Question

Two massive particles of mssses m and M (M>m) are separated by a distance 'l'. They rotate with equal angular velocity under their gravitational attraction. What is the linear speed of the lighter mass?

I don't understand why the masses are rotating due to gravity... Gravity acts on the centers of mass and hence their is no torque. Does the question mean that their revolving like a binary star system?

Also the answer is sqrt{G*(M^2)/((M+m)*l)}...Thanks for any explanations

Answer 1

They are indeed rotating due to gravity. If there was no gravitational attraction they would each fly off away from each other. As you correctly point out, there is no torque and therefore the angular momentum is conserved. Since there is no external forces, the center of mass must be moving at constant velocity and therefore cannot be rotating. The only way this can happen is if the rotation has the center of mass as its axis. Since they are separated a length l, the center of mass would be \(x_{cm}=\frac{ lM }{ m+M }\) from the smaller mass m. Now, there is a constant force on m which is \(F=G \frac{ mM }{ l^2 }\). Let us take a polar coordinate system with origin at the center of mass like the one in the following drawing.Newtons second law then tells us \[F^{\rightarrow}=-\frac{ GMm }{ l ^{2} }r ^{\rightarrow}=m \frac{ d^2 }{ dt^2 }(x_{cm}r^{\rightarrow}) \rightarrow G \frac{ Mm }{ l^2 } =m \frac{ v^2 }{ x_{cm} }\](\(r^{\rightarrow}\) corresponds to the unit vector in the increasing radial direction. The formula it yields is an elementary highschool equation and I don't want to bore you with all of the maths). Solving for v, we conclude\[v=\sqrt{\frac{ GM^2 }{ l(M+m) }}\]

Answer 2

No that this is the v according to a stationary observer with respect to the center of mass.

Answer 3

Note*

Answer 4

Thanks for the time spent in the equation editor... As I understand, you're basically equating the gravitational force at the mass m, to the centripetal force experienced, right?