Calculation beyond computional limits

[short circut]

So i am ttying to calculate a velocity in this problem i dreamt up. Only problem i can't get a computer to solve it. So i was hoping someone here could help

i am trying to find the value of x such that

\frac{\sqrt{1-x^2}}{x}=1.0967*10^{-86}

This arose in me trying to calculate somewhat relativistically how fast a raindrop would have to travel to for into a black hole. My result is very crude.

edit:

I think i have it now. And i can't find delete

 

[rasmhop]

Let a = 1.0967*10^{-86}

Square the equation to get it to get
\frac{1-x^2}{x^2} =a^2
Now solve for x^2 and take the positive square root (x must be positive since a is positive).

x is going to be VERY close to 1 (in fact I expect most software will report x=1 if you ask for a numerical answer).

 

[short circut]

Let a = 1.0967*10^{-86}

Square the equation to get it to get
\frac{1-x^2}{x^2} =a^2
Now solve for x^2 and take the positive square root (x must be positive since a is positive).

x is going to be VERY close to 1 (in fact I expect most software will report x=1 if you ask for a numerical answer).

Yeah everything is reporting 1. Thats what i was trying to get around. I don't think i can get around it. Because that gives me a ratio of sqrt(1/(1.(86 zeros)1)) Which i can't even imagine. So basically you have to get that v/c ratio of the speed of light to turn a raindrop into a black hole semirelativistically. I am not well versed in relativistic fluid dynamics so i imagine this is quite a ways off. Plus not to mention i only have 3 sig figs. So i could never actually measure this anyways.

 

[litup]

Don't you have a computing center in your school? Oh, maybe you are in HS, sorry then. Most colleges have computers that with something like Mathematica, can give quadruple and more precision. At 16 digits per precision, I guess you would need sextuple precision or so, it certainly can be done with the right software.

 

[rasmhop]

Well if you really want a numeric approximation I just asked Maple to compute 250 digits of this. I'm not sure exactly how Maple does its floating-point computation, but I suspect the long string of 0's at the end is a sign that it doesn't handle such high-precision numbers by default (but up to something like 150 digits it seems correct).

0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999398624555000000000000000000000000000000000000000000000000000000000000000000000

I have no idea how this could be of any use however.

 

[uart]

No need need for any computational power here. You can solve it using a binomial expansion to a VERY close approximation as :

x \simeq 1 - \frac{a^2}{2}

where a= 1.0967 \times 10^{-86}