A satellite is launched into a circular orbit, R 4 above the surface of the earth. The time period of revolution is T = 2 π n 3 2 R g . Where R is the radius of the earth and g is the acceleration due to gravity. Then what is value on n ?

T ⁡ = 2 π r ⁡ 3 2 g R ⁡ 2 where r = R + h Then, T ⁡ = 2 π R ⁡ + h ⁡ R ⁡ R ⁡ + h ⁡ g Here, h = R 4 T ⁡ = 2 π R ⁡ + R ⁡ 4 R ⁡ R ⁡ + R ⁡ 4 g = 2 π 1 . 2 5 3 2 R ⁡ g

Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationSatellite MotionEasy2 minPYQ_2020

PhysicsEasynumerical

A satellite is launched into a circular orbit, $\frac{R}{4}$ above the surface of the earth. The time period of revolution is $T = 2 π {(n)}^{\frac{3}{2}} \sqrt{\frac{R}{g}}$ . Where $R$ is the radius of the earth and $g$ is the acceleration due to gravity. Then what is value on $n$ ?

Answer:

1.25

Solution:

$T = \frac{2 π r^{\frac{3}{2}}}{g R^{2}} where r = R + h$

Then, $T = \frac{2 π (R + h)}{R} \sqrt{(\frac{R + h}{g})}$

Here, $h = \frac{R}{4}$

$T = \frac{2 π (R + \frac{R}{4})}{R} \sqrt{(\frac{R + \frac{R}{4}}{g})} = 2 π {(1.25)}^{\frac{3}{2}} \sqrt{\frac{R}{g}}$

Subject:PhysicsTopic:GravitationSubtopic:Satellite Motion

⏱ 2mℹ️ Source: PYQ_2020

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