https://www.physicsforums.com/showpost.php?p=2497098
Why does one need to subtract both accelerations? I just took one side, as I'm only looking at one particle and it's acceleration and not both. I don't understand why you do need to do this. Please enlighten me
To sum it up the value used in the calculations in the other threads are outdated and hence a different value for G has been used.
In 2006 the value was between 6.67361 and 6.67495
Now the value is between 6.67304 and 6.67464
If the value is truly between 6.67304 and 6.67361, then the prior...
As I said G needs to be bigger that 6.674 for the equation to work (approx. 6.67433)
So any true value under 6.674 will be even more off the true answer.
6.67384 will give me a time of 96139.5s
Edit:
Using the min. value at 6.67304e-11 I get 96145.3s
Using the max value at 6.67464e-11 I get...
http://www.wolframalpha.com/input/?i=integrate+sqrt%281%2F4G%29*sqrt%28r%2F%281-r%29%29dr+from+r%3D0+to+1+where+G%3D6.674e-11
I used pi as in the constant in my calculator and wolfram, and that should be quite accurate.
And for G i used 6.674e-11 (actual value is slightly below that, but to get...
I have now looked into the maths required and ran into a different problem:
dt = \sqrt{\frac{s_{0}}{2G(m_{1}+m_{2})}}\sqrt{\frac{s}{s_{0}-s}}ds was the start of the problem.
By substitution:
dt = \sqrt{\frac{s_{0}}{2G(m_{1}+m_{2})}}\frac{2s_{0}u}{(u^2+1)^2}du
Integrating that with the limits...
uO
Thank you very much.
I will need to learn Integration by substitution and read through both threads with this kind of question, until I understand it.
If I in future, by any means, need more help I will contact you via this thread agains, althy I believe this to be unlikely.
I have never in my life done integration by substitution.
I am still doing my AS-levels :D
I guess that's why I don't have a clue what youre doing.
Is there a way of solving this problem without integration by substitution?
I don't not understand where you get that equation from.
And what is u? What actually is c?
And if s0 is 1metre, then what is s?
Sorry, I don't really get what youre telling me :S
I can't make sense of your working when I don't know where it comes from, what it means and what u is.. :S
m = 1, hence the individual acceleration at the starting position is G/s or -G/s depending on which one you're looking at.
I am concentrating on one particle, so the force together is irrelevant.
The problem is, that the magnitude of the force increases the more the particle moves.
Lets all delete our posts? :D
v = ds/dt = 3.267*10-5 = s'(t) is what I did in my previous post.
s(t) = 3.267*10-5t + c
t = (s-c)/(3.267*10-5)
Now what?{s as of the fourth post means the total distance between the particles.
"v2/2 = -G(m + m)/s + c = -2Gm/s + c"
Where did you get the (m+m) from...
What? There is no negative square root when you subtract the second part. The problem is that the acceleration tends to infinity as the distance tends to zero. Thats why I used the lower limit of 0.1.
If I continue looking at it like that:
ds/dt = 3.267*10-5
s = 3.267*10-5t + c
t =...
That would mean:
v²/2 = -G/(1/2) - (-G/0)
v²/2 = -2G + G/0
which is not possible.
Lets say for the ease, or until you tell me a better way of doing it that the lower limit is 0.1 and not 0.
v²/2 = -2G + G/0.1 = 8G
v = √16G = 3.267*10-5
What does that tell me?
So my first attempt was correct, just needed to continue on that.
I got 2s, as each object only moves distance s as you also said. As both object move that distance, the distance between them is 2s.
Integrating would give me [v²/2]: = [-G/s]: (: representing the limits)
or v²/2 + b = -G/s + c...
That seems to be the formula for a uniform acceleration.
True the acceleration is 6.674 EE 10-11 m/S2, but only for the initial moment. As soon as it moves, the gravitation take more effect and the acceleration increases.
Edit:
In my first post I wrote:
t = √(2s/a)
which equals your equation...
There are two point masses in space, each 1kg and one metre apart from each other.
The only Force acting on the masses is the gravitational force between them.
After what time do they collide?
Gravitational Force F = Gm1m2/(2s)2 where m is the mass and s the distance to the center of both...