Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationSatellite MotionEasy2 minPYQ_2017

PhysicsEasysingle choice

A geostationary satellite orbits around the earth in a circular orbit of radius $36000 km$ . Then, the time period of a sky satellite orbiting a few $100 km$ above the earth's surface $(R = 64000 km)$ will approximately be

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Answer:

Solution:

By Kepler's law of planetary motion,

$T^{2} \propto r^{3}$ , hence $T_{1}^{2} \propto r_{1}^{3}$

Similarly $T_{2}^{2} \propto r_{2}^{3}$

$∴ T_{1}^{2} \propto {(36000)}^{3}$

$T_{2}^{2} \propto {(6400 + h)}^{3}$

Therefore,

${(T_{2})}^{2} = T_{1}^{2} {(\frac{6400 + h}{36000})}^{3}$

$> T_{1}^{2} {[\frac{6400}{3600}]}^{3}$

$> {(24)}^{2} {[\frac{8}{45}]}^{3}$

$∴ T_{2} > \frac{24 \times 8 \times \sqrt{8}}{45 \times \sqrt{45}}$

$> 1.8 h$

So, $T_{2} ≅ 2 h$

Stream:BITSATSubject:PhysicsTopic:GravitationSubtopic:Satellite Motion

⏱ 2mℹ️ Source: PYQ_2017

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