Calculating how long it takes for a comet to reach a sun of radius zero

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Homework Help Overview

The discussion revolves around calculating the time it takes for a comet to reach a sun with a radius of zero, starting from a distance R_max. The problem involves gravitational potential energy and kinetic energy, utilizing a specific integral equation related to motion under gravity.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of the equation provided, questioning the correctness of their expressions for energy and potential energy. Some express uncertainty about the units of the time variable and the complexity of the integral.

Discussion Status

There is active engagement with various attempts to clarify the equations and integrals involved. Some participants have offered corrections and suggestions for simplifying the integral, while others are focused on ensuring the physics concepts are correctly applied. No consensus has been reached regarding the best approach to solve the integral.

Contextual Notes

Participants note the potential difficulty of the integral and express concerns about the relevance of the problem to their studies. There is mention of previous coursework and the desire for conceptual understanding rather than just computational results.

xdrgnh
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Homework Statement


Consider the extreme case that the comet is released from rest at a distance R_max from the sun. In this case L is actually zero. Use the technique described in connection with 4.58 to find how long the comet takes to reach the sun. The radius of the sun for now is zero.


Homework Equations


Equation 4.58 is T=∫dr/(E-U(r))^(1/2)

E should be -GM_sM_c/R_max

and U(r) should be -GM_sM_c/r


The Attempt at a Solution



So I intergrate this and my bound are 0 to R_max. I just want to know if I'm on the right track becasue the integral is very messy. Thanks.
 
Last edited:
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Is T supposed to be the time? If so, equation 4.58 doesn't seem to work out unit-wise.
 
Yes it is. Well I forgot to mult that integral by a (m/2)^(1/2). But that doesn't change the difficulty of the intergrand.
 
Without that factor, the units of T are 1/force. Multiplying by mass1/2 won't fix it.

In any case, I don't see why you think the integrand is a mess. It looks like it should succumb to a simple u-substitution.
 
Darn I made another mistake while typing I'll fix the OP re look at it. I added a square root.
 
$$ v = \frac{dr}{dt} \Rightarrow dt = dr/v $$ $$v = \sqrt{2K/m} = \sqrt{2[E - U(r)]/m}$$ $$t = \int\,dt = \int v^{-1} dr = \int \left(2\frac{E - U(r)}{m}\right)^{-1/2}\,dr$$

Is that the equation?
 
Yep.
 
Any word if my E-U(r) is correct?
 
xdrgnh said:
Any word if my E-U(r) is correct?

Yeah, the expression $$- G\frac{mM}{R} + G\frac{mM}{r} $$ looks okay to me. So the integrand would become $$\left[2GM \left(\frac{1}{r} - \frac{1}{R}\right) \right]^{-1/2} $$ Did you have any luck computing it?
 
  • #10
Not much luck. I haven't integrated any complicated integral since I took calc II two year ago. But I feel a trig sub might work. But I was mainly concerned with the physics part rather then the math. I doubt a question like this would appear on the exam and I',m only doing it for my own enrichment. But see how the trig goes if I can figure out how it would work. If not I might try a u sub. I know I should simplify it first.
 
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