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Chapter 6: Problem 778

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) \(6 / \sqrt{2}\)

Short Answer

Expert verified
The time period of another satellite at a height of \(2.5R\) from the surface for earth is calculated to be 12 hours. However, none of the given options are correct.

Step by step solution

01

Find the formula for the time period of orbiting satellites

Using Kepler's Third Law, we can derive an expression for the time period of a satellite in orbit. The law states that the square of the orbital period of a satellite (T) is directly proportional to the cube of its semi-major axis (r). Mathematically, this can be expressed as \[T^2 \propto r^3,\] where r is the total distance (height + Earth's radius) from the center of the Earth to the satellite.
02

Calculate the time period of the geo-stationary satellite

In the case of the geo-stationary satellite, we know that its height is \(6R\) above the Earth's surface. Therefore, its total distance from the Earth's center is \[r_1 = R + 6R = 7R.\] Since the satellite is geo-stationary, we know that its time period \(T_1\) is equal to 24 hours. Therefore, from Kepler's Law, we have \[T_1^2 \propto r_1^3 \Rightarrow (24)^2 \propto (7R)^3.\]
03

Calculate the distance of the second satellite from the Earth's center

We are given that the second satellite is at a height of \(2.5R\) from the Earth's surface. Therefore, its total distance from the Earth's center is \[r_2 = R + 2.5R = 3.5R.\]
04

Apply Kepler's Third Law to the second satellite

Let the time period of the second satellite be \(T_2\). From Kepler's Third Law, we have \[T_2^2 \propto r_2^3 \Rightarrow T_2^2 \propto (3.5R)^3.\]
05

Find the ratio of time periods of the two satellites

Dividing the equations obtained in Step 2 and Step 4, we get \[\frac{T_2^2}{T_1^2} = \frac{(3.5R)^3}{(7R)^3}.\]
06

Solve for the time period of the second satellite

Plugging in the values of \(T_1\) and simplifying, we get \[\frac{T_2^2}{(24)^2} = \frac{(3.5R)^3}{(7R)^3} \Rightarrow T_2 = 24 \times \frac{3.5}{7} = 12\] Since we only have options in multiples of 6, we can compare this value to the options. The closest answer is 10, which corresponds to option (C). However, since we calculated the time period to be 12, none of the given options are correct.

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Most popular questions from this chapter

The escape velocity of a body from earth's surface is Ve. The escape velocity of the same body from a height equal to 7 R from earth's surface will be (A) \((\mathrm{Ve} / \sqrt{2})\) (B) \((\mathrm{Ve} / 2)\) (C) \((\mathrm{Ve} / 2 \sqrt{2})\) (D) \((\mathrm{Ve} / 4)\)

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