Question

A planet is revolving around the sun in an elliptical orbit.
Its closests distance from the sun is rmin. The farthest
distance from the sun is rmax. If the orbital angular velocity
of the planet when it is nearest to the sun is ⍵, then the
orbital angular velocity at the point when it is at the farthest
distance from the sun is
(a) √(rₘᵢₙ / rₘₐₓ ) ⍵ (b) √(rₘₐₓ / rₘᵢₙ ) ⍵
(c) (rₘₐₓ² / rₘᵢₙ ² )⍵
(d) (rₘᵢₙ² / rₘₐₓ² )⍵

Answer 1

Let $\sf{v_1}$ and $\sf{v_2}$ be the linear velocities of the planet at closest and farthest distances respectively.

So according to the question,

• $\sf{v_1=\omega\,r_{min}}$

Let $\omega'$ be the angular velocity of the planet at the farthest distance. Then,

• $\sf{v_2=\omega'\,r_{max}}$

The area swept out by the radius vector of the planet drawn from the sun per unit time at the closest distance will be,

$\longrightarrow\sf{\dfrac{dA_1}{dt}=\dfrac{1}{2}\,v_1\,r_{min}}$

Similarly, area swept out per unit time at the farthest distance will be,

$\longrightarrow\sf{\dfrac{dA_2}{dt}=\dfrac{1}{2}\,v_2\,r_{max}}$

By Kepler's Law of Areas, we have,

$\longrightarrow\sf{\dfrac{dA_1}{dt}=\dfrac{dA_2}{dt}}$

$\longrightarrow\sf{\dfrac{1}{2}\,v_1\,r_{min}=\dfrac{1}{2}\,v_2\,r_{max}}$

$\longrightarrow\sf{v_1\,r_{min}=v_2\,r_{max}}$

$\longrightarrow\sf{\omega\,r_{min}\cdot r_{min}=\omega'\,r_{max}\cdot r_{max}}$

$\longrightarrow\sf{\omega(r_{min})^2=\omega'(r_{max})^2}$

$\longrightarrow\sf{\underline{\underline{\omega'=\omega\left(\dfrac{r_{min}}{r_{max}}\right)^2}}}$