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So I am trying to figure out the solution of differential equation in this case. At $t=0$ a body of mass $m$ is at the origin. Other body of the same mass $m$ is at a distance $x$ from the origin. I want to find out what would be their respective distances from the origin after time $t$. None if the bodies are fixed , so they move due to mutual gravitational attraction. According to me the differential equation would be $\frac{\mathrm{d}^2x}{\mathrm{d}t^2}=G\frac{m}{x^2}$.

But how should I go about solving it ? I can solve 1st order equation and after some more study of differential equations I can solve some 2nd order equations.

But this equation doesn't seem to be easy.

Wolfram Alpha shows a long weird solution and that too without any $t$ in the solution.

How should I solve it and interpret the solution ?

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  • $\begingroup$ Is the problem in one dimension, or is $x$ just the distance from the origin in 3-D space? $\endgroup$ – preferred_anon Sep 6 '16 at 15:14
  • $\begingroup$ @DanielLittlewood No it's 1 D only ,x is distance along the x axis $\endgroup$ – Shashaank Sep 6 '16 at 15:18
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    $\begingroup$ I don't know if your equation is correct, but to solve, multiply both sides of the equation by $x'$ and integrate. You get $$\frac{1}{2} (x')^{2} = -G \frac{m}{x} + C$$ You can then separate and integrate again, though I think the solution will be in terms of elliptic functions unless $C = 0$. I actually did a very similar problem here. $\endgroup$ – Mattos Sep 6 '16 at 15:24
  • $\begingroup$ @Mattos That's a great way to solve the equation. $\endgroup$ – Shashaank Sep 6 '16 at 15:58
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If one of the bodies is fixed at the origin, then your equation is correct. One way to solve it is to multiply through by $\dot{x}$ and notice that the LHS is the derivative of $\frac{1}{2}\dot{x}^{2}$, and the RHS is the derivative of $-Gm/x$ (both differentiated wrt $t$). Finding the exact solution is then straightforward (although might look nasty).

On the other hand, if both the bodies are free, then they might move on complicated paths. But all is not lost: write $x_{1}$ for the position of the first mass, and $x_{2}$ for the second. Newton says $$\ddot{x_{1}}=\frac{Gm}{|x_{1}-x_{2}|^{2}}=-\ddot{x_{2}}$$ which suggests that we consider the motion of the quantity $y=\frac{1}{2}(x_{1}+x_{2})$ (in the more general case where the two particles have different masses, this is equivalent to considering the motion of their centre of mass). By the equation above, $\ddot{y}=0$, and if both particles start from rest this implies that $y$ remains fixed at the same spot forever.

How does that help? Well write $z=x_{1}-y=\frac{1}{2}(x_{1}-x_{2})$. Then $\ddot{z}=\ddot{x_{1}}-\ddot{y}=\ddot{x_{1}}$. So, from above, replacing $x_{1}-x_{2}$ by $2z$, $$\ddot{z}=\frac{Gm}{(x_{1}-x_{2})^{2}}=\frac{Gm}{4z^{2}}$$ So the relative separation follows the equation you wanted to solve in the first place!

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  • $\begingroup$ A perfect answer. Could you please just explain the last steps. z=(x1-y)=1/2(x1-x2) . So d^2z/dt^2 = 1/2(d^2 (x1)/dt^2 - d^2(x2)/dt^2) ? And particularly the last step. $\endgroup$ – Shashaank Sep 6 '16 at 16:06
  • $\begingroup$ Thank you! What you write is true, but what I have written is different (and also true). We go from $z=x_{1}-y$ to $\ddot{z}=\ddot{x_{1}}-\ddot{y}$ by differentiating twice, and then using the fact that $\ddot{y}=0$. The last step is got by replacing $x_{1}-x_{2}$ in the application of Newton's Second Law above, by $2z$. I will add this in (and correct my arithmetical mistake!) $\endgroup$ – preferred_anon Sep 6 '16 at 20:07
  • $\begingroup$ Thank-you ! And from here on I can solve the last differential equation like the way you wrote in the 1st few lines $\endgroup$ – Shashaank Sep 6 '16 at 20:13
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Your equation is incorrect; that equation assumes that one of the bodies are fixed (the one-body problem). What you are looking at is the two-body problem.

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  • $\begingroup$ Anyone care to explain the downvote? $\endgroup$ – Bobson Dugnutt Sep 6 '16 at 15:16
  • $\begingroup$ Not the downvoter, but one $m$ has clearly been cancelled from Newton's Second Law. $\endgroup$ – preferred_anon Sep 6 '16 at 15:17
  • $\begingroup$ The R.H.S is the acceleration not the force ! $\endgroup$ – Shashaank Sep 6 '16 at 15:17
  • $\begingroup$ @DanielLittlewood D'oh! Thanks to Shashaank as well! $\endgroup$ – Bobson Dugnutt Sep 6 '16 at 15:19
  • $\begingroup$ @Lovsovs Isn't there any way to solve the above equation without using centre of mass concet $\endgroup$ – Shashaank Sep 6 '16 at 15:21

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