Recent content by mr_trollface_-

since centripetal accelration is : v^2 / R. Where v = w*R. we get ((w*R)^2)/R . and that simplified is w*R^2. and angular velocity = (pi*2*R)/T since it is distance divided by time. putting all of that into one equatin and solving for T we get : T= (4*pi*R^(3/2))/( sqrt( G*( 4*M + n ) ) )...

Mhmm, So I calculated the centripetal acceleration to be: (G*M)/(R^2) + ( G*m )/((2*R)^2). am I right? and also by which equation could I relate the centripetal acceleration to the period ? p.s. sorry if some of what I'm writing doesn't make much sense , It's really late at night and I'm...

I did think of that, but then dissmissed because I thought that Newton's third law would cause them to cancel out. So in this case, the gravitational force exerted on one of the moons by the other is: (G*M*m)/(2*R). right? . But I have no idea how it will contribute to the period of the moon

The formula is Kepler's 3rd law. And the big M there is the mass of the planet.

1. Homework Statement Two identical moons of mass m maintain opposite positions in the same circular orbit of radius R around a planet of mass M. Find T2 the square of the orbital period. 2. Homework Equations T2=(4*pi2*R3)/ ( G*M ) [b]3. The Attempt at a Solution Hi...