Okay there was this dude on the interwebz who was like:
>if you're stranded 10 meters away from the ISS… wouldnt it pull you in, with gravity?
So i was like "sure, maybe i can calculate how long it would take to pull you back".
But I've spent hours with this and i think i still have wrong answer.
So yeah, I KNOW THIS IS WRONG, BUT I DON'T KNOW IN WHICH PART I'VE MADE MISTAKE.
Pls help red /sci/onymous.
At first i thought it will be simple:
I'm just going to take equation for gravitational force F = G * ( m1 * m2 / r^2) and then you put this force into newton's second law of motion a = F / m and then calculate the time by using t = sqrt (2s / a)
But then i realized that it would be inaccurate, because as you get closer that gravitational force between you and space station would increase.
I can't do any higher level mathematics for solving this accurately, so i cheated. I said: Let's make it much more accurate by dividing this path into more chunks, calculate everything for those chunks separately and then add those times together. It is cheating but it's best i can do.
Problem is that i don't know if i used enough of these segments.
And another big problem is that ISS is not just mass point, but it has some volume and as you get closer the gravitational force disperse to the sides. I have not accounted for that.
Also i didn't count gravitational pull the other way that you have on ISS, because i've tried to calculate it and it looks negligible.
So this is only rough approximation.
The mass of ISS is about 450t according to wikipedia. Let's say your mass is 80kg. Gravitational constant is 6,67·10^-11 [of bullshit units].
Also we must assume there is absolutely 0 relative velocity between you and ISS.
First segment:
F = k * ( m1 * m2 / r^2)
F = 6,67*10^-11 * ((450 000 * 80) / 10^2)
F = 0.000024012 N
a = F / m
a = 0.000024012 / 80
a = 3.0015*10^-7
t = sqrt (2s / a)
h will be 2m because second segment will be 8m so we calcullate it only for 2m
t = sqrt (2*2 / 3.0015*10^-7)
t = 3650.57 s
Second segment:
F = 6,67*10^-11 * ((450 000 * 80) / 8^2)
F = 0.00003751875
a = F / m
a = 0.00003751875 / 80
a = 0.000000468984375
third segment will be 5m se we calculate only (8-5) m = 3m
But there is a problem: we already have some velocity, so i can't use t = sqrt (2s / a).
I think i should use this formula s = v0 * t + 0,5 * a*(t^2) and somehow get t from it.
And v0 calculate as v = v0(of previous segment) + a * t
v0 of previuos segment is in this case 0 but in the next segment it should be this v0
I'll be honest, i cheated and used this online calculators:
http://planetcalc.com/981/
and
http://www.calculatorsoup.com/calculators/physics/velocity_a_t.php
And this is what i got:
v0 = 0.0010957185855 m/s
t = 1935.89 s
Third segment:
F = 6,67*10^-11 * (36000000 / 5^2)
F = 0.000096048
a = 0.000096048 / 80
a = 0.0000012006
Next segment is 2m so we put here (5-2) = 3m
Time for online calculator again…
v0 = 0.0020036207472187
t = 1120.87 s
4th segment:
F = 6,67*10^-11 * (36000000 / 2^2)
F = 0.0006003
a = 0.0006003 / 80
a = 0.00000750375
length of segment is 2 - 1 = 1
Calc time…
v0 = 0.0033493372692187
t = 236.12 s
5th segment:
F = 6,67*10^-11 * (36000000 / 1^2)
F = 0.0024012
a = 0.0024012 / 80
a = 0.000030015
length is 1 - 0,5 = 0,5
now calc
v0 = 0.0051211227192187
t = 79.24 s
6th segment
F = 6,67*10^-11 * (36000000 / 0,5^2)
F = 0.0096048
a = 0.0096048 / 80
a = 0.00012006
length 0,5 because that is final segment
calc pls
v0 = 0.0074995113192187
t = 48.13 s
Now just add up all times together and bam 48.13 + 79.24 + 236.12 + 1120.87 + 1935.89 + 3650.57 = 7070.82 s
7070.82s = 117.847 min = almost 2 hours
But this doesn't make a sense. Only 2 hours is a bullshit.
So what the heck did i do wrong?
I know, you can solve this with your high level math, but i would appritiate if someone use the same method as me and tell me in which part i failed.