Two planets connected by a spring no gravity

[physicsishard]

hello I am confusing about a physics problem right now.
instead of gravity, two planets are connected by a spring with spring constant K. And since the mass of one planet is much bigger than that of the other. So, lighter planet does circular motion around the heavier planet. Then, it asks ω (angular velocity) of this circular motion (no gravity between these two objects). I know that the equation for this problem is that
mv^2/R=kx (centripetal force is supplied by the spring) but, my TA said that x=R in equilibrium position. And the ω=(k/m)^(0.5) I don't understand it !
Thank you very much if you can clarify this point !

 

[ehild]

hello I am confusing about a physics problem right now.
instead of gravity, two planets are connected by a spring with spring constant K. And since the mass of one planet is much bigger than that of the other. So, lighter planet does circular motion around the heavier planet. Then, it asks ω (angular velocity) of this circular motion (no gravity between these two objects). I know that the equation for this problem is that
mv^2/R=kx (centripetal force is supplied by the spring) but, my TA said that x=R in equilibrium position. And the ω=(k/m)^(0.5) I don't understand it !

It is not clear from the text, but the relaxed length of the spring is taken zero. So x, the change of length is equal to the distance of the smaller planet from the other one. You can write the equation for te centripetal force as mv^2/R=kR. As it is circular motion and the radius is R how is the linear speed v related to the angular speed ω ?

ehild