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Chapter 6: Problem 793

A geostationary satellite is orbiting the earth at a height of \(5 \mathrm{R}\) above that of surface of the earth. \(\mathrm{R}\) being the radius of the earth. The time period of another satellite in hours at a height of $2 \mathrm{R}\( from the surface of earth is \)\ldots \ldots \ldots .$ hr (A) 5 (B) 10 (C) \(6 \sqrt{2}\) (D) \(6 / \sqrt{2}\)

Short Answer

Expert verified
The short answer based on the provided solution is: The time period of the satellite at a height of 2R from the surface of the Earth is 12 hours. However, the correct answer is not among the available options. The closest option would be (B) 10 hours, but it's not the exact answer.

Step by step solution

01

Understand the formula needed for the problem

We need to use the formula for the period of a satellite orbiting around a planet (in our case, the Earth). The formula is given by Kepler's Third Law: \[T = 2 \pi \sqrt{\frac{a^3}{GM}}\] where \(T\) is the time period of the satellite, \(a\) is the orbital radius (which is the sum of the Earth's radius and the satellite's height from the surface), \(G\) is the gravitational constant, and \(M\) is the mass of the Earth.
02

Find the time period of the geostationary satellite

We know that a geostationary satellite orbits at a height of 5R above the surface of the Earth. Therefore, its orbital radius is: \[a_1 = R + 5R = 6R\] Now, we can plug in values to find the time period of the geostationary satellite: \[T_1 = 2 \pi \sqrt{\frac{(6R)^3}{GM}}\]
03

Express the time period of the given satellite in terms of the geostationary satellite's time period

The given satellite orbits at a height of 2R from the surface of the Earth, so its orbital radius is: \[a_2 = R + 2R = 3R\] The time period of this satellite is given by: \[T_2 = 2 \pi \sqrt{\frac{(3R)^3}{GM}}\] To find out the time period of the given satellite in terms of the period of the geostationary satellite, we can write: \[\frac{T_2}{T_1} = \frac{2 \pi \sqrt{\frac{(3R)^3}{GM}}}{2 \pi \sqrt{\frac{(6R)^3}{GM}}}\]
04

Solve for the time period of the given satellite

By simplifying the above equation, we have: \[\frac{T_2}{T_1} = \frac{\sqrt{(3R)^3}}{\sqrt{(6R)^3}} = \frac{3R}{6R} = \frac{1}{2}\] \[\Rightarrow T_2 = \frac{1}{2} T_1\] We are given that the time period of the geostationary satellite is 24 hours (one orbit in a day), so: \[T_2 = \frac{1}{2} \cdot 24 = 12\] So, the time period of the given satellite is 12 hours. Since the correct answer is not among the available options, there might be a mistake in the problem statement or the options provided. The closest available option would be (B) 10 hours, but it's not the exact answer.

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