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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 find the time period of rotation of a satellite orbiting close to the surface of the Earth, we can derive the formula step by step. ### Step 1: Understand the parameters - Let \( R \) be the radius of the Earth. - Let \( g \) be the acceleration due to gravity at the surface of the Earth. - The satellite is orbiting at a distance \( r \) from the center of the Earth, which is approximately equal to \( R \) when it is close to the surface. ### 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 gravitational constant and \( M \) is the mass of the Earth. When the satellite is very close to the Earth's surface, we can approximate \( r \) as \( R \): \[ v = \sqrt{\frac{GM}{R}} \] ### Step 3: Relate velocity to the time period The time period \( T \) of the satellite's orbit is related to the orbital velocity and the circumference of the orbit: \[ T = \frac{2\pi r}{v} \] Substituting \( r \) with \( R \) and the expression for \( v \): \[ T = \frac{2\pi R}{\sqrt{\frac{GM}{R}}} \] ### Step 4: Simplify the expression Rearranging the equation gives: \[ T = 2\pi R \cdot \sqrt{\frac{R}{GM}} = 2\pi \sqrt{\frac{R^3}{GM}} \] ### Step 5: Use the relationship between \( g \), \( G \), and \( M \) We know that the acceleration due to gravity \( g \) can be expressed as: \[ g = \frac{GM}{R^2} \] From this, we can express \( GM \) as: \[ GM = gR^2 \] Substituting this back into the time period formula gives: \[ T = 2\pi \sqrt{\frac{R^3}{gR^2}} = 2\pi \sqrt{\frac{R}{g}} \] ### Step 6: Calculate the time period Using the known values: - \( R \approx 6.4 \times 10^6 \, \text{m} \) - \( g \approx 9.8 \, \text{m/s}^2 \) Substituting these values into the formula: \[ T = 2\pi \sqrt{\frac{6.4 \times 10^6}{9.8}} \] Calculating this will yield a time period of approximately \( 84.6 \) minutes. ### Conclusion Thus, the correct option for the time period of rotation of a satellite orbiting close to the surface of the Earth is: **(b) 84.6 minutes.** ---
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