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CAF123
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Homework Statement
Two uniform lead spheres each have mass 5000kg and radius 47cm. They are released from rest with their centres 1m apart and move under their mutual gravitational attraction. Show that they will collide in less than 425s.
Homework Equations
By Gauss' Law, force on a sphere exerted by the other is ##F = - m^2 G/R^2##.
The Attempt at a Solution
The acceleration of the spheres as they approach each other increases as an inverse power of R. To get an upper bound of the collision time, assume constant acceleration throughout and use standard kinematical equations. This gave me the solution. I'm interested in solving for the time more analytically by solving a DE.
The spheres approach each other in a straight line so force on one from other is $$F = - \frac{m^2 G}{R^2} = m \frac{dv}{dt} \Rightarrow - \frac{m G}{R^2} = \frac{dv}{dt} = \frac{dv}{dR}v$$ Solving for v gives $$ \frac{1}{2} v^2 = mG \left(\frac{1}{R} - \frac{1}{R_0}\right)$$ where at ##R_0## the configuration is at rest. Then $$\frac{dR}{dt} = \sqrt{2mG} \sqrt{ \frac{R_0-R}{RR_0}}$$ so need to then solve $$\sqrt{R_0} \int_{R_0}^R \frac{\sqrt{R'}}{\sqrt{R_0-R'}} dR' = \sqrt{2mG} \int_0^T dt$$ where T is the collision time. The integral on lhs can be solved by ##R' = R_0 \sin^2 \theta## where ##R_0 = 0.06 m## and ##R=0.03m##. (The system will collide at the CoM) . For these values of R and R_0, and choosing the integration range to be ##[-\pi/2, -0.253], | \sin \theta| = - \sin \theta## and ##|\cos \theta| = \cos \theta##. This gives $$T \approx -1.12 \frac{1}{\sqrt{mG}}$$ while in the solution assuming constant acceleration it was $$T \approx 0.245 \frac{1}{\sqrt{mG}}$$ so the numbers are off in my analysis.
Thanks for any pointers!