Solve the following :

Question:

Calculate the ratio of the angular momentum of the earth about its axis due to its spinning motion to that about the sun due to its orbital motion. Radius of the earth $=6400 \mathrm{~km}$ and radius of the orbit of the earth about the sun $=1.5 \times 10^{8} \mathrm{~km}$.

Solution:

$L_{\text {spinning }}=\mid \omega$

$=\left(\frac{2}{5} M_{e} R_{e}^{2}\right)\left(\frac{2 \pi}{T_{e}}\right)$

$=\left(\frac{2}{5} \times M_{e} \times 6400\right)^{2}\left(\frac{2 \pi}{24 \times 60 \times 60}\right)-(\mathrm{i})$

$L_{\text {orbital }}=\mid \omega$

$=\left(M_{e} R_{0}^{2}\right)\left(\frac{2 \pi}{T}\right)$

$=\left(M_{e} \times 1.5 \times 10^{8}\right)^{2}\left(\frac{2 \pi}{365 \times 24 \times 60 \times 60}\right)-($ ii $)$

Divide (i) by (ii)

$\frac{L_{s}}{L_{0}}=2.65 \times 10^{7}$

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