Gravitation 4 Question 5
6. The ratio of earth’s orbital angular momentum (about the sun) to its mass is $4.4 \times 10^{15} m^{2} / s$. The area enclosed by earth’s orbit is approximately $m^{2}$.
(1997C, 1M)
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Answer:
Correct Answer: 6. $6.94 \times 10^{22}$
Solution:
- Areal velocity of a planet round the sun is constant and is given by
$$ \frac{d A}{d t}=\frac{L}{2 m} $$
where, $L=$ angular momentum of planet (earth) about sun and $\quad m=$ mass of planet (earth).
Given,
$$ \frac{L}{m}=4.4 \times 10^{15} m^{2} / s $$
$\therefore$ Area enclosed by earth in time $T$ ( 365 days ) will be
$$ \begin{aligned} \text { Area } & =\frac{d A}{d t} \cdot T=\frac{L}{2 m} \cdot T \\ & =\frac{1}{2} \times 4.4 \times 10^{15} \times 365 \times 24 \times 3600 m^{2} \end{aligned} $$
$$ \text { Area } \approx 6.94 \times 10^{22} m^{2} $$
