Gravitation Ques 49
49. The distance of Neptune and Saturn from the sun is nearly $10^{13}$ and $10^{12}$ meter respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio
[1994]
(a) $10$
(b) $100$
(c) $10 \sqrt{10}$
(d) $1000$
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Answer:
Correct Answer: 49.(c)
Solution:
- (c) $T^{2} \propto R^{3}$ (According to Kepler’s law)
$ T_1^{2} \propto(10^{13})^{3} \text{ and } T_2^{2} \propto(10^{12})^{3} $
$\therefore \quad \frac{T_1^{2}}{T_2^{2}}=(10)^{3}$ or $\frac{T_1}{T_2}=10 \sqrt{10}$
BETA
