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With usual notation, the time period of ...

With usual notation, the time period of rotation of a satellite orbiting close to the surface of earth
(a)`T=2pisqrt(R/g)` ,(b)84.6 minutes ,(c)`T=2pisqrt(R^3/(GM))` , (d)`T=2pisqrt((GM)/R)`

A

a is only correct

B

a & b are correct

C

a,b & c are correct

D

a,b & d are correct

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the time period of rotation of a satellite orbiting close to the surface of the Earth, we can follow these steps: ### Step 1: Understand the problem We need to find the time period \( T \) of a satellite that is orbiting very close to the surface of the Earth. We will derive the formula for the time period using gravitational concepts. ### Step 2: Use the formula for orbital velocity The orbital velocity \( v \) of a satellite is given by the formula: \[ v = \sqrt{\frac{GM}{R}} \] where: - \( G \) is the universal gravitational constant, - \( M \) is the mass of the Earth, - \( R \) is the radius of the Earth. ### Step 3: Relate time period to orbital velocity The time period \( T \) can be expressed in terms of the orbital radius \( R \) and the velocity \( v \): \[ T = \frac{2\pi R}{v} \] Substituting the expression for \( v \) from Step 2: \[ T = \frac{2\pi R}{\sqrt{\frac{GM}{R}}} \] ### Step 4: Simplify the expression Now, we can simplify the expression for \( T \): \[ T = 2\pi R \cdot \sqrt{\frac{R}{GM}} = 2\pi \sqrt{\frac{R^3}{GM}} \] ### Step 5: Use the relation for acceleration due to gravity We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by: \[ g = \frac{GM}{R^2} \] From this, we can express \( GM \) as: \[ GM = gR^2 \] ### Step 6: Substitute \( GM \) back into the time period equation Substituting \( GM \) into the time period equation gives: \[ T = 2\pi \sqrt{\frac{R^3}{gR^2}} = 2\pi \sqrt{\frac{R}{g}} \] ### Conclusion Thus, the time period of a satellite orbiting close to the surface of the Earth is given by: \[ T = 2\pi \sqrt{\frac{R}{g}} \] This corresponds to option (a) \( T = 2\pi\sqrt{\frac{R}{g}} \). ### Final Answer The correct answer is (a) \( T = 2\pi\sqrt{\frac{R}{g}} \). ---
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