How fast is a comet moving when it crosses Earth's orbit?

  • Is it about the same as Earth's orbital speed?


    It depends. An object's speed is related to the length of it's orbit, and this varies greatly. See Kepler's Laws.

  • Walter

    Walter Correct answer

    8 years ago

    1. Comets don't cross Earth's orbit really. Orbits are one-dimensional objects and their chance of crossing in 3D space is 0. Henceforth, I consider a comet at distance 1AU from the Sun.


    2. What's the maximum speed of a returning comet at 1AU from the Sun? This can be easily worked out from the orbial energy
      $$
      E = \frac{1}{2}v^2 - \frac{GM_\odot}{r},\qquad\qquad(*)
      $$
      which is conserved along the orbit ($v$ and $r$ the Heliocentric speed and distance). For a returning comet, $E<0$ and the speed cannot exceed the escape speed (which occurs for $E=0$)
      $$
      v_{\rm escape}^2 = 2\frac{GM_\odot}{r}.
      $$
      The speed of the Earth can be worked out from the Virial theorem, according to which the orbital averages of the kinetic and potential energies, $T=\frac{1}{2}v^2$ and $W=-GM_\odot/r$, satisfy
      $
      2\langle T\rangle + \langle W\rangle=0.
      $
      For a (near-)circular orbit (such as Earth's), $r$ is constant and we have
      $
      v^2_{\rm Earth} = GM_\odot/r.
      $
      Thus, at $r$=1AU
      $$
      v_{\rm escape} = \sqrt{2} v_{\rm Earth}
      $$
      as already pointed out by Peter Horvath. Non-returning comets have local speed exceeding the escape speed.


    3. Can a comet near Earth have a speed similar to Earth's orbital speed?. Let's assume a comet with the same speed as Earth at $r$=1AU and work out the consequences. Such a comet must have the same orbital energy as the Earth and, since
      $$
      E = -\frac{GM_\odot}{2a}\qquad\qquad(**)
      $$
      with $a$ the orbital semimajor axis, must also have $a=1AU$ and the same orbital period as Earth, i.e. one year. Moreover, the comet's apohelion satisfies
      $$
      r_{\rm apo}\le 2a = 2{\rm AU}.
      $$
      Such comets don't exist AFAIK. Most returning comets have much longer periods than 1 year.


    4. What's the typical speed of a returning comet when at distance 1AU from the Sun? To work out this question, let's parameterise the comet's orbit by its period $P=2\pi\sqrt{a^3/GM_\odot}$. From this relation we immediately get
      $$
      \frac{a_{\rm comet}}{\rm AU} = \left(\frac{P_{\rm comet}}{\rm yr}\right)^{2/3}.
      $$
      From equations ($*$) and ($**$), we can then find
      $$
      v_{\rm comet}(r=1{\rm AU}) = \sqrt{2-\left(\frac{P_{\rm comet}}{\rm yr}\right)^{-2/3}}\,v_{\rm Earth}.
      $$
      In the limit of $P\to\infty$, this recovers our previous result $v_{\rm comet}\to v_{\rm escape}$. For typical period of $\sim$70yr, the speed of the comet is close to this number.


    5. Finally, I shall coment that all this only relates to the magnitude of the orbital velocity (speed), but not to its direction. Comets are typically on highly eccentric orbits and, when at $r$=1AU, move in quite a different direction than Earth, even if their speed is only slightly larger. So the relative speed $|\boldsymbol{v}_{\rm comet}-\boldsymbol{v}_{\rm Earth}|$ of a comet with respect to Earth can be anyting between about 10 and 70 km/s.



    This is a great answer! You're better than Wolfram! This is exactly the information and tools that I was looking for. Thanks very much. Doug.

    who is Wolfram?

    @Walter the answer to "**Who** is Wolfram" is either Stephen Wolfram or perhaps Wolfram Research. But of course user38715 was referring to wolframalpha.com

License under CC-BY-SA with attribution


Content dated before 7/24/2021 11:53 AM

Tags used