Can ISON hit the Earth? WARNING math inside

Here's the real math on how much of the comet's mass it would take to propel the remaining mass to earth. Let's be super generous, let us assume that the volatiles of the comet will jet out in exactly the right direction to send the maximum possible remaining mass to earth, thus the comet's entire mass of volatile ices can be used as "fuel" to provide the thrust needed to send it to earth. That is the ultimate in generous assumptions. The question is then how much would remain to reach earth if that were the case. It would take over 57000 m/s delta-V, 57 km/s, to send any fragment of ISON to earth at perihelion. Furthermore, the delta-V would have have to push the comet in a sunward direction.
[link to imageshack.us]
That is the delta-V that must be achieved the volatile ices in the comet. Given an "exhaust velocity" of around 900 m/s for a comet (Combi et al 1997 [link to link.springer.com] ) we can calculate how much of the comet would have to act as "volatile fuel" and how much of it would remain to reach earth. Let's be extremely generous, ridiculously so. Let's say ISON is on the "friggen huge" range of comet nucleus sizes and has a diameter of 50 km, similar to Hale-Bopp. It's not that big, but for the sake of being generous to this example, let's just pretend it is. Let's also assume it is at the top end of the range of comet density, 1.2g/cm^3. It would therefore have a mass of about 7.854 x 10^16 kilograms. In order to deliver 57 km/sec of delta-V, all but about 0.0000000000245 kg of the comet would have to be "fuel" and would never reach earth. That is because delta V = Exhaust velocity * ln(total mass/final[payload] mass) The "payload fraction" that reaches earth would have a whopping mass of 0.0000000000245 kg.

In terms of more realistic numbers, comet separation velocities of major fragments are in the tens of m/s, orders of magnitude too small to send any to earth:
[link to www.lpi.usra.edu]

Astro! Finally a man with knowledge.
Here is my question for you.
So we can't use kinetic energy of the comet. We agree that the material of the comet can propel the comet or part of it. Let's assume for simplicity that the comet is divided to two parts. One of mass m1 (that will fly down relative to the pass in order to hit the Earth), another of mass m2 that will propel m1. To keep the momentum

m1 x V1 = m2 x V2
and also
m1 + m2 = m

we know V1 = 5.7 x 10e4
V2 = 9 x 10e2

thus:

m1 = m2 x V2 / V1 = 0.01578 x m2
m2 = 63 x m1

In other words in order to throw one tone chunk with a speed 57 km/s you need 63 tons of material. To propel it. Not sure how you came up to your tiny number.

Incorrect. You're not using the right formula for the numbers you're plugging in. To achieve a given delta-V (change in velocity) using a given portion of the mass of an object you need to apply the Tsiolkovsky rocket equation. delta V = Exhaust velocity * ln(total mass/final[payload] mass) That's the natural log of total mass/final mass. A simple diagram of the situation:
[link to upload.wikimedia.org]
It's the same situation as your comet question, only we're talking about a comet vs a rocket, but the basic principle is the same; sending some unknown fraction of the mass of the original object one way at a given velocity to propel the remaining mass the other way at a given velocity.

Thank you, that makes sense. But then:

V1 = V2 x (m1+m2)/m1
m1xV1 = (m1+m2) x V2

m2 = m1 x (V1+V2)/V2
m2 = 64 m1

You still don't appear to be using the Tsiolkovsky rocket equation? Solving for the remaining mass of the comet the formula looks like this:
m1 (final mass) = m0 (total mass)*e^-(delta V/exhaust velocity)
Therefore m1 = 0.000000000024534302 kg

What's about you?

You are spamming not only this thread with your ridiculous drivel.
ADHD?

.

well, the Russians will straighten this whole mess out,
when they published their findings on Friday.

What's about you?

You are spamming not only this thread with your ridiculous drivel.
ADHD?

.

Wishful thinking? - Dream on.

.