1
$\begingroup$

Would a short-period comet, with a 2-mile nucleus, be able to come within 200,000 miles of Earth without breaking up?

$\endgroup$
  • $\begingroup$ James K, Rob Jefferies, and Ken Draco. Your answers are all very helpful, perhaps you will help with the rest of my problem. I need for the comet to pass by, causing minor damage, perhaps such as can come from a meteor shower. I need for this comet to return in 30 years and seriously damage planet Earth. $\endgroup$ – BoomerChick Sep 17 '17 at 8:49
  • $\begingroup$ Then you need a collision with Earth (make the comet smaller). Read about the Tunguska event including accounts of witnesses. Apparently it was a small comet or part of a comet (Don’t trust Wiki too much. Anybody edits anybody and posts anything nowadays). If we say roughly Tunguska was 50 meters in size, then if you have a comet 10 times as large, consider explosion and damage to be roughly 10$^3$=1000 times $\endgroup$ – user18491 Sep 17 '17 at 12:47
  • $\begingroup$ as devastating. So, raise to the cube to get a rough idea without misrepresenting things too much. A larger comet will also burst/disintegrate but may leave a crater. Use something like this purdue.edu/impactearth to get a rough idea or impact.ese.ic.ac.uk/ImpactEffects However, all this is still EXTREMELY rough approximations. Set density to 0.5 Hope it all helps and will suffice for the book. $\endgroup$ – user18491 Sep 17 '17 at 12:48
2
$\begingroup$

It wouldn't break up.

As my Granddad used to say "Close is only good in horseshoes". If the comet doesn't impact the Earth will be able to pass by unchanged. 200000 miles is a long way, about the distance of the moon from Earth. Asteroids have come this close before (eg 2011 XC2).

Comets can be broken up by close encounters with the sun, but the sun is rather hotter than the Earth [citation needed]. 200000 miles is also well outside the Roche limit for the Earth-comet system, and at the speed that comets travel relative to the Earth, tidal effects won't have enough time to act.

So the comet will pass the Earth quite happily. It is a rather small comet, and even the comet's coma won't interact with the Earth. After closest approach it is possible that the Earth could pass through the comet's tail, which would cause a meteor storm.

Even at 100000 miles, there is not much effect on satellites nearly all of which orbit at 22000 miles or less (22000 is geostationary orbit) It is plausible that they could be damaged by meteoroids.

The dust that comes of comets is all quite small. If you are looking for a "near miss that still causes damage" try this. The comet makes a close pass of Jupiter that causes it to split into two parts. The larger part misses Earth by 100000 miles but the smaller part (a 300m chunk that wasn't noticed until too late because story) hits Merchtem 10km outside Brussels

$\endgroup$
  • $\begingroup$ Would the same apply if the comet came within 100,000 miles of Earth? Might either one damage satellites? And might the meteor storm include fragments large enough to constitute minor, but not catastrophic damage from falling space debris? $\endgroup$ – BoomerChick Sep 16 '17 at 7:21
2
$\begingroup$

No, that isn't close enough. The Roche Limit for break up of a rigid body, caused by tidal forces is given by $$d = 1.26 R_C \left(\frac{M_E}{M_C}\right)^{1/3},$$ where $M_E$ is the mass of the Earth, $M_C$ and $R_C$ are the mass and radius of the comet, and $d$ is how close the comet can come to the Earth before tidal forces disrupt it.

If you have a comet with a 1.6 km radius and assume it has a density of 0.6 g/cc, then $M_C \simeq 10^{13}$ kg.

Putting this into the Roche Limit formula, we get $d = 17,000$ km. So the comet would have to get within 10,000 km of the surface of the Earth before it broke up.

$\endgroup$
  • $\begingroup$ While the formula is a good one, the calculations and assumptions make it look like we arrived at some exact figure (17,000). This is an extreme approximation with a lot more variables in fact. Such a clear-cut answer, in my opinion, is very misleading for the folks who are gonna read or rely on it. Hence, I'm sorry, but I'm voting it down. $\endgroup$ – user18491 Sep 16 '17 at 19:49
  • $\begingroup$ The calculation is as precise as the clearly stated assumptions. $\endgroup$ – Rob Jeffries Sep 16 '17 at 20:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.