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Chapter 6: Problem 804

The earth revolves round the sun in one year. If distance between then becomes double the new period will be years. (A) \(0.5\) (B) \(2 \sqrt{2}\) (C) 4 (D) 8

Short Answer

Expert verified
The new period will be \(2 \sqrt{2}\) years.

Step by step solution

01

Write down the given information

The Earth revolves around the Sun in one year, so the period T₁ = 1 year. We need to find the new period, T₂, when the distance between them becomes double.
02

Apply Kepler's Third Law

According to Kepler's Third Law, the square of the period of a planet is directly proportional to the cube of the semi-major axis of its orbit: \(T^2 \propto a^3\) Where T is the period of revolution and a is the semi-major axis.
03

Set up a proportion

Since we're given that the distance between the Earth and the Sun doubles, we can set up a proportion using Kepler's Third Law: \(\frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3}\) We know that \(a_2\) is double that of \(a_1\), so we can substitute \(2a_1\) for \(a_2\): \(\frac{T_1^2}{a_1^3} = \frac{T_2^2}{(2a_1)^3}\)
04

Substitute given values

Now, we can substitute the given values: \(T_1 = 1\) year and \(a_2 = 2a_1\): \(\frac{1^2}{a_1^3} = \frac{T_2^2}{(2a_1)^3}\)
05

Simplify and solve for T₂

To solve for T₂, we can simplify the equation: \(\frac{1}{a_1^3} = \frac{T_2^2}{8a_1^3}\) Now, we can cross multiply: \(8a_1^3 = T_2^2 \cdot a_1^3\) Notice that both sides have the same \(a_1^3\) term, which we can divide to cancel it out: \(8 = T_2^2\) To solve for T₂, we take the square root of both sides: \(T_2 = \sqrt{8}\)
06

Write the final answer

By solving, we find: \(T_2 = 2 \sqrt{2}\) Thus, the new period will be \(2 \sqrt{2}\) years, which corresponds to the option (B).

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