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marmoset
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Homework Statement
Two bodies mass [itex]m_1[/itex] and [itex]m_2[/itex] are a distance [itex]x[/itex] apart. How long do they take to collide as a result of their gravitational attraction to each other.
Homework Equations
This question was asked in another thread which I found in a search. However the final answer was never posted, and I am trying to work it out.
This is a post from the thread.
D H said:Here are two ways to arrive at a description of the time evolution of the separation distance.
Non-inertial frame
Choose one of the masses -- say m1 -- as the origin of a reference frame. The goal is to describe the location of the other mass (m2) as a function of time. Denote the x-hat axis of our frame as pointing toward mass m2 and denote x as the distance between the two masses. The equation of motion in this accelerating frame is
[tex]a \equiv \frac {d^2 x \hat x}{dt^2}
= \frac {F_{\text{net}}}{m_2} - \ddot R[/tex]
where F_net is the net force acting on mass m2 and R-double-dot is the frame acceleration. By Newton's law of gravitation and Newton's third law,
[tex]
\begin{aligned}
F_{\text{net}} &= -\,\frac{Gm_1m_2}{x^2} \hat x \\
\ddot R &= - \, \frac {F_{\text{net}}}{m_1} = \frac{Gm_2}{x^2} \hat x
\end{aligned}
[/tex]
While the reference frame is accelerating, it is not rotating. The unit vector x-hat thusly is constant. Combining the above results,
[tex]a =
-\,\frac{Gm_1}{x^2} - \frac{Gm_2}{x^2} = -\,\frac{G(m_1+m_2)}{x^2}[/tex]
Define [itex]v[/itex] as the first time derivative of [itex]x[/itex]. Then
[tex]a = \frac {dv}{dt} = \frac{dv}{dx}\,\frac {dx}{dt} = v\frac{dv}{dx}[/tex]
Combining with the previous result,
[tex] v\frac{dv}{dx} = -\,\frac{G(m_1+m_2)}{x^2}[/tex]
This is a separable and easily differential equation, giving v as a function of x. Integrating velocity with respect to time will yield the desired time desciption of the displacement vector.
The Attempt at a Solution
I then get that:
[tex] \int v dv = -G(m_1+m_2) \int x^{-2} dx\\ [/tex]
[tex]\frac{v^2}{2} = \frac{G(m_1+m_2)}{x} + c\\ [/tex]
[tex]\frac{dx}{dt} = \sqrt{\frac{2G(m_1+m_2)}{x} + c}
[/tex]
However from here I am not sure how to get x as a function of t. I think it is explained later in the thread (for reference, https://www.physicsforums.com/showthread.php?p=1791110#post1791110) but I do not understand.
I tried to integrate both sides with respect to t, but then I couldn't rearrange it to get x on its own on one side.
Thanks for the help :)
(edited many times to get latex right, first time I've done it)
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