The orbital velocity of an artificial satellite in a circular orbit very close to Earth is $v$ . The velocity of a geosynchronous satellite orbiting in a circular orbit at an altitude of $6 R$ from Earth's surface will be

\frac{v}{\sqrt{7}}

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\frac{v}{\sqrt{6}}

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v

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\sqrt{6} v

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Solution

The correct option is B $\frac{v}{\sqrt{7}}$
$v_{1} \propto \frac{1}{\sqrt{R}}, v_{2} \propto \frac{1}{\sqrt{7 R}}$
$\frac{v_{2}}{v_{1}} = \frac{1}{\sqrt{7}} \Rightarrow v_{2} = \frac{v_{1}}{\sqrt{7}} = \frac{v}{\sqrt{7}}$

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The orbital velocity of an artificial satellite in a circular orbit very close to Earth is v. The velocity of a geosynchronous satellite orbiting in a circular orbit at an altitude of 6R from Earth's surface will be

The correct option is B v√7v1∝1√R,v2∝1√7Rv2v1=1√7⇒v2=v1√7=v√7

The orbital velocity of an artificial satellite in a circular orbit very close to Earth is $v$ . The velocity of a geosynchronous satellite orbiting in a circular orbit at an altitude of $6 R$ from Earth's surface will be

The correct option is B $\frac{v}{\sqrt{7}}$
$v_{1} \propto \frac{1}{\sqrt{R}}, v_{2} \propto \frac{1}{\sqrt{7 R}}$
$\frac{v_{2}}{v_{1}} = \frac{1}{\sqrt{7}} \Rightarrow v_{2} = \frac{v_{1}}{\sqrt{7}} = \frac{v}{\sqrt{7}}$