A) 1 : 1
B) \[\sqrt{3}\,:1\]
C) 3 : 1
D) 9 : 1
Correct Answer: B
Solution :
Orbital velocity \[v=\sqrt{\frac{GM}{r}}\] where, M = mass of the planet r = radius of the orbit Orbital velocity is independent of the mass of the orbiting body and is always along the tangent of the orbit. For a given planet (here earth), greater the radius of orbital, lesser will be the orbital velocity of the satellite \[\left( v\propto \frac{1}{\sqrt{r}} \right)\] \[\therefore \] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\frac{{{r}_{2}}}{{{r}_{1}}}}=\sqrt{\frac{3}{1}}\]You need to login to perform this action.
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