Closest approach from initial velocity and impact parameter

  • #1
Kelli Van Brunt
11
3
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"?
 

Answers and Replies

  • #2
TSny
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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.
 
  • #3
Kelli Van Brunt
11
3
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?
 
  • #4
TSny
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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.
 
  • #5
Kelli Van Brunt
11
3
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?
 
  • #6
TSny
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##\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.
 
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