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,
$L_1=M_e{v}_1{r}_1=M_e\cdot \frac{2\pi r_1}{T_1}\cdot r_1$
$L_1=2\times 3.14\times 6.0\times 10^{24}\times \frac{{\left(1.6\times 10^{11}\right)}^2}{365\times 24\times 60\times 60}$
$(\because$ 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.