A planet moving along an elliptical orbit is closest to the sun at a distance $ {r_1} $ and farthest away at a distance $ {r_2} $. If $ {v_1} $ and $ {v_2} $ are the linear velocities at these points respectively, then the ratio $ {{{v_1}} \mathord{\left/
 {\vphantom {{{v_1}} {{v_2}}}} \right.} {{v_2}}} $ is?
(A) $ {{{r_1}} \mathord{\left/ {{{{r_1}} {{r_2}}}} \right.} {{r_2}}} $
(B) $ {{{r_2}} \mathord{\left/{{{{r_2}} {{r_1}}}} \right.} {{r_1}}} $
(C) $ {{({r_1}} \mathord{\left/{{{({r_1}} {{r_2}}}} \right.} {{r_2}}}{)^2} $
(D) $ {{({r_2}} \mathord{\left/{{{({r_2}} {{r_1}}}} \right.} {{r_1}}}{)^2} $

Hint
A planet moving in an orbit around the sun obeys the principle of conservation of angular momentum. Equate its angular momentum at distance $ {r_1} $ with its angular momentum at distance $ {r_2} $. Rearrange the resulting equation.
Formula used: The principle of conservation of angular momentum $ {m_1}{v_1}{r_1} = {m_2}{v_2}{r_2} $ where $ m $ is mass, $ v $ is linear velocity and $ r $ is distance.

Complete step by step answer
From Kepler's law, we know that all planets orbit their star in an elliptical orbit (and not circular) with the star at one focus of the ellipse. This allows differences in the distance of the planet to its star throughout its entire orbit. Now, just like the law of conservation of linear momentum applies to linear motion, the law of conservation of angular momentum applies to bodies in orbital motions. In statement, the law (or principle) of angular momentum states that the total angular momentum of a system is conserved at any time or space in as much as there is no net torque applied to the system.
In mathematical terms,
 $ {m_1}{v_1}{r_1} = {m_2}{v_2}{r_2} $
Where $ m $ is mass, $ v $ is linear velocity and $ r $ is distance.
We would apply this to our question above. At distance $ {r_1} $, the angular momentum $ L $ is:
 $ {L_1} = {m_1}{v_1}{r_1} $ where $ {v_1} $ is linear velocity.
Similarly, at a distance $ {r_2} $ ,
 $ {L_2} = {m_2}{v_2}{r_2} $
Thus, applying the conservation relation, we have
 $ kg{m^2}/s $
Since $ {m_1} = {m_2} = m $ (because the planet’s mass remains the same), we can cancel it out on both sides. So,
 $ {v_1}{r_1} = {v_2}{r_2} $
Rearranging, by dividing both sides by $ {v_2} $ and $ {r_1} $ we have
 $ \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{r_2}}}{{{r_1}}} $
 $ \therefore \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{r_2}}}{{{r_1}}} $
Hence, the correct option is B.

Note
A common error is incorrectly replacing $ v $ with $ \omega $ or vice versa in the formula for angular momentum. This results in a different arrangement which will result in a different incorrect answer. For example, replacing $ v $ with $ \omega $ in $ L = mvr $ or replacing $ \omega $ with $ v $ in $ L = mw{r^2} $ which is an alternate form of angular momentum formula. To circumvent this, we can perform a quick unit check. For example, $ L = mw{r^2} $ or $ L = mvr $ will yield $ kg{m^2}/s $ while any other one yields a separate incorrect result.