Gravitation Question 140
Question: The escape velocity for a rocket from earth is 11.2 km/sec. Its value on a planet where acceleration due to gravity is double that on the earth and diameter of the planet is twice that of earth will be in km/sec[NCERT 1983; CPMT 1990; MP PMT 2000; UPSEAT 1999]
Options:
A) 11.2
B) 5.6
C) 22.4
D) 53.6
Show Answer
Answer:
Correct Answer: C
Solution:
$ [\frac{v_{p}}{v_{e}}=\sqrt{\frac{g_{p}}{g_{e}}\times \frac{R_{p}}{R_{e}}}] $ = $ [\sqrt{2\times 2}=2] $ Þ $ [v_{p}=2\times v_{e}=2\times 11.2=22.4\ km/s] $
BETA
