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

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

Walter Correct answer

8 years agoComets 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.

**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.**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.**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.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?

http://www.wolframalpha.com/input/?i=how+fast+is+a+comet+moving+when+it+crosses+earth%27s+orbit%3F :-)

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

stevenvh 8 years ago

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