A planet moving around sun sweeps area A1 in 2 days, A2 in 3 days and A3 in 6 days. Then the relation between A1, A2 and A3 is

Q: A planet moving around sun sweeps area A1 in 2 days, A2 in 3 days and A3 in 6 days. Then the relation between A1, A2 and A3 is

Numerical

(a) 3A1 = 2A2 = A3

(b) 2A1 = 3A2 = 6A3

(c) 3A1 = 2A2 = 6A3

(d) 6A1 = 3A2 = 2A3

Click to See Answer :
Ans: (a)
Sol: From Kepler’s 2nd Law (Law of Area )

$\displaystyle \frac{dA}{dt} = constant$

$\displaystyle \frac{A_1}{2} = \frac{A_2}{3} = \frac{A_3}{6} $

On multiplying by 6 ,

A1 = 2A2 = A3

 

The orbital angular momentum of a satellite revolving at a distance r from the center is L . If the distance is increase to 16r , then the new angular momentum will be

Q: The orbital angular momentum of a satellite revolving at a distance r from the center is L . If the distance is increase to 16r , then the new angular momentum will be

(a) 16 L

(b) 64 L

(c) L/4

(d) 4 L

Click to See Answer :
Ans: (d)

 

A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is R0 and mass of the earth M , the angular momentum about the center of the earth is

Q: A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is R0 and mass of the earth M , the angular momentum about the center of the earth is

(a) $m \sqrt{G M R_0} $

(b) $ m \sqrt{G m R_0} $

(c) $ m \sqrt{\frac{G M}{R_0}} $

(d) $ M \sqrt{\frac{G m}{R_0}} $

Click to See Answer :
Ans: (a)
Sol: L = m v r

$\displaystyle L = m \sqrt{\frac{G M}{R_0} } \; R_0$

$\displaystyle L = \sqrt{G M R_0}$