A geo-stationary satellite is orbiting the earth of a height of 6 R above the surface of earth,

Name: Problem 778, Chapter 6: JEE Physics
Uploaded: 2022-02-25T15:12:32-08:00
Duration: 47 s
Channel: Hast Aggarwal
Description: A geo-stationary satellite is orbiting the earth of a height of 6 R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 2.5 R from the surface for earth is =…………(A) 6 (B) 6 √(2) (C) 10 (D) 6 / √(2)

step 1 So the question is, periodic time of a satellite revolving above surface and I equals to R, wher

Question

A geo-stationary satellite is orbiting the earth of a height of $6 \mathrm{R}$ above the surface of earth, $\mathrm{R}$ being the radius of earth. The time period of another satellite at a height of $2.5 \mathrm{R}$ from the surface for earth is $=\ldots \ldots \ldots \ldots$ (A) 6 (B) $6 \sqrt{2}$ (C) 10 (D) $6 / \sqrt{2}$

   A geo-stationary satellite is orbiting the earth of a height of $6 \mathrm{R}$ above the surface of earth, $\mathrm{R}$ being the radius of earth. The time period of another satellite at a height of $2.5 \mathrm{R}$ from the surface for earth is $=\ldots \ldots \ldots \ldots$
(A) 6
(B) $6 \sqrt{2}$
(C) 10
(D) $6 / \sqrt{2}$

JEE Physics

StemEZ 1st Edition

Chapter 6, Problem 778

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Solved by Expert Hast Aggarwal

02/25/2022

Step 1

The formula for the orbital period $ T $ of a satellite is given by: \[ T = 2\pi \sqrt{\frac{r^3}{GM}} \] where $ r $ is the distance from the center of the Earth to the satellite, $ G $ is the gravitational constant, and $ M $ is the mass of the Earth. Show more…

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A geo-stationary satellite is orbiting the earth of a height of $6 \mathrm{R}$ above the surface of earth, $\mathrm{R}$ being the radius of earth. The time period of another satellite at a height of $2.5 \mathrm{R}$ from the surface for earth is $=\ldots \ldots \ldots \ldots$ (A) 6 (B) $6 \sqrt{2}$ (C) 10 (D) $6 / \sqrt{2}$

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Key Concepts

Orbital Dynamics

Orbital dynamics explains how the gravitational force acting on a satellite provides the necessary centripetal force to maintain its orbit. This interplay determines the satellite's speed and period, and is key to understanding how variations in orbital radius affect the time it takes to complete one orbit.

Kepler's Third Law

Kepler's Third Law establishes that the square of the orbital period (T²) of a satellite is proportional to the cube of the orbit's radius (r³). This law is fundamental in orbital mechanics and allows the comparison of the periods of satellites at different altitudes based on their distances from the center of the orbiting body.

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