Hale Bopp has a period of 2400 days, with a mean distance from the sun at 174 AU and 1AU away in perihelion and 357 AU away in aphelion, what is its speed at the perihelion?

what formulas do you need in order to get this? We have not even tackled kinetic and potential energy in class but that was the topic i found when i searched it up


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  • $\begingroup$ The title asks for speed at aphelion, but the text asks for speed at perihelion. Which is it? $\endgroup$ – Mike Feb 3 '18 at 14:47
  • $\begingroup$ perihelion sorry $\endgroup$ – SuperMage1 Feb 3 '18 at 15:04
  • $\begingroup$ Approximately twice the mean speed? $\endgroup$ – Pieter Feb 3 '18 at 15:07
  • $\begingroup$ What have you studied? Kepler's Laws? $\endgroup$ – Mike Feb 3 '18 at 15:07
  • 1
    $\begingroup$ @sammygerbil Certainly not with a deep understanding of the physics. But in practice, a lot of students will study Kepler's Laws (including basic facts about ellipses, areal velocity versus angular velocity, etc.) before energy. And I think my answer shows that that is enough to tackle the problem. After all, Kepler himself predates our notions of kinetic and potential energy. $\endgroup$ – Mike Feb 5 '18 at 17:46

Given the numbers you quoted, you can easily calculate(*) the semimajor axis $a$ and the semiminor axis $b$. From these, you should know that the total area of the ellipse that the comet follows is $A = \pi\, a\, b$. Now, Kepler's second law says that the rate at which area is swept out by the line between the Sun and the comet is constant. We can calculate the rate over an entire period $P$ of the comet: \begin{equation} \frac{dA}{dt} = \frac{\pi\, a\, b} {P}. \end{equation} Again, this quantity is constant, no matter whether it's over some infinitesimal amount of time, or the whole orbit. Now, at perihelion (or aphelion), there's no radial motion — it's all angular. So if the comet moves some angle $d\theta$ while it's at radius $r$, the area it sweeps out is $dA = \frac{1}{2} r^2\, d\theta$. But we also know that the velocity is just $r\, d\theta/dt$, so we have \begin{equation} \frac{dA}{dt} = \frac{1}{2} r^2 \frac{d\theta}{dt} = \frac{1}{2}r\, v. \end{equation} You know what $r$ is, so just solve for $v$, and you've got your answer.

The Wikipedia link I gave has some more details about these steps, if you're wondering.

(*) The semimajor axis is the (arithmetic) mean of the distance at perihelion $r_\mathrm{per}$ and the distance at aphelion $r_\mathrm{ap}$: \begin{equation} a = \frac{r_\mathrm{per} + r_\mathrm{ap}} {2}. \end{equation} Meanwhile, the semiminor axis is the geometric mean of those distances: \begin{equation} b = \sqrt{r_\mathrm{per}\, r_\mathrm{ap}}. \end{equation}

  • $\begingroup$ isn't the semi major axis already the mean distance? $\endgroup$ – SuperMage1 Feb 4 '18 at 2:26
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    $\begingroup$ It depends on how you take the mean. See here for more detail. $\endgroup$ – Mike Feb 4 '18 at 13:24

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