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(A) By the law of conservation of angular momentum,the angular momentum $L$ of the planet remains constant at all points in its orbit.At the perihelio

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$\frac{v_P}{v_A}=\frac{1+e}{1-e}$

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(A) By the law of conservation of angular momentum,the angular momentum $L$ of the planet remains constant at all points in its orbit.At the perihelion $P$ and aphelion $A$,the velocity vector is perpendicular to the position vector,so the angular momentum is given by $L = mvr$.Since $L_P = L_A$,we have $m v_P r_P = m v_A r_A$,which implies $\frac{v_P}{v_A} = \frac{r_A}{r_P}$.For an ellipse with semi-major axis $a$ and eccentricity $e$,the distance of the closest point (perihelion) is $r_P = a(1-e)$ and the distance of the farthest point (aphelion) is $r_A = a(1+e)$.Substituting these values,we get $\frac{v_P}{v_A} = \frac{a(1+e)}{a(1-e)} = \frac{1+e}{1-e}$.

Column-$I$	Column-$II$
$(1)$ Kepler's first law	$(a)$ Law of periods
$(2)$ Kepler's second law	$(b)$ Law of orbits
$(3)$ Kepler's third law	$(c)$ Law of areas

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Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Kepler's first law $(a)$ Law of periods

$(2)$ Kepler's second law $(b)$ Law of orbits

$(3)$ Kepler's third law $(c)$ Law of areas

Assertion $(A)$: Angular speed,linear speed,and kinetic energy change with time,but angular momentum remains constant for a planet orbiting the sun.
Reason $(R)$: Angular momentum is constant as no external torque acts on the planet.

State Kepler's law of periods (Kepler's third law) for planetary motion.

The Earth is revolving around the Sun. If the distance of the Earth from the Sun is reduced to $\frac{1}{4}$ of the present distance,then the length of the present day will be reduced by a factor of:

If $r_p, v_p, L_p$ and $r_a, v_a, L_a$ are radii,velocities and angular momenta of a planet at perihelion and aphelion of its elliptical orbit around the Sun respectively,then

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