Question

If the earth is a point mass of $6\times {{10}^{24}}kg$ revolving around the sun at a distance of $1.5\times {{10}^{8}}km$ and in time of $T=3.14\times {{10}^{7}}s$, then the angular momentum of the earth around the sun is
A. $1.2\times {{10}^{18}}kg{{m}^{2}}{{s}^{-1}}$
B. $1.8\times {{10}^{20}}kg{{m}^{2}}{{s}^{-1}}$
C. $1.5\times {{10}^{37}}kg{{m}^{2}}{{s}^{-1}}$
D. $2.7\times {{10}^{40}}kg{{m}^{2}}{{s}^{-1}}$

Answer 1

Hint: Use the formula for the moment of inertia of a point mass around an axis rotating in a circular path and also the formula for the angular velocity in terms of time period of revolution. After calculating these two, substitute these in the formula for angular momentum.

Formula used:
$L=I\omega $
where L is the angular momentum of a body, with moment of inertia I and angular velocity $\omega $ around the axis of rotation.
$I=m{{r}^{2}}$
where I is a moment of inertia of a point mass m revolving around an axis in circular path of radius r.
$\omega =\dfrac{2\pi }{T}$
where T is the time period of revolution.

Complete step by step answer:
It is said to assume that the earth is a point mass. Then this means that the angular moment of the earth can be written as $L=I\omega $ …. (i)
In this case, the L is the angular moment of the earth around the sun, I is the moment of inertia of earth and $\omega $ is the angular velocity of the earth around the sun.
The moment of inertia of the earth around the sun is $I=m{{r}^{2}}$ …. (ii).
In this case, the m is the mass of the earth and r is the distance of the earth from the earth if we assume that the earth is revolving in a circular path with the sun at its centre.
It is given that $m=6\times {{10}^{24}}kg$, $r=1.5\times {{10}^{8}}km=1.5\times {{10}^{11}}m$.
Substitute these values in (ii).
$\Rightarrow I=6\times {{10}^{24}}\times {{\left( 1.5\times {{10}^{11}} \right)}^{2}}kg{{m}^{2}}$
Now, the angular velocity of the earth around the sun is $\omega =\dfrac{2\pi }{T}=\dfrac{2\pi }{3.14\times {{10}^{7}}}{{s}^{-1}}$.
Substitute the known values in equation (i).
$\Rightarrow L=6\times {{10}^{24}}\times {{\left( 1.5\times {{10}^{11}} \right)}^{2}}\times \dfrac{2\pi }{3.14\times {{10}^{7}}}\\
\therefore L =2.7\times {{10}^{40}}kg{{m}^{2}}{{s}^{-1}}$
Therefore, the angular momentum of the earth around the sun is $2.7\times {{10}^{40}}kg{{m}^{2}}{{s}^{-1}}$.

Hence, the correct option is D.

Note:For a point mass the angular momentum is also given as $L=mvr$, where v is the orbital speed of the earth.For a point mass rotating with constant speed in circular path of radius r, its orbital speed is given as $v=\omega r$. We can also use these formulae and calculate the moment of inertia of the earth around the sun.