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enter image description hereA comet, traveling with speed $V$, is at a very great distance from a star. If the path of the comet were unaffected by the presence of the star it's closest distance of approach would be $d$. However, the comet experiences an attractive force $\frac{\gamma}{r^2}$ per unit mass directed towards the star. Calculate the actual distance of the closest approach.

The first part of this exercise was to prove that $$\frac{1}{2}mv^2+\int{F(r)dr}=\ const.$$ which I proved.
May I use this instead of Binet's equation?

I've done this:
$\vec r=r\vec{e_r}$ - position vector
$\dot{\vec r}=\dot r\vec{e_r}+r\dot\theta\vec{e_\theta}$ - velocity
$F(r)=\frac{\gamma m}{r^2}$
$\int F(r)dr=\int \frac{\gamma m}{r^2}dr=-\frac{\gamma m}{r}$
Hence,
$$\frac{1}{2}mv^2-\frac{\gamma m}{r}=\frac{1}{2}mV^2-\frac{\gamma m}{d}$$
Since $v=0$ at an apse, I have $$-\frac{\gamma m}{r}=\frac{1}{2}mV^2-\frac{\gamma m}{d}$$ at the closest approach.
From this I have that a distance of the closest approach is $$r=\frac{2\gamma d}{dV^2+2\gamma}$$

dfeuer
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gov
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    Are you sure that $v=0$ at an apse ? – Tony Piccolo Aug 27 '13 at 22:06
  • Thanks, I see that mistake now. $\dot r=0$ at an apse. And $\dot{\vec r}^2=\dot r^2+r^2\dot\theta^2$. So, if there's no force, the closest distance would be $d$ and velocity would be $V$, perpendicular to position vector. It means that $r\dot\theta=V$ and $r=d$, so $r^2\dot\theta=dV$. I can express $r^2\dot\theta^2$ from this. But can I do it this way or I have to use Binet's equation? – gov Aug 27 '13 at 22:12
  • Can you give more context ?: where did you find the exercise and is your question the exact statement ? – Tony Piccolo Aug 27 '13 at 22:24
  • Mary Lunn - "A First Course in Mechanics". My question is if I can use energy equation instead of Binet's equation. – gov Aug 27 '13 at 22:28
  • Since this is the mathematics stack exchange, you should explain any physics concepts you use, including specifically "energy equation" and "Binet's equation". At the very least, please link to a readable explanation of these. Also, what does "ate an apse" mean? – dfeuer Aug 28 '13 at 02:10
  • "ate an apse" means that I made a typing mistake :D I will write explanations as soon as possible. – gov Aug 28 '13 at 09:46
  • There are many problems in your approach. e.g. 1) If you have an attractive potential, the sign infront of $\frac{\gamma m}{r}$ is negative. 2) At periapsis, $v$ itself doesn't vanish, only the radial component vanishes. – achille hui Aug 28 '13 at 10:02
  • @achillehui I know that $V=-\int F(r) dr$ but I used equation $$\frac{1}{2}mv^2+\int{F(r)dr}=\ const.$$ so I didn't need to write $-$ in front of integral. And I am pretty sure that $$\int F(r)dr=\int -\frac{\gamma m}{r^2}dr=\frac{\gamma m}{r}$$ is correct. Also, I corrected myself about velocity in comment, but I haven't corrected it yet in question. I will do that soon. – gov Aug 28 '13 at 10:18
  • $$\begin{align} & \frac{d}{dt}\big[\frac12 m v^2\big] = \vec{F}(r(t))\cdot\vec{v}(t) = F_r(r(t)) \frac{dr(t)}{dt}\ \implies & \big[ \frac12 m v^2\big]_d^r \color{red}{-} \int_d^r F_r(s) ds = 0\ \implies & \frac12 m v^2 \color{red}{-} \int^r \big( -\frac{\gamma m}{s^2} \big) ds = \frac12 m v^2 \color{red}{-} \frac{\gamma m}{r} = \text{constant.} \end{align}$$ where $F_r$ is the radial component of the force and are considered to be positive when it is pointing outward. – achille hui Aug 28 '13 at 10:44
  • I have this:$$\frac{d}{dt}(\frac{1}{2}m\vec v^2)=\frac{d}{dt}(\frac{1}{2}m\vec v\vec v)=m\vec v\frac{d\vec v}{dt}=m\frac{d \vec v}{dt}\vec v=\vec F \vec v=\vec F\frac{d\vec r}{dt}$$ $$\implies d(\frac{1}{2}m\vec v^2)=\vec F d\vec r$$ $$\vec r=r\vec e_r\implies d\vec r=dr\vec e_r+rd\theta \vec e_\theta$$ $$\vec F d\vec r=-F(r)\vec e_r(dr\vec e_r+rd\theta \vec e_\theta)=-F(r)dr=-d\int F(r)dr(=\frac{d}{dt}(\frac{1}{2}m\vec v^2))$$ $$d(\frac{1}{2}m\vec v^2+\int F(r)dr)=d(\frac{1}{2}m\vec v^2)+d\int F(r)dr=-d\int F(r)dr+d\int F(r)dr=0$$ $$\implies \frac{1}{2}m\vec v^2+\int F(r)dr= const.$$ – gov Aug 28 '13 at 11:03
  • In your derivation, $\vec{F} = -F(r)\hat{e}_r$. This means $F(r) = |\vec{F}|$ and hence equal to $\frac{\gamma m}{r^2}$ instead of $-\frac{\gamma m}{r^2}$. – achille hui Aug 28 '13 at 11:18
  • Yes, you are right. I corrected it. Is it ok now? Is my proof correct? – gov Aug 28 '13 at 11:32
  • Your proof is still incorrect. At periapsis, only the radial component of $v$ vanishes, the tangential component didn't. You should use conservation of angular momentum to determine the correct tangent component of $v$ there. This is essentially what Tony's answer doing. – achille hui Aug 28 '13 at 17:42
  • @achillehui I asked if my proof for $$\frac{1}{2}mv^2+\int{F(r)dr}=\ const.$$ was correct. – gov Aug 29 '13 at 13:28
  • @gov, that part of proof is correct. – achille hui Aug 29 '13 at 14:39

1 Answers1

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In effect you can bypass Binet.

Start from the conservation laws $$r^2 \dot \theta=dV$$$$\dot r^2+r^2 \dot \theta^2=V^2+2 \gamma \left (\frac 1{r}- \frac 1{d} \right)$$Obtain $\dot \theta$ from the first and substitute in the second.
Then set $\,\dot r=0$ and solve with respect to $r$.
The solutions are $$r=d \quad ,\quad r= \frac {d^2V^2}{2\gamma-dV^2}$$If $\,V<\sqrt \frac {\gamma}{d}$, then $$\frac {d^2V^2}{2\gamma-dV^2}<d$$

Tony Piccolo
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