Gravitation Ques 38
38. The period of revolution of planet $A$ around the Sun is $8$ times that of $B$. The distance of $A$ from the Sun is how many times greater than that of $B$ from the Sun?
[1997]
(a) $2$
(b) $3$
(c) $4$
(d) $ 5$
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Answer:
Correct Answer: 38.(a)
Solution:
- (a) Let $T_A$ and $T_B$ be time period of $A$ and $B$ about sun.
$T_A=8 T_B$
$\frac{T_A}{T_B}=8$
According to Kepler’s Law, $T^{2} \propto r^{3}$
$\frac{T_A}{T_B}=\frac{(r_A)^{3}}{(r_B)^{3}} \Rightarrow(\frac{r_A}{r_B})^{3}=8 \quad \Rightarrow \frac{r_A}{r_B}=2$
BETA
