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

• user38715

8 years ago

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.

• 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