Question

If A is aerial velocity of a planet of mass M its angular momentum is 

(A) M 

(B) 2 MA 

(C) A²M 

(D) AM²

Answer 1

Correct option: (B) 2 MA

Explanation:

Areal Velocity A = (πR² / T) [as Area velocity = {(area by radius vector) / (time)}]

and T = (2π / ω)

hence A = [(πR²) / 2π] × ω = (ωR² / 2)  

i.e. ω = (2A / R²)

angular momentum  = Iω

= MR² × (2A / R²)

= 2 AM