Question

(a) Calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun. Is it reasonable to model it as a particle? (b) Calculate the magnitude of the angular momentum of the earth due to its rotation around an axis through the north and south poles, modeling it as a uniform sphere. Consult Appendix E and the astronomical data in Appendix F.

Answer 1

Answer:

(a) $L=2.6742\times 10^{40}\, kg.m^2.s^{-1}$

(b) $L_s=7.07 \times 10^{23} \, kg.m^2.s^{-1}$

Explanation:

For Earth we have:

• mass of earth, $m_E=5.97\times 10^{24}\, kg$

• radius of earth, $R_E=6.38\times 10^6m$

• orbital radius, $r=1.5 \times 10^{11}m$

• period of rotation, $t_{rot}=24h=86400\, s$

• period of revolution, $t_{rev}= 1 yr=3.156\times 10^7 s$

(a)

Angular momentum, $L=?$

∵$L=I.\omega$...............................(1)

For a particle of mass m moving in a circular path at a distance r from the axis,

$I=m.r^2$

& $v=r.\omega$

Putting respecstive values in eq. (1)

$L=m_E\times r^{2}\times \omega$

$L=5.97\times 10^{24}\times (1.5 \times 10^{11})^2\times \frac{2\pi}{3.156\times 10^7}$

$L=2.6742\times 10^{40}\, kg.m^2.s^{-1}$

To model earth as a particle is  reasonable because the distance between the sun and the earth is very large s compared to the radius if the earth.

(b)

For a uniform sphere of mass M and radius R and an axis through its center,  $I=\frac{2}{5}M.R^2$

using eq. (1)

$L_s=(\frac{2}{5}m_E.R_E^2) \omega$

$L_s=\frac{2}{5} \times 5.97\times 10^{24} \times 6.38\times 10^6\times \frac{2\pi}{86400}$

$L_s=7.07 \times 10^{23} \, kg.m^2.s^{-1}$