What is the period of revolution of an Earth | vedclass

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(C) The period of revolution $T$ of a satellite orbiting very close to the Earth's surface is given by the formula:$T = 2 \pi \sqrt{\frac{R_E}{g}}$Giv

Vedclass · Accepted Answer

$83.73$ minutes

Vedclass · Answer

(C) The period of revolution $T$ of a satellite orbiting very close to the Earth's surface is given by the formula:$T = 2 \pi \sqrt{\frac{R_E}{g}}$Given values:$R_E = 6400 \ km = 6.4 	imes 10^6 \ m$$g = 10 \ ms^{-2}$$\pi = 3.14$Substituting these values into the formula:$T = 2 	imes 3.14 	imes \sqrt{\frac{6.4 	imes 10^6}{10}}$$T = 6.28 	imes \sqrt{6.4 	imes 10^5} = 6.28 	imes \sqrt{64 	imes 10^4}$$T = 6.28 	imes 800 = 5024 \ s$To convert the time into minutes:$T = \frac{5024}{60} \ min \approx 83.73 \ min$Therefore, the period of revolution is $83.73$ minutes.

What is the period of revolution of an Earth satellite? Ignore the height of the satellite above the surface of the Earth.
Given:
$(1)$ The value of gravitational acceleration $g = 10 \ ms^{-2}$.
$(2)$ Radius of Earth $R_E = 6400 \ km$. Take $\pi = 3.14$.

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What is the period of revolution of an Earth satellite? Ignore the height of the satellite above the surface of the Earth.Given:$(1)$ The value of gravitational acceleration $g = 10 \ ms^{-2}$.$(2)$ Radius of Earth $R_E = 6400 \ km$. Take $\pi = 3.14$.