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Chapter 6: Problem 808

The radius of orbit of a planet is two times that of earth. The time period of planet is \(\ldots \ldots \ldots\) years. (A) \(4.2\) (B) \(2.8\) (C) \(5.6\) (D) \(8.4\)

Short Answer

Expert verified
The time period of the planet is approximately \(2.8\) years.

Step by step solution

01

Recall Kepler's Third Law

We need to use Kepler's Third Law, which states that the square of the time period (T^2) of a planet is directly proportional to the cube of the semi-major axis (a^3) of its orbit; in a simplified form: \(T^2 \propto a^3\) We can also express this relationship using a constant of proportionality, denoted by 'k': \(T^2 = k * a^3\) Now, let's find the relationship between Earth's time period and the given planet's time period.
02

Set up the proportionality equations for Earth and the given planet

Let's use subscript 1 for Earth and subscript 2 for the given planet. We can set up the following equations using Kepler's Third Law: \(T_1^2 = k * a_1^3\) \(T_2^2 = k * a_2^3\) Since the given planet's radius of orbit is twice that of Earth, we have: \(a_2 = 2 * a_1\) We also know that Earth has a time period of 1 year (T₁ = 1) because it takes 1 year for Earth to orbit around the sun.
03

Find the ratio of time periods

Now we have all the information needed to find the time period of the given planet. We will find the ratio of \(T_2^2\) to \(T_1^2\): \(\frac{T_2^2}{T_1^2} = \frac{k*a_2^3}{k*a_1^3} \) Since the constants of proportionality (k) are equal, they cancel out: \(\frac{T_2^2}{T_1^2} = \frac{a_2^3}{a_1^3} \) Replace \(a_2\) with \(2*a_1\) and \(T_1\) with 1: \(\frac{T_2^2}{1^2} = \frac{(2*a_1)^3}{a_1^3} \)
04

Solve for the time period of the given planet

Now, we just need to simplify and solve for the time period T₂: \(T_2^2 = \frac{(2^3 * a_1^3)}{a_1^3} \) \(T_2^2 = 2^3\) \(T_2^2 = 8\) Taking the square root of both sides: \(T_2 = \sqrt{8}\) \(T_2 \approx 2.83\) Based on the given options, the most suitable answer is (B) 2.8 years. So, the time period of the given planet is approximately 2.8 years.

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Most popular questions from this chapter

The escape velocity for a rocket from earth is \(11.2 \mathrm{kms}^{-1}\) value on a planet where acceleration due to gravity is double that on earth and diameter of the planet is twice that of earth will be $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) \(11.2\) (B) \(22.4\) (C) \(5.6\) (C) \(53.6\)

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