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Chapter 5: Problem 586

If distance of the earth becomes three times that of the present distance from the sun then number of days in one year will be .... \(\\{\mathrm{A}\\}[365 \times 3]\) \(\\{\mathrm{B}\\}[365 \times 27]\) \(\\{\mathrm{C}\\}[365 \times(3 \sqrt{3})]\) \(\\{\mathrm{D}\\}[365 /(3 \sqrt{3})]\)

Short Answer

Expert verified
\(T_2 = 365 \sqrt{27}\)

Step by step solution

01

Understand Kepler's Third Law

Kepler's Third Law states that the square of the time period (T) of a planet's orbit is proportional to the cube of the semi-major axis (a) of its orbit. Mathematically, it is represented as: \(T^2 \propto a^3\) For any two orbits, we can write the equation as: \(\frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3}\) Here, \(T_1\) and \(T_2\) represent the time periods of the two orbits, and \(a_1\) and \(a_2\) are their respective semi-major axes.
02

Set up the equation with the given information

In this case, the first orbit is the present orbit of the Earth with a distance of a, and the second orbit is when the Earth's distance becomes three times the present distance, i.e., 3a. We are given that the number of days in one year currently is 365 days, i.e., \(T_1 = 365\). We have to find the number of days (\(T_2\)) in one year when the Earth's distance is three times the present distance. Using the relation from Kepler's Third Law: \(\frac{365^2}{T_2^2} = \frac{a^3}{(3a)^3}\)
03

Solve for \(T_2\)

Now, we need to solve this equation for \(T_2\). We can start by simplifying the equation and canceling out the terms: \(\frac{365^2}{T_2^2} = \frac{a^3}{27a^3}\) From this, we can observe that \(a^3\) will cancel out on both sides: \(\frac{365^2}{T_2^2} = \frac{1}{27}\) Next, we need to isolate \(T_2\) on one side of the equation, so we can cross-multiply: \(T_2^2 = 27 \cdot 365^2\) To find the value of \(T_2\), we can take the square root of both sides: \(T_2 = \sqrt{27 \cdot 365^2}\) Now, we can compute the value of \(T_2\): \(T_2 = 365 \sqrt{27}\)
04

Match the answer with the given options

The value of \(T_2\) that we found is \(365 \sqrt{27}\). If we look at the answer choices, we can see that the correct option is: \({\mathrm{B}}[365 \times 27]\)

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Most popular questions from this chapter

A gramophone record of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) is rotating with angular speed \(\mathrm{W}\). If two pieces of wax each of mass \(\mathrm{M}\) are kept on it at a distance of \(R / 2\) from the centre on opposite side then the new angular velocity will be..... \(\\{\mathrm{A}\\}(\omega / 2)\) \(\\{B\\}[m \omega /(M+m)\) \(\\{C\\}[M \omega /(M+m)]\) \(\\{\mathrm{D}\\}[(\mathrm{M}+\mathrm{m}) / \mathrm{M} \omega]\)

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