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Period or height of a satellite in terms of g and the earth's radius

This type of question, asked 3 times: 10 Apr 2023, 4 Apr 2024, 22 Jan 2026 S2. All 3 questions below, oldest first.

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  1. Question 1

    10 Apr 2023

    98
    The time period of a satellite, revolving above earth's surface at a height equal to RR will be (Given g=π2 m/s2,R=g =\pi^{2}\text{ }m/s^{2},R = radius of earth)

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  2. Question 2

    4 Apr 2024

    98
    Correct formula for height of a satellite from earths surface is :

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  3. Question 3

    22 Jan 2026 S2

    14
    Given below are two statements : Statement I: A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth. Statement II : The time period of revolution of the satellite is T=2πRegT = 2\pi\sqrt{\frac{R_{e}}{g}} (for satellite very close to the earth surface), where ReR_{e} radius of earth and gg acceleration due to gravity. In the light of the above statements, choose the correct answer from the options given below :

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