t₂ can have negative values as one or both bodies may have already passed the orbit intersection point when the two bodies approach each other with the minimum distance, i.e., when t = 0. In the middle panel of Figure 2, body 2 is ${\rm{\Delta }}t$ ahead of the orbit intersection point when body 1 is passing the intersection ($t={t}_{1}$). If body 2 were farther away from the intersection at $t={t}_{1}$, the pair would have a larger minimum distance than ${D}_{\min }.$ Likewise, if body 2 were located near the orbit intersection point at $t={t}_{1}$, the minimum distance would be smaller than ${D}_{\min }.$ Collision is possible if ${D}_{\min }$ is smaller than a specified collision radius τ, i.e., ${\rm{\Delta }}t\lt {\rm{\Delta }}{t}_{\mathrm{col}}$, where ${\rm{\Delta }}{t}_{\mathrm{col}}$ is given by

$$\begin{eqnarray}&&{\rm{\Delta }}{t}_{\mathrm{col}}=\displaystyle \frac{\tau U}{| {{\boldsymbol{v}}}_{1}\times {{\boldsymbol{v}}}_{2}| }.\end{eqnarray} \tag{ 6 }$$

The collision radius τ would be the sum of the physical radii of the two bodies, if we neglect their mutual gravity. If we wish to take account of their mutual gravity, we can multiply by the gravitational focusing factor.

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Thus, if body 2 passes the orbit intersection point later than body 1, collision would occur if, and only if, at time $t={t}_{1}$ body 2 is within ${\rm{\Delta }}{t}_{\mathrm{col}}$ of reaching the intersection point. Conversely, if body 2 passes the orbit intersection point ahead of body 1, collision would occur if and only if body 2 already has passed within the time interval ${\rm{\Delta }}{t}_{\mathrm{col}}$ before body 1 passes the intersection point. Therefore, whenever body 1 passes the intersection point, collision would occur if body 2 is within the time interval $2{\rm{\Delta }}{t}_{\mathrm{col}}$ of reaching the intersection point.

Considering that body 1 passes the intersection point once per its orbital revolution, the probability P₁ that it collides with body 2 in any single orbital period is

$$\begin{eqnarray}&&{P}_{1}=\displaystyle \frac{2{\rm{\Delta }}{t}_{\mathrm{col}}}{{T}_{2}}.\end{eqnarray} \tag{ 7 }$$

Thus, the collision probability per unit time, P, for body 1 to collide with body 2 is P₁ divided by the orbital period of body 1:

$$\begin{eqnarray}&&P=\displaystyle \frac{2{\rm{\Delta }}{t}_{\mathrm{col}}}{{T}_{1}{T}_{2}}=\displaystyle \frac{2\tau U}{| {{\boldsymbol{v}}}_{1}\times {{\boldsymbol{v}}}_{2}| {T}_{1}{T}_{2}}.\end{eqnarray} \tag{ 8 }$$

Note that P is the collision probability per unit time when two orbits exactly intersect. The general case of non-intersecting orbits will be considered in the next section.

Even when two orbits do not intersect, collision is still possible if the distance between the closest points of the two orbits, so-called "minimum orbit intersection distance," or MOID, is less than the collision radius τ. We note that, in general, there exist multiple local minima of the distance between two orbits (Gronchi 2005), and it is possible that more than one of these could be less than τ. The collision probabilities introduced by those local minima can be easily integrated, if necessary, as they have the same functional form as that of the MOID. In the case study that we describe in Section 5, we include contributions by all the multiple local minima.

The calculation of P for the non-intersecting case and its averaging for $\mathrm{MOID}\lt \tau$ is described well in Greenberg (1982). Here, we derive the same equation with an alternative approach. First, we note that two skewed lines, i.e., a pair of lines not intersecting nor being parallel, have a unique minimum distance, and the vector along the minimum distance is normal to both the lines. This can be generalized as the following theorem.

Theorem 2. Two smooth curves have the local minimum distance along the line orthogonal to the tangents of both curves.

Stated in other words, the normal planes of two smooth curves at a certain point on each curve should coincide with each other if there exists a local minimum distance at the given points.

Then, as before, we linearize the motion of the two bodies in the neighborhood of the MOID. Figure 3 illustrates the situation when the minimum non-zero distance between two orbits, denoted by ${\boldsymbol{s}}$, is along the Z-axis. By Theorem 2, both bodies move normal to the Z-axis in the vicinity of the MOID location. Without loss of generality, we choose the Y-axis to be parallel to the motion of body 2. The path of body 2, shifted by $-{\boldsymbol{s}}=-(0,0,{\rm{s}})$, intersects the path of body 1 at the origin, as in Section 2.1. Therefore, from Equations (2) and (4),

$$\begin{eqnarray}&&{{\boldsymbol{r}}}_{1}+{t}_{1}{{\boldsymbol{v}}}_{1}={{\boldsymbol{r}}}_{2}-{\boldsymbol{s}}+{t}_{2}{{\boldsymbol{v}}}_{2},\end{eqnarray} \tag{ 9 }$$

$$\begin{eqnarray}&&({{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}+{\boldsymbol{s}})\times {\boldsymbol{U}}=({t}_{1}-{t}_{2})({{\boldsymbol{v}}}_{1}\times {{\boldsymbol{v}}}_{2}).\end{eqnarray} \tag{ 10 }$$

As before, we consider body 1 and body 2 to have a minimum distance ${D}_{\min }$ at ${{\boldsymbol{r}}}_{1}$ and ${{\boldsymbol{r}}}_{2}$ at t = 0. In Equation (10), the resultant direction of the right-hand side is parallel to the Z-axis. As ${\boldsymbol{U}}$ is on the XY plane, ${{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}+{\boldsymbol{s}}$ should also lie on the XY plane and its magnitude is

$$\begin{eqnarray}&&| {{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}+{\boldsymbol{s}}| =\sqrt{{D}_{\min }^{2}-{s}^{2}},\end{eqnarray} \tag{ 11 }$$

from the Pythagorean theorem (Figure 3). Interestingly, ${\boldsymbol{U}}$ is normal to both ${\boldsymbol{s}}$ and ${{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2};$ the latter is normal to ${\boldsymbol{U}}$ according to Theorem 1. Therefore, ${{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}+{\boldsymbol{s}}$ is also normal to ${\boldsymbol{U}}$, which gives

$$\begin{eqnarray}&&{\rm{\Delta }}{t}_{\mathrm{col}}=\displaystyle \frac{\tau U}{| {{\boldsymbol{v}}}_{1}\times {{\boldsymbol{v}}}_{2}| }\sqrt{1-\displaystyle \frac{{s}^{2}}{{\tau }^{2}}}\end{eqnarray} \tag{ 12 }$$

for collision radius τ.

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For two fixed orbits having $0\lt s\leqslant \tau$, we can use Equation (12) without any modification. However, in Monte-Carlo-type numerical simulations, the averaged value of P, within $0\lt s\leqslant \tau$, is more useful as we can get a better estimate of impact flux from the same number of test projectiles.

In the small vicinity where collisions are allowed, s can be assumed to be uniformly distributed. (This follows from the argument that, for a random distribution of lines with fixed directions in space, the fraction of cases with a minimum distance less than s is proportional to s; this is a good assumption for $\tau \ll$ heliocentric distance.) Averaging ${\rm{\Delta }}{t}_{\mathrm{col}}$ within $0\lt s\leqslant \tau$, Equation (12) gives $\pi /4$ times ${\rm{\Delta }}{t}_{\mathrm{col}}$ of Equation (6). Therefore, the averaged collision probability per unit time when two bodies have s random in the range $(0,\tau )$ becomes

$$\begin{eqnarray}&&P=\displaystyle \frac{\pi \tau U}{2| {{\boldsymbol{v}}}_{1}\times {{\boldsymbol{v}}}_{2}| {T}_{1}{T}_{2}}.\end{eqnarray} \tag{ 13 }$$

In Wetherill's (1967) derivation, his Equation (7) gives the probability of collision per unit time for two bodies on intersecting orbits. Expressing his equation with our notation gives

$$\begin{eqnarray}&&P=\displaystyle \frac{\pi {\eta }_{1}}{2| {{\boldsymbol{v}}}_{1}| {T}_{1}{T}_{2}},\end{eqnarray} \tag{ 14 }$$

where

$$\begin{eqnarray}&&{\eta }_{1}=\displaystyle \frac{\tau U}{\sqrt{{U}^{2}-{({U}_{x}\cos {\alpha }_{1}+{U}_{y}\sin {\alpha }_{1})}^{2}}}.\end{eqnarray} \tag{ 15 }$$

In the above equation, U_x and U_y are the X and Y components of ${\boldsymbol{U}}$, and ${\alpha }_{1}$ is the angle between the X-axis and the velocity vector of body 1. The Sun is located in the −X direction. The physical meaning of ${\eta }_{1}$ is the maximum distance from the orbit intersection point where body 1 should be present to allow collision to occur when body 2 is at the orbit intersection point. Derivation of Equations (14) and (15) is lengthy and complicated in Wetherill (1967), and his equation does not look commutative between body 1 and body 2 at first glance. Below, we prove that his equation is the same as our simplified derivation of the collision probability per unit time, Equation (13), which is clearly symmetric in body 1 and body 2.

The term, ${U}_{x}\cos {\alpha }_{1}+{U}_{y}\sin {\alpha }_{1}$, in the denominator of Equation (15) is the projection of the encounter velocity ${\boldsymbol{U}}$ along the velocity direction of body 1. Therefore, the denominator is equal to the component of ${\boldsymbol{U}}$ normal to ${{\boldsymbol{v}}}_{1}$. Thus,

$$\begin{eqnarray}&&{\eta }_{1}=\displaystyle \frac{\tau U| {{\boldsymbol{v}}}_{1}| }{| {\boldsymbol{U}}\times {{\boldsymbol{v}}}_{1}| }=\displaystyle \frac{\tau U| {{\boldsymbol{v}}}_{1}| }{| {{\boldsymbol{v}}}_{1}\times {{\boldsymbol{v}}}_{2}| },\end{eqnarray} \tag{ 16 }$$

which gives Equation (13) when it is substituted in Equation (14).

It is evident that the right-hand side of Equation (13) is singular when ${{\boldsymbol{v}}}_{1}$ and ${{\boldsymbol{v}}}_{2}$ are parallel. This situation is illustrated in Figure 4. The origin of this singularity lies in our linear approximation for the motion of bodies near the collision point. To resolve this singularity, we can use a better approximation of the true motion of the two bodies, namely parabolic motion instead of linear motion,

$$\begin{eqnarray}&&{{\boldsymbol{\rho }}}_{i}(t)={{\boldsymbol{r}}}_{i}+t{{\boldsymbol{v}}}_{i}+\displaystyle \frac{1}{2}{(t-{t}_{i})}^{2}{\boldsymbol{g}}-\displaystyle \frac{1}{2}{{t}_{i}}^{2}{\boldsymbol{g}},\end{eqnarray} \tag{ 17 }$$

where ${\boldsymbol{g}}$ is the gravitational acceleration due to the Sun, assumed to be constant in the vicinity of the encounter. The constant vector ${{\boldsymbol{v}}}_{i}$ is the velocity of body i at the orbit intersection point, and body i passes the orbit intersection point at time t_i. At this time, the direction of the velocity vector of both bodies is the same. The encounter geometry is illustrated in Figure 4 in a coordinate system with the origin at the orbit intersection point, the Sun is located on the negative X-axis, the orbital poles of the two bodies are aligned to the Z-direction, and the motion of the two bodies is approximated as parabolic paths. (Note that the time epochs, t₁ and t₂, when body 1 and body 2 pass the origin, have negative values in the case illustrated in Figure 4.) We denote by α the angle between the X-axis and the common direction of the velocity vectors; the range of α is 0 to π. The time t = 0 is defined as the epoch when the two bodies approach the minimum distance, ${D}_{\min };$ at this time their position vectors are ${{\boldsymbol{r}}}_{1}$ and ${{\boldsymbol{r}}}_{2}$, and the encounter velocity is given by the relative velocity at t = 0:

$$\begin{eqnarray}&&{{\boldsymbol{U}}}_{0}={{\boldsymbol{v}}}_{1}-{{\boldsymbol{v}}}_{2}-({t}_{1}-{t}_{2}){\boldsymbol{g}}.\end{eqnarray} \tag{ 18 }$$

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We define the outer orbit, the one having smaller curvature, as body 1 and the inner orbit as body 2. Because the outer body should have a higher velocity than the inner body at t = 0, we introduce a velocity ratio

$$\begin{eqnarray}&&k={v}_{2}/{v}_{1},\qquad -1\lt k\lt 1,\end{eqnarray} \tag{ 19 }$$

where the negative values occur when body 2 orbits in the opposite direction of body 1. For simplicity, in the following, we consider only the case of positive k for the derivation of P. The derivation for the negative k is similar, and we note that the final results (Equations (27), (29), (36), and (37)) hold true for both positive and negative k. The general derivation with an alternative approach is also provided in the Appendix.

The path of body 2 from the origin to point B is slightly longer than the path of body 1 from the origin to point A in the case illustrated in Figure 4. Because the Y components of the velocities remain constant, the ratio of the travel time for $\overline{{OB}^{\prime} }$ for body 2 and $\overline{{OA}}$ for body 1 is equal to the inverse of the ratio of initial velocity, $1/k\gt 1$. (Here, $B^{\prime}$ is the location of body 2 having the same Y component as point A.) The relation between the times t₁ and t₂ (when body 1 and body 2 pass the origin) can be written as

$$\begin{eqnarray}&&{t}_{2}=\displaystyle \frac{{t}_{1}}{k}-\delta t,\end{eqnarray} \tag{ 20 }$$

where the small time interval, $\delta t$, for body 2 to go from $B^{\prime}$ to B is given by

$$\begin{eqnarray}&&\delta t\simeq \displaystyle \frac{{D}_{\min }}{{v}_{2}\tan \alpha }.\end{eqnarray} \tag{ 21 }$$

Note that $\delta t$ is positive when $0\lt \alpha \lt \pi /2$ and negative for $\pi /2\lt \alpha \lt \pi$.

To solve for t₁ and t₂, we take a similar approach as in Section 2. Thus, similar to Equation (2), we can set up the following relation for the positions of the two bodies using the orbit intersection point:

$$\begin{eqnarray}&&{{\boldsymbol{r}}}_{1}+{t}_{1}{{\boldsymbol{v}}}_{1}-\displaystyle \frac{1}{2}{{t}_{1}}^{2}{\boldsymbol{g}}={{\boldsymbol{r}}}_{2}+{t}_{2}{{\boldsymbol{v}}}_{2}-\displaystyle \frac{1}{2}{{t}_{2}}^{2}{\boldsymbol{g}}.\end{eqnarray} \tag{ 22 }$$

By taking the vector product with the encounter velocity, ${{\boldsymbol{U}}}_{0}$ (Equation (18)), on both sides of Equation (22), and using ${{\boldsymbol{v}}}_{1}\times {{\boldsymbol{v}}}_{2}=0$ (for tangential encounters), we get

$$\begin{eqnarray}&&({{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2})\times {{\boldsymbol{U}}}_{0}=\displaystyle \frac{1}{2}({{t}_{1}}^{2}-{{t}_{2}}^{2}){\boldsymbol{g}}\\ &&\quad \times {\boldsymbol{U}}+({t}_{1}-{t}_{2}){\boldsymbol{g}}\times ({t}_{2}{{\boldsymbol{v}}}_{2}-{t}_{1}{{\boldsymbol{v}}}_{1}),\end{eqnarray} \tag{ 23 }$$

where ${\boldsymbol{U}}={{\boldsymbol{v}}}_{1}-{{\boldsymbol{v}}}_{2}$. With the use of Theorem 1, the above equation simplifies to the following:

$$\begin{eqnarray}&&{D}_{\min }{U}_{0}=\left[\displaystyle \frac{1}{2}({{t}_{2}}^{2}-{{t}_{1}}^{2})U+({t}_{2}-{t}_{1})\delta t\,{v}_{1}\right]g\sin \alpha .\end{eqnarray} \tag{ 24 }$$

Because $\delta t$ is a small value, we neglect the term with $\delta t$ for the moment (but we return to it below). Also, considering that $| {t}_{1}-{t}_{2}| g$ is much smaller than $| {{\boldsymbol{v}}}_{1}-{{\boldsymbol{v}}}_{2}|$, ${\boldsymbol{U}}$ can be approximated as ${{\boldsymbol{U}}}_{0}$. Then, we find the minimum distance of approach of the two bodies is related to their times of passage at the origin as follows:

$$\begin{eqnarray}&&{D}_{\min }=\displaystyle \frac{1}{2}({{t}_{2}}^{2}-{{t}_{1}}^{2})g\sin \alpha .\end{eqnarray} \tag{ 25 }$$

By using ${t}_{1}\simeq {{kt}}_{2}$ and rearranging Equation (25), we obtain

$$\begin{eqnarray}&&{t}_{2}=\mp \sqrt{\displaystyle \frac{2{D}_{\min }}{(1-{k}^{2})g\sin \alpha }},\end{eqnarray} \tag{ 26 }$$

where the minus sign is for the case when the bodies have already passed the origin when the minimum distance is achieved (as in Figure 4), whereas the plus sign is for the opposite case.

Analogous to the derivation in Section 2, we can define the time interval ${\rm{\Delta }}{t}_{\mathrm{col}}=| {t}_{2}-{t}_{1}|$ for collision to occur for a given collision radius τ:

$$\begin{eqnarray}&&{\rm{\Delta }}{t}_{\mathrm{col}}=\sqrt{\displaystyle \frac{2(1-k)\tau }{(1+k)g\sin \alpha }}.\end{eqnarray} \tag{ 27 }$$

Thus, if body 2 has passed the origin more than ${\rm{\Delta }}{t}_{\mathrm{col}}$ earlier than body 1, body 1 cannot catch up to collide with body 2. On the other hand, unlike Figure 4, in order for the collision to occur before the bodies reach the orbit intersection point, the expected t₁ should not be more than ${\rm{\Delta }}{t}_{\mathrm{col}}$ earlier than the expected t₂.

If we wish to be more accurate, Equations (20) and (25) give the following better estimate of ${\rm{\Delta }}{t}_{\mathrm{col}}$:

$$\begin{eqnarray}&&{\rm{\Delta }}{t}_{\mathrm{col}}\simeq \sqrt{\displaystyle \frac{2(1-k)\tau }{(1+k)g\sin \alpha }}\pm \displaystyle \frac{k\delta t}{1+k},\end{eqnarray} \tag{ 28 }$$

where the sign order is the same as in Equation (26).

In other words, regardless of the choice between the above two equations, if body 2 passes the origin within a time interval $\sim 2{\rm{\Delta }}{t}_{\mathrm{col}}$ of when body 1 passes the origin, the collision would occur. (Note that the $\delta t$ term in Equation (28) cancels out when we add the time intervals with the plus and minus signs.) Thus, the collision probability per unit time of these two bodies whose velocity vectors are parallel near the collision point is given by

$$\begin{eqnarray}&&P=\displaystyle \frac{2{\rm{\Delta }}{t}_{\mathrm{col}}}{{T}_{1}{T}_{2}}=\displaystyle \frac{1}{{T}_{1}{T}_{2}}\sqrt{\displaystyle \frac{8(1-k)\tau }{(1+k)g\sin \alpha }}.\end{eqnarray} \tag{ 29 }$$

This equation may appear counterintuitive at first glance because P decreases as the velocity ratio k approaches 1. It is true that the path lengths OA and OB increase as k approaches 1. However, as the two velocities become identical, it takes longer for body 1 to catch up with body 2. Consequently, when body 1 is at the origin, body 2 should be located within a smaller distance to have a chance to collide with body 1.

Now we consider a more general collision geometry where the orbits do not intersect but their minimum approach distance, s, is small enough to allow collision. We define a coordinate system that has the origin as the point on body 1's orbit where the MOID, $0\lt s\leqslant \tau$, occurs. Again, the Sun is located along the −X direction, and the orbit of body 1 is on the XY plane. We assume that the variation of the vertical velocity component of body 2 is negligible in the small region where collisions are possible.

According to Theorem 2, at the location where MOID occurs, the position vector of body 2 should be normal to the velocity direction of both bodies, $(\cos \alpha ,\sin \alpha ,0)$. Thus, the position of body 2 at $t={t}_{2}$ can be expressed as follows:

$$\begin{eqnarray}&&{\boldsymbol{s}}=s(-\cos \beta \sin \alpha ,\cos \beta \cos \alpha ,\sin \beta ),\end{eqnarray} \tag{ 30 }$$

where β is the angle between the position vector of body 2 and the XY plane.

Figure 5 shows the orbit of body 1 on the XY plane with the projection of the orbit of body 2 on the XY plane. As before, when body 1 passes point A, it can barely touch body 2 located at B. Because the path of body 2 shifted by $-{\boldsymbol{s}}$ intersects the path of body 1 at the origin, from Equations (22) and (23), we get

$$\begin{eqnarray}&&{{\boldsymbol{r}}}_{1}+{t}_{1}{{\boldsymbol{v}}}_{1}-\displaystyle \frac{1}{2}{{t}_{1}}^{2}{\boldsymbol{g}}={{\boldsymbol{r}}}_{2}-{\boldsymbol{s}}+{t}_{2}{{\boldsymbol{v}}}_{2}-\displaystyle \frac{1}{2}{{t}_{2}}^{2}{\boldsymbol{g}},\end{eqnarray} \tag{ 31 }$$

$$\begin{eqnarray}&&({{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}+{\boldsymbol{s}})\times {{\boldsymbol{U}}}_{0}=\displaystyle \frac{1}{2}({{t}_{1}}^{2}-{{t}_{2}}^{2}){\boldsymbol{g}}\\ &&\quad \times {\boldsymbol{U}}+({t}_{1}-{t}_{2}){\boldsymbol{g}}\times ({t}_{2}{{\boldsymbol{v}}}_{2}-{t}_{1}{{\boldsymbol{v}}}_{1}).\end{eqnarray} \tag{ 32 }$$

As in Section 3.1, we approximate ${{\boldsymbol{U}}}_{0}\simeq {\boldsymbol{U}}$ and ${t}_{2}\simeq {{kt}}_{1}$. As ${{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}\,\perp \,{{\boldsymbol{U}}}_{0}$ (Theorem 1) and ${\boldsymbol{s}}\,\perp \,{\boldsymbol{U}}$ (Theorem 2), ${{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}+{\boldsymbol{s}}$ is also close to orthogonal to ${{\boldsymbol{U}}}_{0}$ and ${\boldsymbol{U}}$; therefore, we get

$$\begin{eqnarray}&&| {{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}+{\boldsymbol{s}}| =\displaystyle \frac{1}{2}({{t}_{2}}^{2}-{{t}_{1}}^{2})g\sin \alpha .\end{eqnarray} \tag{ 33 }$$

The Z-directional motion of body 2 is negligible in the small vicinity of the MOID location, so body 2 is moving in the plane parallel to the orbital plane of body 1. Thus, as illustrated in Figure 6, ${{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}+{\boldsymbol{s}}$ lies in the XY plane and has the length

$$\begin{eqnarray}&&| {{\boldsymbol{r}}}_{2}-{{\boldsymbol{r}}}_{1}-{\boldsymbol{s}}| ={D}_{\min }\cos \gamma -s\cos \beta ,\end{eqnarray} \tag{ 34 }$$

where γ is the angle between the line corresponding to the MOID and the XY plane. By the law of sines, the angle γ is given by

$$\begin{eqnarray}&&{D}_{\min }\sin \gamma =s\sin \beta .\end{eqnarray} \tag{ 35 }$$

As is the case in Section 3.1, t₂ is not exactly the same as ${t}_{1}/k$, but the difference almost cancels with the time difference on the opposite side, i.e., the case where the minimum distance is achieved before the bodies reach the origin. Thus, from Equation (33) through (35), we obtain, for a collision radius of τ, the time interval ${\rm{\Delta }}{t}_{\mathrm{col}}$ to be

$$\begin{eqnarray}&&{\rm{\Delta }}{t}_{\mathrm{col}}=\sqrt{\displaystyle \frac{2(1-k)\tau }{(1+k)g\sin \alpha }}{\left(\sqrt{1-\displaystyle \frac{{s}^{2}}{{\tau }^{2}}{\sin }^{2}\beta }-\displaystyle \frac{s}{\tau }\cos \beta \right)}^{1/2}.\end{eqnarray} \tag{ 36 }$$

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Assuming a uniform distribution within the parameter space $0\lt s\leqslant \tau$ and $-\pi /2\leqslant \beta \leqslant \pi /2$, we can obtain the average value of ${\rm{\Delta }}{t}_{\mathrm{col}}$. (Care must be taken over the range of β, as the orbit of the faster body should always be outside of the slower body in the vicinity of the MOID.) With numerical integration we found the average value of the factor ${(\sqrt{1-\tfrac{{s}^{2}}{{\tau }^{2}}{\sin }^{2}\beta }-\tfrac{s}{\tau }\cos \beta )}^{1/2}$ is 0.61. Thus, we obtain the collision probability per unit time,

$$\begin{eqnarray}&&P=\displaystyle \frac{1.7}{{T}_{1}{T}_{2}}\sqrt{\displaystyle \frac{(1-k)\tau }{(1+k)g\sin \alpha }}.\end{eqnarray} \tag{ 37 }$$

In the Appendix, we present fully analytic derivations that prove that Equations (28) and (36) are valid for $\tau /r\ll 1-k$. This condition is well satisfied in solar system applications of the impact rate of asteroids on planets since planet sizes are much smaller than their heliocentric distances. An example where this condition may be violated is for large planets orbiting close to their host star. For example, in the case of hot Jupiter WASP-18b orbiting with a semimajor axis of 0.02 au, the physical radius is 2.7% of the semimajor axis (Southworth et al. 2009). Considering the gravitational focusing factor, we calculate $\tau /r$ of WASP-18b to be equal to $1-k$ when k = 0.7. For the more common planetary targets with a larger semimajor axis and smaller radius than WASP-18b, Equations (28) and (36) are valid approximations unless the encounter velocity is vanishingly small.

Let us set up a coordinate system with the origin at the point of intersection of the two orbits, the XY plane is the common orbital plane of the two orbits, and the X direction is radially outward from the Sun as in Section 2.1. In general, the minimum distance between the center-of-figure of the two bodies is not zero and it does not occur at the point of intersection of the two orbits. Unlike Section 3.1, let us assume that body 2 passes the origin at time t = 0 with velocity ${{\boldsymbol{v}}}_{2}$ and body 1 passes the origin a time $t={\rm{\Delta }}t$ with velocity ${{\boldsymbol{v}}}_{1}$. (Note that ${\rm{\Delta }}t$ here may be negative, if body 1 passes the origin before body 2.) At some time $t={t}_{* }$, the two bodies achieve a minimum mutual distance. For tangential encounters, we can write (without loss of generality)

$$\begin{eqnarray}&&{{\boldsymbol{v}}}_{2}=k{{\boldsymbol{v}}}_{1}={{kv}}_{1}(\cos \alpha ,\sin \alpha ),\qquad -1\lt k\lt 1.\end{eqnarray} \tag{ 40 }$$

In the vicinity of the origin, we can approximate the motion of body 1 and body 2 as follows:

$$\begin{eqnarray}&&{{\boldsymbol{r}}}_{1}=(t-{\rm{\Delta }}t){{\boldsymbol{v}}}_{1}-\displaystyle \frac{1}{2}g{(t-{\rm{\Delta }}t)}^{2}\hat{{\boldsymbol{x}}},\end{eqnarray} \tag{ 41 }$$

$$\begin{eqnarray}&&{{\boldsymbol{r}}}_{2}=t{{\boldsymbol{v}}}_{2}-\displaystyle \frac{1}{2}{{gt}}^{2}\hat{{\boldsymbol{x}}}={kt}{{\boldsymbol{v}}}_{1}-\displaystyle \frac{1}{2}{{gt}}^{2}\hat{{\boldsymbol{x}}}.\end{eqnarray} \tag{ 42 }$$

Then the square of the distance between the two bodies can be expressed as a function of time:

$$\begin{eqnarray}| {{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}{| }^{2} & = & {[(1-k)t-{\rm{\Delta }}t]}^{2}{v}_{1y}^{2}\\ & & +{\left\{[(1-k)t-{\rm{\Delta }}t]{v}_{1x}+\left(t-\displaystyle \frac{1}{2}{\rm{\Delta }}t\right)g{\rm{\Delta }}t\right\}}^{2}.\end{eqnarray} \tag{ 43 }$$

The minimum of the mutual distance occurs at $t={t}_{* }$, which can be obtained by the condition $\partial | {{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}{| }^{2}/\partial t=0$.

$$\begin{eqnarray}\displaystyle \frac{\partial | {{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}{| }^{2}}{\partial t} & = & 2{(1-k)}^{2}{v}_{1}^{2}[1+2\varepsilon \cos \alpha +{\varepsilon }^{2}]t\\ & & -2(1-k){v}_{1}^{2}\left[1+\displaystyle \frac{1}{2}(3-k)\varepsilon \cos \alpha \right.\\ & & \left.+\displaystyle \frac{1}{2}(1-k){\varepsilon }^{2}\right]{\rm{\Delta }}t,\end{eqnarray} \tag{ 44 }$$

where ε is defined as

$$\begin{eqnarray}&&\varepsilon =\displaystyle \frac{g{\rm{\Delta }}t}{(1-k){v}_{1}}.\end{eqnarray} \tag{ 45 }$$

Then the minimum distance occurs at

$$\begin{eqnarray}&&{t}_{* }={\rm{\Delta }}t\displaystyle \frac{1+\tfrac{1}{2}(3-k)\varepsilon \cos \alpha +\tfrac{1}{2}(1-k){\varepsilon }^{2}}{(1-k){ \mathcal Q }},\end{eqnarray} \tag{ 46 }$$

where

$$\begin{eqnarray}&&{ \mathcal Q }=1+2\varepsilon \cos \alpha +{\varepsilon }^{2}.\end{eqnarray} \tag{ 47 }$$

It is useful to compute

$$\begin{eqnarray*}&&(1-k){t}_{* }-{\rm{\Delta }}t=-\displaystyle \frac{1+k}{2}\displaystyle \frac{{\rm{\Delta }}t}{{ \mathcal Q }}(\cos \alpha +\varepsilon )\varepsilon ,\end{eqnarray*}$$

$$\begin{eqnarray*}&&{t}_{* }-\displaystyle \frac{1}{2}{\rm{\Delta }}t=\displaystyle \frac{1}{2}\displaystyle \frac{1+k}{1-k}\displaystyle \frac{{\rm{\Delta }}t}{{ \mathcal Q }}(1+\varepsilon \cos \alpha ).\end{eqnarray*}$$

Then, setting $t={t}_{* }$ in Equation (43), we find

$$\begin{eqnarray}&&| {{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}{| }_{\min }^{2}={({\rm{\Delta }}t)}^{4}\displaystyle \frac{{(1+k)}^{2}{g}^{2}{\sin }^{2}\alpha }{4{(1-k)}^{2}{ \mathcal Q }}.\end{eqnarray} \tag{ 48 }$$

For collision to occur, we must have $| {{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}{| }_{\min }\leqslant \tau$, where τ is the collision radius. Thus, collision will occur provided $| {\rm{\Delta }}t| \leqslant {\rm{\Delta }}{t}_{\mathrm{col}}$, where ${\rm{\Delta }}{t}_{\mathrm{col}}$ is given by

$$\begin{eqnarray}&&{({\rm{\Delta }}{t}_{\mathrm{col}})}^{4}{(1+k)}^{2}{g}^{2}{\sin }^{2}\alpha \\ &&\quad =4{\tau }^{2}{(1-k)}^{2}(1+2\varepsilon \cos \alpha +{\varepsilon }^{2}).\end{eqnarray} \tag{ 49 }$$

Recall that $\varepsilon \propto {\rm{\Delta }}t$ (Equation (45)); therefore, Equation (49) presents a quartic equation for ${\rm{\Delta }}{t}_{\mathrm{col}}$, whose exact analytic solution is possible but tedious. Provided that $| \varepsilon | \ll 1$, we obtain the leading order solution:

$$\begin{eqnarray}&&{\rm{\Delta }}{t}_{\mathrm{col}}\simeq \sqrt{\displaystyle \frac{2(1-k)\tau }{(1+k)g\sin \alpha }}.\end{eqnarray} \tag{ 50 }$$

A better approximation can be achieved by plugging the first approximation, Equation (50), into (49), to obtain the leading order correction; this yields

$$\begin{eqnarray}&&{\rm{\Delta }}{t}_{\mathrm{col}}\simeq \sqrt{\displaystyle \frac{2(1-k)\tau }{(1+k)g\sin \alpha }}\pm \displaystyle \frac{\tau }{(1+k){v}_{1}\tan \alpha }.\end{eqnarray} \tag{ 51 }$$

The two equations above are equivalent to Equations (27) and (28).

The condition $| \varepsilon | \ll 1$ requires that the change, $g{\rm{\Delta }}{t}_{\mathrm{col}}$, in the heliocentric velocity of the bodies over the time ${\rm{\Delta }}{t}_{\mathrm{col}}$ is much smaller than the encounter velocity, ${\rm{\Delta }}v={v}_{1}-{v}_{2}$. We note that, using this approximate solution in Equation (45), we have

$$\begin{eqnarray}&&\varepsilon \simeq \sqrt{\displaystyle \frac{2g\tau }{(1-{k}^{2}){v}_{1}^{2}\sin \alpha }}=\sqrt{\displaystyle \frac{\tau /r}{\langle {v}_{T}\rangle {\rm{\Delta }}v/{v}_{c}^{2}}},\end{eqnarray} \tag{ 52 }$$

where $\langle {v}_{T}\rangle =({v}_{1}+{v}_{2})\sin \alpha /2$ is the average transverse velocity of the two bodies at the collision point, and we used $g={{GM}}_{\odot }/{r}^{2}={v}_{c}^{2}/r$ (r is the heliocentric distance at the collision point, and ${v}_{c}=\sqrt{{{GM}}_{\odot }/r}$ is the heliocentric circular velocity). Thus, the condition $| \varepsilon | \ll 1$ is equivalent to

$$\begin{eqnarray}&&\sqrt{\displaystyle \frac{\tau /r}{{\rm{\Delta }}v/{v}_{c}}}\ll 1\qquad \qquad (\mathrm{for}\,\ \ \mathrm{prograde}),\end{eqnarray} \tag{ 53 }$$

$$\begin{eqnarray}&&\sqrt{\displaystyle \frac{\tau /r}{2\langle {v}_{T}\rangle /{v}_{c}}}\ll 1\qquad \qquad (\mathrm{for}\,\ \ \mathrm{retrograde}).\end{eqnarray} \tag{ 54 }$$

In other words, we can state the condition as: the collision radius (as a fraction of the heliocentric distance) is much smaller than the velocity difference $(1-| k| ){v}_{1}$ (as a fraction of the orbital velocity). In physical terms, we can describe this as the condition that the collision radius τ should be much smaller than the heliocentric distance when the velocity difference is not a vanishingly small fraction of the orbital velocity.

For the case of non-intersecting orbits, we note that at the location of the MOID, the tangent to each orbit is normal to the relative distance vector, ${\boldsymbol{s}}$. We assume that the minimum mutual distance occurs in the vicinity of the MOID location. Let us choose as the origin the point on body 1's orbit where the MOID occurs, and let us choose the plane of body 1 as the XY plane.

Let us assume that body 2 passes the MOID location, ${\boldsymbol{s}}$, at time t = 0 with velocity ${{\boldsymbol{v}}}_{2}$ and that body 1 passes the origin at some time later, $t={\rm{\Delta }}t$, with velocity ${{\boldsymbol{v}}}_{1}$. Then, for near-tangential encounters we can write

$$\begin{eqnarray}&&{{\boldsymbol{v}}}_{{\boldsymbol{i}}}={{\rm{v}}}_{{\rm{i}}}(\cos \alpha ,\sin \alpha ,0),\end{eqnarray} \tag{ 55 }$$

$$\begin{eqnarray}&&{{\boldsymbol{v}}}_{2}=k{{\boldsymbol{v}}}_{1},\qquad -1\lt k\lt 1.\end{eqnarray} \tag{ 56 }$$

We can also express the time-dependent positions of body 1 and body 2 as follows:

$$\begin{eqnarray}&&{{\boldsymbol{r}}}_{1}=(t-{\rm{\Delta }}t){{\boldsymbol{v}}}_{1}-\displaystyle \frac{1}{2}g{(t-{\rm{\Delta }}t)}^{2}\hat{{\boldsymbol{x}}},\end{eqnarray} \tag{ 57 }$$

$$\begin{eqnarray}&&{{\boldsymbol{r}}}_{2}={\boldsymbol{s}}+t{{\boldsymbol{v}}}_{2}-\displaystyle \frac{1}{2}{{gt}}^{2}\hat{{\boldsymbol{x}}}={\boldsymbol{s}}+{kt}{{\boldsymbol{v}}}_{1}-\displaystyle \frac{1}{2}{{gt}}^{2}\hat{{\boldsymbol{x}}}.\end{eqnarray} \tag{ 58 }$$

Then, the distance between the two bodies is

$$\begin{eqnarray}&&| {{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}{| }^{2}={s}_{z}^{2}+{\{[(1-k)t-{\rm{\Delta }}t]{v}_{1y}-{s}_{y}\}}^{2}\\ &&\quad +{\{[(1-k)t-{\rm{\Delta }}t]{v}_{1x}+\left(t-\displaystyle \frac{1}{2}{\rm{\Delta }}t\right)g{\rm{\Delta }}t-{s}_{x}\}}^{2}\\ &&\quad ={[(1-k)t-{\rm{\Delta }}t]}^{2}{v}_{1}^{2}+{\left(t-\displaystyle \frac{1}{2}{\rm{\Delta }}t\right)}^{2}{(g{\rm{\Delta }}t)}^{2}\\ &&\quad +2\left(t-\displaystyle \frac{1}{2}{\rm{\Delta }}t\right)[(1-k)t-{\rm{\Delta }}t]{v}_{1x}g{\rm{\Delta }}t\\ &&\quad -2\left(t-\displaystyle \frac{1}{2}{\rm{\Delta }}t\right){s}_{x}g{\rm{\Delta }}t+{s}^{2}.\end{eqnarray} \tag{ 59 }$$

The minimum of the mutual distance occurs at $t={t}_{* }$, which can be obtained by the condition $\partial | {{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}{| }^{2}/\partial t=0$. First, we derive

$$\begin{eqnarray}&&\displaystyle \frac{\partial }{\partial t}| {{\boldsymbol{r}}}_{2}-{{\boldsymbol{r}}}_{1}{| }^{2}=2[(1-k)t-{\rm{\Delta }}t](1-k){v}_{1}^{2}\\ &&\quad +2\left(t-\displaystyle \frac{1}{2}{\rm{\Delta }}t\right){(g{\rm{\Delta }}t)}^{2}+2\phantom{\rule{}{3.1ex}}[(1-k)t\\ &&\quad \left.-{\rm{\Delta }}t+(1-k)\left(t-\displaystyle \frac{1}{2}{\rm{\Delta }}t\right)\right]{v}_{1x}g{\rm{\Delta }}t-2{s}_{x}g{\rm{\Delta }}t\end{eqnarray} \tag{ 60 }$$

$$\begin{eqnarray}&&=2{(1-k)}^{2}{v}_{1}^{2}(1+2\varepsilon \cos \alpha +{\varepsilon }^{2})t\\ &&\quad -2(1-k){v}_{1}^{2}\left[1+\displaystyle \frac{1}{2}(3-k)\varepsilon \cos \alpha \right.\\ &&\quad \left.+\displaystyle \frac{1}{2}(1-k){\varepsilon }^{2}+\displaystyle \frac{{{gs}}_{x}}{(1-k){v}_{1}^{2}}\right]{\rm{\Delta }}t\end{eqnarray} \tag{ 61 }$$

where ε is given by Equation (45). Then, we find

$$\begin{eqnarray}{t}_{* } & = & \displaystyle \frac{{\rm{\Delta }}t}{(1-k){ \mathcal Q }}\left[1+\displaystyle \frac{1}{2}(3-k)\varepsilon \cos \alpha \right.\\ & & \left.+\displaystyle \frac{1}{2}(1-k){\varepsilon }^{2}+\displaystyle \frac{{{gs}}_{x}}{(1-k){v}_{1}^{2}}\right].\end{eqnarray} \tag{ 62 }$$

It is useful to compute

$$\begin{eqnarray*}&&(1-k){t}_{* }-{\rm{\Delta }}t=-\displaystyle \frac{1+k}{2}\displaystyle \frac{{\rm{\Delta }}t}{{ \mathcal Q }}\left[\varepsilon \cos \alpha +{\varepsilon }^{2}-\displaystyle \frac{2{{gs}}_{x}}{(1-{k}^{2}){v}_{1}^{2}}\right],\end{eqnarray*}$$

$$\begin{eqnarray*}&&{t}_{* }-\displaystyle \frac{1}{2}{\rm{\Delta }}t=\displaystyle \frac{1}{2}\displaystyle \frac{1+k}{1-k}\displaystyle \frac{{\rm{\Delta }}t}{{ \mathcal Q }}\left[1+\varepsilon \cos \alpha +\displaystyle \frac{2{{gs}}_{x}}{(1-{k}^{2}){v}_{1}^{2}}\right].\end{eqnarray*}$$

We use Equation (62) in Equation (59) to find that the minimum distance

$$\begin{eqnarray}&&| {{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}{| }_{\min }^{2}={s}^{2}+\displaystyle \frac{{(1+k)}^{2}{({\rm{\Delta }}t)}^{2}{v}_{1}^{2}}{4{{ \mathcal Q }}^{2}}\{{(\varepsilon \cos \alpha +{\varepsilon }^{2}-2\lambda )}^{2}\\ &&\quad +{(1+\varepsilon \cos \alpha +2\lambda )}^{2}{\varepsilon }^{2}-2(1+\varepsilon \cos \alpha +2\lambda )\\ &&\quad \times (\varepsilon \cos \alpha +{\varepsilon }^{2}-2\lambda )\varepsilon \cos \alpha \\ &&\quad -4\lambda (1+\varepsilon \cos \alpha +2\lambda )(1+2\varepsilon \cos \alpha +{\varepsilon }^{2})\},\end{eqnarray} \tag{ 63 }$$

where

$$\begin{eqnarray}&&\lambda =\displaystyle \frac{{{gs}}_{x}}{(1-{k}^{2}){v}_{1}^{2}}.\end{eqnarray} \tag{ 64 }$$

In a similar way to Equation (52), the small parameter λ can be understood as

$$\begin{eqnarray}&&\lambda =\displaystyle \frac{{s}_{x}/r}{2\langle v\rangle {\rm{\Delta }}v/{{v}_{c}}^{2}},\end{eqnarray} \tag{ 65 }$$

where $\langle v\rangle =({v}_{1}+{v}_{2})/2$ is the average velocity of the two bodies. Because ${s}_{x}\leqslant \tau$, we see that λ is of order ${\varepsilon }^{2}$.

Setting $| {{\boldsymbol{r}}}_{1}-{{\boldsymbol{r}}}_{2}{| }_{\min }^{2}={\tau }^{2}$, we obtain a polynomial equation for ${\rm{\Delta }}{t}_{\mathrm{col}}$:

$$\begin{eqnarray}&&\displaystyle \frac{1}{4}{(1+k)}^{2}{({\rm{\Delta }}t)}^{2}{v}_{1}^{2}\{{(\varepsilon \cos \alpha +{\varepsilon }^{2}-2\lambda )}^{2}\\ &&\quad +{(1+\varepsilon \cos \alpha +2\lambda )}^{2}{\varepsilon }^{2}-2(1+\varepsilon \cos \alpha +2\lambda )\\ &&\quad \times (\varepsilon \cos \alpha +{\varepsilon }^{2}-2\lambda )\varepsilon \cos \alpha -4\lambda (1+\varepsilon \cos \alpha +2\lambda )\\ &&\quad \times (1+2\varepsilon \cos \alpha +{\varepsilon }^{2})\}=({\tau }^{2}-{s}^{2}){(1+2\varepsilon \cos \alpha +{\varepsilon }^{2})}^{2}.\end{eqnarray} \tag{ 66 }$$

This is a polynomial of the 6th degree in ${\rm{\Delta }}t$. Keeping only the leading order terms, we need only solve a quadratic in ${({\rm{\Delta }}t)}^{2}$:

$$\begin{eqnarray}&&\displaystyle \frac{1}{4}\displaystyle \frac{{(1+k)}^{2}{g}^{2}{\sin }^{2}\alpha }{{(1-k)}^{2}}{({\rm{\Delta }}t)}^{4}-\displaystyle \frac{(1+k){{gs}}_{x}}{1-k}{({\rm{\Delta }}t)}^{2}\\ &&\quad -({\tau }^{2}-{s}^{2})=0.\end{eqnarray} \tag{ 67 }$$

Only one of the two solutions is physical; we find

$$\begin{eqnarray}&&{\rm{\Delta }}{t}_{\mathrm{col}}=\sqrt{\displaystyle \frac{2\tau (1-k)}{g\sin \alpha (1+k)}}{\left(\sqrt{1-\displaystyle \frac{{s}^{2}}{{\tau }^{2}}{\sin }^{2}\beta }-\displaystyle \frac{s}{\tau }\cos \beta \right)}^{1/2},\end{eqnarray} \tag{ 68 }$$

which is the same as Equation (36).