### 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 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

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