Question

The average radii of orbits of mercury and earth around the sun are $ 6 \times {10^7}km $ and $ 1.5 \times {10^8}km $ respectively. The ratio of their orbital speed will be:-
(A) $ \sqrt 5 :\sqrt 2 $
(B) $ \sqrt 2 :\sqrt 5 $
(C) none of these
(D) all of the above

Answer 1

Hint
We can find the relation between the orbital speed of a planet and the radius of the orbit from the equation, $ \dfrac{{m{v^2}}}{R} = G\dfrac{{Mm}}{{{R^2}}} $ , where we get $ v \propto \dfrac{1}{{\sqrt R }} $ . So, by finding the ratio of the radii of the orbits in form of the equation, we can find the velocity.
To solve this problem, we use the condition for the revolution of a planet in an orbit around the sun,
 $\Rightarrow \dfrac{{m{v^2}}}{R} = G\dfrac{{Mm}}{{{R^2}}} $
where $ m $ is the mass of the planet and $ M $ is the mass of the sun, $ v $ is the orbital velocity of the planet, $ R $ is the radius of the orbit and $ G $ is the gravitational constant.

Complete step by step answer
 For any planet revolving around the sun in a circular orbit, the centripetal force which keeps the planet in the orbit is given by,
 $\Rightarrow F = \dfrac{{m{v^2}}}{R} $ .
And from Newton’s law of gravitation, the gravitational force between the sun and any planet is given by,
  $\Rightarrow F = G\dfrac{{Mm}}{{{R^2}}} $
 where $ m $ is the mass of the planet.
Now equating these two forces we get
 $\Rightarrow \dfrac{{m{v^2}}}{R} = G\dfrac{{Mm}}{{{R^2}}} $
Now by cancelling $ m $ from the numerator of both sides and $ R $ from the denominator of both the sides we get,
 $\Rightarrow {v^2} = \dfrac{{GM}}{R} $
Taking square root over both sides we get,
 $\Rightarrow v = \sqrt {\dfrac{{GM}}{R}} $
Now, for the case of Mercury, orbital velocity is $ {v_m} $ and the radius of the orbit is $ {R_m} $ .
Therefore, we can write the equation as,
 $\Rightarrow {v_m} = \sqrt {\dfrac{{GM}}{{{R_m}}}} $
And for earth the orbital velocity is $ {v_e} $ and the radius of the orbit is $ {R_e} $ .
 $ \therefore {v_e} = \sqrt {\dfrac{{GM}}{{{R_e}}}} $
 $ M $ is the mass of the sun, the same in both cases and $ G $ is universal constant.
So, now the ratio of the velocity of mercury to that of earth is
$\Rightarrow \dfrac{{{v_m}}}{{{v_e}}} = \dfrac{{\sqrt {\dfrac{{GM}}{{{R_m}}}} }}{{\sqrt {\dfrac{{GM}}{{{R_e}}}} }} $
Now, cancelling $ \sqrt {GM} $ from both the numerator and denominator we get,
 $\Rightarrow \dfrac{{{v_m}}}{{{v_e}}} = \dfrac{{\sqrt {\dfrac{1}{{{R_m}}}} }}{{\sqrt {\dfrac{1}{{{R_e}}}} }} = \dfrac{1}{{\sqrt {{R_m}} }} \times \sqrt {{R_e}} $
Now, from the question we have, $ {R_m} = 6 \times {10^7}km $ and $ {R_e} = 1.5 \times {10^8}km $ . So, by substituting the values we get,
 $\Rightarrow \dfrac{{{v_m}}}{{{v_e}}} = \sqrt {\dfrac{{1.5 \times {{10}^8}}}{{6 \times {{10}^7}}}} $
By cancelling $ {10^7} $ from numerator and denominator, we have
 $\Rightarrow \dfrac{{{v_m}}}{{{v_e}}} = \sqrt {\dfrac{{1.5 \times 10}}{6}} = \dfrac{{\sqrt {15} }}{{\sqrt 6 }} = \dfrac{{\sqrt 3 \times \sqrt 5 }}{{\sqrt 3 \times \sqrt 2 }} $
By cancelling $ \sqrt 3 $ from numerator and denominator,
 $\Rightarrow \dfrac{{{v_m}}}{{{v_e}}} = \dfrac{{\sqrt 5 }}{{\sqrt 2 }} $
Therefore, the answer, $ {v_m}:{v_e} = \sqrt 5 :\sqrt 2 $ .
So, option (A) is correct.

Note
We can also solve this problem alternatively by the following method,
From Kepler's third law of planetary motion, we know that
 $\Rightarrow {T^2} \propto {R^3} $ where $ T $ is the time period of the orbital motion and $ R $ is the radius of the orbit.
 $ \therefore T \propto {R^{{\raise0.7ex\hbox{ $ 3 $ } \!\mathord{\left/
 {\vphantom {3 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}}} $
Now for a planet, the orbital velocity is given by,
 $\Rightarrow v = \dfrac{{2\pi R}}{T} $
 $\Rightarrow v \propto \dfrac{R}{T} $
So in place of $ T $ we can write $ {R^{{\raise0.7ex\hbox{ $ 3 $ } \!\mathord{\left/
 {\vphantom {3 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}}} $
 $ \therefore v \propto \dfrac{R}{{{R^{{\raise0.7ex\hbox{ $ 3 $ } \!\mathord{\left/
 {\vphantom {3 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}}}}} $
 $ \Rightarrow v \propto \dfrac{1}{{{R^{{\raise0.7ex\hbox{ $ 1 $ } \!\mathord{\left/
 {\vphantom {1 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}}}}} $
So now for mercury and earth,
 $\Rightarrow {v_m} \propto \dfrac{1}{{{R_m}^{{\raise0.7ex\hbox{ $ 1 $ } \!\mathord{\left/
 {\vphantom {1 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}}}} $ and $ {v_e} \propto \dfrac{1}{{{R_e}^{{\raise0.7ex\hbox{ $ 1 $ } \!\mathord{\left/
 {\vphantom {1 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}}}} $
Therefore, taking the ratio,
 $\Rightarrow \dfrac{{{v_m}}}{{{v_e}}} = \dfrac{{{R_e}^{{\raise0.7ex\hbox{ $ 1 $ } \!\mathord{\left/
 {\vphantom {1 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}}}}{{{R_m}^{{\raise0.7ex\hbox{ $ 1 $ } \!\mathord{\left/
 {\vphantom {1 2}}\right.}
\!\lower0.7ex\hbox{ $ 2 $ }}}}} $
Substituting the values we get,
 $\Rightarrow \dfrac{{{v_m}}}{{{v_e}}} = \sqrt {\dfrac{{1.5 \times {{10}^8}}}{{6 \times {{10}^7}}}} $ , by doing similar calculation as given previously, we get
  $\Rightarrow {v_m}:{v_e} = \sqrt 5 :\sqrt 2 $.