Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Advanced Physics Homework Help
Closest approach from initial velocity and impact parameter
Reply to thread
Message
[QUOTE="Kelli Van Brunt, post: 6240506, member: 653509"] [B]Homework Statement:[/B] 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. [B]Relevant Equations:[/B] 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"? [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Advanced Physics Homework Help
Closest approach from initial velocity and impact parameter
Back
Top