To find the time period of a satellite at a distance \( 4R \) from the center of the Earth, we can use Kepler's third law of planetary motion, which states:

\[
T^2 \propto r^3
\]

1. Given:
- Time period at Earth's surface (\( T_0 \)): \( 1.4 \) hours (or \( T_0 = 1.4 \times 3600 \) seconds).
- Radius of the Earth: \( R \).

2. At Distance \( 4R \):
- The new radius \( r = 4R \).

3. Using the relationship:
\[
\frac{T^2}{T_0^2} = \frac{r^3}{R^3}
\]

Substituting \( r = 4R \):
\[
\frac{T^2}{T_0^2} = \frac{(4R)^3}{R^3} = \frac{64R^3}{R^3} = 64
\]

4. Solving for \( T \):
\[
T^2 = 64 T_0^2 ; T = 8 T_0
\]

5. Substituting \( T_0 \):
\[
T = 8\sqrt{2}, \text{hours} \]

According to Kepler's Third Law, \(T^2\) is proportional to the cube of the semi-major axis. Since the major axis is given as \(b\), the semi-major axis is \(b/2\), making \(T^2 \propto b^3\).

Areal velocity is given by \(\frac{dA}{dt} = \frac{L}{2m} = \frac{vr}{2} = \frac{\sqrt{GMr}}{2}\). Since areal velocity is proportional to \(\sqrt{r}\), the ratio is \(\sqrt{\frac{R}{4R}} = \frac{1}{2}\) or 1 : 2.

According to Kepler's second law, the areal velocity of the planet is constant and is given by \(\frac{dA}{dt} = \frac{L}{2m}\). Integrating over one time period \(T\) gives \(A = \frac{L}{2m} T ⇒ L = \frac{2mA}{T}\).

Assertion (A) is true. For a satellite in circular orbit, orbital speed \( v = \sqrt{\frac{GM}{r}} \) implying \( r \propto \frac{1}{v^2} \). The period is \( T = \frac{2\pi r}{v} \). Substituting \( r \), we get \( T \propto \frac{1/v^2}{v} \propto \frac{1}{v^3} \).

Reason (R) is true. The formula \( T = \frac{2\pi r}{v} \) is the correct definition for the period of uniform circular motion. However, (R) is a kinematic definition and does not explain the dynamic relationship between \( T \) and \( v \) for a satellite, which requires considering gravity. Thus, (R) is not the correct explanation of (A).

Assertion (A) is true. Gravitational force is a central force, so the torque about the Sun is zero, leading to angular momentum conservation. Reason (R) is also true. Gravitational force is conservative, so total mechanical energy of the system is conserved. However, energy conservation does not explain angular momentum conservation, as they are distinct conservation laws.

Assertion (A): This statement is Kepler's Second Law, which is a direct consequence of angular momentum conservation. So (A) is true.nReason (R): For a central force, like gravity, the force acts along the radius vector, meaning no transverse force component exists. Thus, transverse acceleration is zero. So (R) is true.n(R) explains (A) because zero transverse acceleration leads to conservation of angular momentum, which implies Kepler's Second Law.

Assertion (A) is true because Kepler's second law represents the conservation of angular momentum. Reason (R) is false because areal velocity is \(\frac{dA}{dt} = \frac{L}{2m}\), which is proportional to angular momentum \(L\) but not equal to it.