# Maximum velocity of ISON and some other stuff

1. Nov 29, 2013

### ilikescience94

Hey I'm trying to find out the maximum velocity that ISON had, should I use Newtonian physics or general relativity equations to find it? What are the values I should assign for mass and distance? Would I need to use differential equations because ISON melted? I don't want anyone giving me the answer, I want to struggle around a bit so I can understand it better, thanks.

On an unrelated note, I was wondering if dark matter worked like matter in how 99.999% of it is space, or if a different value is assigned to it, or do we not know anything about dark matter yet? Thanks again

2. Nov 29, 2013

### Student100

Well, it reached it's maximum velocity yesterday I believe. What you will need is observation data.

3. Nov 29, 2013

### AlephZero

Work it out using Newtonian physics, and compare what you get to the speed of light. Then decide if you know your input data accurately enough for relativity to make a meaningful difference to the answer.

4. Dec 1, 2013

### Chalnoth

And for comparison's sake, you usually need velocities greater than around ten percent of the speed of light for relativity to make very much difference. Obviously this threshold depends a bit upon how accurate you want your measurement to be. But the rule of thumb is that the correction is about $v^2/2c^2$ for lower velocities. So if the velocity is 10% of the speed of light, the correction is about 0.5%.

5. Dec 3, 2013

### Dotini

Spaceweather.com reported T-1 (Nov 27) that ISON was hurtling toward the sun at
240,000 mph.

6. Dec 3, 2013

### ilikescience94

http://www.fallofathousandsuns.com/size-of-comet-ison.html

"The studies determined that Comet ISON is expelling roughly 130 pounds (60 kg) of water and 112,000 pounds (51,000 kg) of dust every minute."

Because of this, will I need to do a differential equation?

Other things I know: Mass of sun(1.989E^30kg), G, and if I knew Ison's distance from the sun as well as the mass at that moment, then what else would I need to know?

Last edited: Dec 3, 2013
7. Dec 3, 2013

### Staff: Mentor

The expulsion rate of matter is totally irrelevant here - not sure why you think it would matter.

The easiest way to calculate it - depending on what starting data you are allowed - is to take the distance at perihelion and use it to calculate the sun's escape speed at that altitude. The difference between the comet's extremely elliptical orbit and an escape trajectory would be negligible.

For a hint, the day of perihelion I was watching a tracker website a little too late and an hour or two after perihelion it was said to be traveling at 850,000 mph and decelerating rapidly. So you should be looking for an answer on the order of a million miles per hour.

8. Dec 3, 2013

### ilikescience94

wouldn't I be using F=G(M1M2/(R)^2), so the decay of ISON would change the M1 value, or would it not matter in the long run?

9. Dec 3, 2013

### Chronos

Expulsion of matter should have a small braking effect.

10. Dec 3, 2013

### ilikescience94

OK, is that the equation i should use as well?

11. Dec 3, 2013

### Chronos

Not unlike comet Kahoutek, ISON is another 'dud of the century'. Yes, I was around for Kahoutek, it too failed to live up to its billing as 'comet of the century'. Then again, that was last century. I'm confident another dud is in the offing.

12. Dec 3, 2013

### Staff: Mentor

It doesn't matter because the change in M1 is identical to the change in m in f=ma, so they cancel out. If you combine those two equations and integrate over infinite distance, you get....
No. I said use the escape velocity equation.

13. Dec 4, 2013

### ilikescience94

Oh so the maximum speed would have been achieved during exit

14. Dec 4, 2013

### Staff: Mentor

No, the maximum speed is at perihelion. Escape speed is a reversible concept: it gives you the maximum/final speed on plunge into an object from infinity, starting at low speed.

So have you found the equation/done the calculation yet? You should be able to find what you need by googling the key terms we have been discussing.

Last edited: Dec 4, 2013
15. Dec 4, 2013

### ilikescience94

So perihelion distance is rc/(1-e), where rc=(h^2)/GM, so what does e equal, and what should I place for h and M?

Or should I use Vescape=√(2GM/R), or a combination of the two?

Last edited: Dec 4, 2013
16. Dec 4, 2013

### Staff: Mentor

Just use the escape velocity equation. You have to know perihelion distance already to use it.

17. Dec 4, 2013

### ilikescience94

OK, so mass of sun is 1.989*10^30kg, the radius is 695,500,000m, G equals 6.67*10^-11, and got a value of 617,656m/s, or 2,223,560 km/hr, which I feel is a bit steep

18. Dec 4, 2013

### Staff: Mentor

Use perihelion distance for R. You just calculated the SURFACE escape velocity. Ison did a flyby; it didn't reach the surface.

19. Dec 4, 2013

### ilikescience94

What would e equal, and what would i use for h?

20. Dec 4, 2013

### Staff: Mentor

You can't calculate perihelion distance, you have to look it up. You can't solve a problem like this unless you have at least some information to start with!

21. Dec 4, 2013

### ilikescience94

ok, I found the perihelion to be 1.861*10^9m, and plugged that in to find 377,592m/s, or about 1,360,000km/s, which is still awful steep, was terminal velocity reached, did I screw up, or is there something I'm not accounting for?

22. Dec 4, 2013

### Staff: Mentor

Well in post #7 I predicted you'd find the value to be about a million mph and that is 840,000, so it is lower than I expected - and lower than I said I saw a few hours after perihelion in that post. Since it is lower than the surface escape velocity, which you correctly calculated (albeit accidentally), you are definitely in the ballpark...

But your perihelion distance is wrong. It is 10^6 km, not 10^9.
http://en.wikipedia.org/wiki/C/2012_S1
So please try again -- though if that is just a typo in the post, your calculation may be right..

As for "terminal velocity" - there is no such thing in this context.

Last edited: Dec 4, 2013
23. Dec 4, 2013

### ilikescience94

i said 10^9m, which is 10^6km, and I was just going off the spoiler of 244,000mph, which apparently is wrong

24. Dec 5, 2013

### Staff: Mentor

Right, my mistake.
Well, 244,000 mph may have been the speed the day before, but it accelerated very rapidly.
 so I did the calc and got 1.36 million km/hr or 843,000 mph. Close enough.

Last edited: Dec 5, 2013
25. Dec 5, 2013

### ilikescience94

Awesome thank youo for all of your help