In the following four periods:(i) Time of rev | vedclass

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(D) $(i)$ For a satellite orbiting just above the Earth's surface,the time period is $T_{st} = 2\pi \sqrt{\frac{R}{g}} \approx 84.6 \; 	ext{min}$.$(i

Vedclass · Accepted Answer

${T_{st}} = {T_{ma}} = {T_{sp}} = {T_{is}}$

Vedclass · Answer

(D) $(i)$ For a satellite orbiting just above the Earth's surface,the time period is $T_{st} = 2\pi \sqrt{\frac{R}{g}} \approx 84.6 \; 	ext{min}$.$(ii)$ For a mass oscillating in a tunnel through the Earth's diameter,the motion is simple harmonic with $T_{ma} = 2\pi \sqrt{\frac{R}{g}} \approx 84.6 \; 	ext{min}$.$(iii)$ For a simple pendulum of length $l=R$ in a uniform field $g$,$T_{sp} = 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{R}{g}} \approx 84.6 \; 	ext{min}$.$(iv)$ For an infinite length simple pendulum in the Earth's real gravitational field,the period is $T_{is} = 2\pi \sqrt{\frac{R}{g}} \approx 84.6 \; 	ext{min}$.Thus,all four periods are equal: ${T_{st}} = {T_{ma}} = {T_{sp}} = {T_{is}}$.

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