# Closest approach from initial velocity and impact parameter

• Kelli Van Brunt
In summary, the conversation discusses the process of finding the minimum value of r for a particle in a two-body orbit around a planet, taking into account energy and angular momentum conservation. The method of taking the kinetic energy to be maximum at the instant of closest approach is found to be incorrect, and instead, it is determined that the value of dr/dt at the point of closest approach is needed. The final solution for r is determined to be r = (-GM - √((GM)^2 + v_0^4b^2))/v_0^2, and the eccentricity of the orbit is found to be e = √(1 + 2EL^2/(GMm)^2). However, when the
Kelli Van Brunt
Homework Statement
A particle with speed v0 and impact parameter b starts far away from a planet of mass M. Starting from scratch, find the distance of closest approach to the planet.
Relevant Equations
E = (mv^2)/2 + L^2/2mr^2 - GMm/r
Here were my assumptions: Energy and angular momentum are both conserved because the only force acting here is a central force. The initial angular momentum of this particle is ##L = mv_0b## and we can treat E as a constant in the homework equation given above. I solved for the KE (1/2 mv^2) in the above equation: $$\frac{mv^2}{2} = E + \frac{GMm}{r} - \frac{L^2}{2mr^2}$$
Then I set the derivative with respect to r equal to zero in order to find r at the instant at which KE is maximum, ie. the instant at which the particle is closest to the planet. This left me with: $$\frac{L^2}{mr^3} = \frac{GMm}{r^2}$$
When this is simplified and L replaced with the expression for L given above, I got my final answer, $$r = \frac{(v_0b)^2}{GM}$$
My book does not technically give a solution to this problem, but in the second part of the exercise, it asks to show that my expression for r is equivalent to ##\frac{k}{e+1}##, where ##k = \frac{L^2}{GMm^2}## and e = eccentricity of the orbit. Plugging in what I had for L, I got ##\frac{(v_0b)^2}{GM(e+1)}##. This is only equal to my answer if the eccentricity is zero, ie. if the orbit is circular, which is clearly not the case, since the particle is coming in from far away. Where did I make my mistake here? How would I implement the eccentricity of the orbit into my original solution "starting from scratch"?

Kelli Van Brunt said:
Homework Equations: E = (mv^2)/2 + L^2/2mr^2 - GMm/r
This equation isn't correct if v represents the speed of the particle. However, I think the equation would be correct if v is replaced by dr/dt. Check the references where you got this equation.

Note that the angular momentum term comes from ##\frac{1}{2} m v^2 =\frac{1}{2}m \left( \dot r^2 + r^2 \dot \theta^2 \right)## and then expressing ##r^2 \dot \theta^2 ## in terms of the angular momentum.

TSny said:
This equation isn't correct if v represents the speed of the particle. However, I think the equation would be correct if v is replaced by dr/dt. Check the references where you got this equation.

Note that the angular momentum term comes from ##\frac{1}{2} m v^2 =\frac{1}{2}m \left( \dot r^2 + r^2 \dot \theta^2 \right)## and then expressing ##r^2 \dot \theta^2 ## in terms of the angular momentum.

dr/dt is correct, sorry - that was what I intended v to mean but I didn't clarify that, so thank you. Does this mean that my method of taking the KE to be maximum at the instant of closest approach is incorrect, since ##\frac{1}{2} m \dot r^2## only accounts for the radial KE of the particle?

Kelli Van Brunt said:
Does this mean that my method of taking the KE to be maximum at the instant of closest approach is incorrect, since ##\frac{1}{2} m \dot r^2## only accounts for the radial KE of the particle?
Yes, that won't work. ##\frac{d}{dr} \left( \dot r^2 \right )## is not equal to zero at the instant of closest approach.

No calculus is needed if you think about the value of ##\dot r## at the point of closest approach.

TSny said:
Yes, that won't work. ##\frac{d}{dr} \left( \dot r^2 \right )## is not equal to zero at the instant of closest approach.

No calculus is needed if you think about the value of ##\dot r## at the point of closest approach.

##\dot r## seems it would equal zero at the point of closest approach, since that will also be a turning point for the particle, so the radial distance is not changing at that instant. I assume this is true for all basic two-body orbits?

So, using ##E = \frac{m \dot r^2}{2} + \frac{L^2}{2mr^2} - \frac{GMm}{r}## and substituting ##E = \frac{mv_0^2}{2}## and ##\dot r = 0##, after multiplying through by ##r^2## and using the quadratic formula, we get $$r = \frac{-GM - \sqrt{(GM)^2+v_0^4b^2}}{v_0^2}$$

I assume we are taking the negative root of the solution here, since we want the minimum value of r. The eccentricity of this orbit is given by $$e = \sqrt{1 + \frac{2EL^2}{(GMm)^2}}$$

Substituting the expressions for E and L, we get $$e = \frac{\sqrt{(GM)^2+v_0^4b^2}}{GM}$$

Combining the expression for e with the expression for r gives, finally, $$r = \frac{GM(e+1)}{v_0^2}$$

This is closer to the actual answer in that I managed to get eccentricity in there, and at least the units are correct, but it is not equivalent to the given ##\frac{(v_0b)^2}{GM(1+e)}##. I can't tell where I'm making a mistake here, unless my initial assumption of ##\dot r = 0## at the closest approach is incorrect?

Kelli Van Brunt said:
##\dot r## seems it would equal zero at the point of closest approach, since that will also be a turning point for the particle, so the radial distance is not changing at that instant. I assume this is true for all basic two-body orbits?
Yes

$$r = \frac{-GM - \sqrt{(GM)^2+v_0^4b^2}}{v_0^2}$$

I assume we are taking the negative root of the solution here, since we want the minimum value of r.
The polar coordinate ##r## is always nonnegative. Once you fix that, you should be able to get the desired result.

Kelli Van Brunt and hutchphd

## What is closest approach from initial velocity and impact parameter?

Closest approach from initial velocity and impact parameter is a calculation used in physics to determine the closest distance between an object's initial velocity and its impact parameter. This is often used in predicting the path of a projectile or determining the likelihood of a collision between two objects.

## How is closest approach calculated?

The closest approach is calculated by using the initial velocity and impact parameter of an object in a mathematical equation. This equation takes into account the object's velocity, angle of trajectory, and the distance from its initial position to the point of impact. The resulting value is the closest distance between the object's initial velocity and impact parameter.

## What is the significance of closest approach?

The closest approach is significant because it helps scientists and researchers understand the potential trajectory of an object and its likelihood of colliding with another object. This information is crucial in predicting the behavior of objects in space, such as asteroids, comets, and other celestial bodies.

## Can closest approach be used to determine the outcome of a collision?

Yes, closest approach can be used to determine the outcome of a collision between two objects. By calculating the closest distance between the initial velocity and impact parameter of two objects, scientists can determine if a collision is likely to occur and the potential severity of the impact.

## Are there any limitations to using closest approach for predicting collisions?

While closest approach is a valuable tool for predicting collisions, it does have limitations. This calculation assumes that both objects are moving in a straight line and do not take into account any external forces or factors that may influence the trajectory of the objects. Additionally, the accuracy of the calculation may be affected by errors in measuring the initial velocity and impact parameter of the objects.

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