## Tuesday, April 24, 2007

### Two Questions (MCQ) on Angular Momentum

The following two questions are similar in that both require the calculation of angular momentum in central field motion under inverse square law forces.

(1) An artificial satellite of mass ‘m’ is orbiting the earth of mass ‘M’ in a circular orbit of radius ‘r’. If ‘G’ is the gravitational constant, the orbital angular momentum of the satellite is

(a) [GMm2r]1/2 (b) [GMmr]1/2 (c) [GMm/r]1/2 (d) [GMm2/r2]1/2 (e) [GMm2r3]1/2

The orbital angular momentum of a satellite is mvr where ‘v’ is the orbital speed. [Angular momentum = Iω = mr2ω = mr2v/r = mvr where ‘I’ is the moment of inertia and ‘ω’ is the angular velocity of the satellite].

The centripetal force required for the circular motion of the satellite is supplied by the gravitational pull so that we have

mv2/r = GMm/r2

From this, m2v2r2 = GMm2r so that angular momentum mvr = [GMm2r]1/2

(2) In a hydrogen atom in its ground state, the electron of mass ‘m’ is moving round the proton in a circular orbit of radius ‘r’. The orbital angular momentum of the electron is (with usual meaning for symbols)

(a) [m2e2r/4πε0]1/2 (b) [m2er/4πε0]1/2 (c) [me2r2/4πε0]1/2

(d) [me2r/4πε0]1/2 (e) [me2r3/4πε0]1/2

The steps for finding the orbital angular momentum of the electron are similar to those in question No.1, with the difference that the centripetal force is supplied in this case by the electrostatic attractive force between the proton and the electron.

We have mv2/r = (1/4πε0)e2/r2 from which m2v2r2 = (1/4πε0) ×me2r, so that orbital angular momentum, mvr = [me2r/4πε0]1/2