Gravitation Question 306
The distance of Neptune and Saturn from the sun is nearly $ [10^{12} and 10^{11}] $ meter respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio
Options:
10
100
C) $ [10\sqrt{10}] $
D) 1000.0
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Answer:
Correct Answer: C
Solution:
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$ [T^{2}\propto R^{3}] $ (According to Kepler’s law) $ [T_{1}^{2}\propto {{\left( 10^{13} \right)}^{3}}and,,T_{2}^{2}\propto {{\left( 10^{12} \right)}^{3}}] $
$ [\therefore \frac{T_{1}^{2}}{T_{2}^{2}}={{\left( 10 \right)}^{3}}or,\frac{T_{1}}{T_{2}}=10\sqrt{10}] $
BETA
