Question

The distance between Sun and Earth is $1.6 \times 10^{11}\, m$ and the radius of Earth is $6.4 \times 10^{6} \,m$. The ratio of the angular momentum of Earth around the Sun to the angular momentum around its own axis is approximately (Assume Earth as a solid sphere with uniform mass density and rotates around the Sun in a circular path.)

• A. $2.0 \times 10^{2}$
• B. $5.1 \times 10^{8}$
• C. $4.3 \times 10^{6}$
• D. $8.7 \times 10^{12}$

Answer 1

Answer: (C)

Given, the distance between the Sun and Earth, $r_{1}=1.6 \times 10^{11}\, m$
Radius of earth, $R_{e}=6.4 \times 10^{6} \,m$
Angular momentum of Earth around the Sun,
$L_{1} =M_{e} \,v_{1}\, r_{1}=M_{e} \cdot \frac{2 \,\pi\, r_{1}}{T_{1}} \cdot r_{1} $
$L_{1} =2 \times 3.14 \times 6.0 \times 10^{24} \times \frac{\left(1.6 \times 10^{11}\right)^{2}}{365 \times 24 \times 60 \times 60}$
$\left(\because\right.$ Mass of earth, $M_{e}=6.0 \times 10^{24} \,kg$ and $T_{1}=365$ days)
$L_{1}=3.06 \times 10^{40}\, kg - m ^{2} / s$
Angular momentum of Earth about its own axis, where, moment $L_{2}=I \omega$ of inertia of Earth (uniform solid sphere) around its own axis, $I=\frac{2}{5} M_{e} R_{e}^{2}$.
Or $I =\frac{2}{5} \times 6.0 \times 10^{24} \times\left(6.4 \times 10^{6}\right)^{2} $
$ \simeq 9.8 \times 10^{37} \,kg - m ^{2}$
The earth's angular velocity, $\omega=\frac{2 \pi}{1 \text { day }}$
$=\frac{2 \pi}{24 \times 60 \times 10} \simeq 7.3 \times 10^{-5} s ^{-1}$
$\therefore L_{2}=9.8 \times 10^{37} \times 7.3 \times 10^{-5} \approx 7.2 \times 10^{33} \,kg - m ^{2} s ^{-1}$
From Eqs. (i) and (ii), we get
$ \frac{L_{1}}{L_{2}}=\frac{3.06 \times 10^{40}}{7.2 \times 10^{33}}=4.3 \times 10^{6}$
The ratio of angular momentum of Earth around the Sun to the angular momentum around its own axis is $4.3 \times 10^{6}$ approximately.