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Chapter 6: Problem 812

Suppose the gravitational force varies inversely as the nth power of distance the time period of planet in circular orbit of radius \(\mathrm{R}\) around the sun will be proportional to (A) \(\mathrm{R}^{[(\mathrm{n}+1) / 2]}\) (B) \(\mathrm{R}^{[(\mathrm{n}-1) / 2]}\) (C) \(\mathrm{R}^{\mathrm{n}}\) (D) \(\mathrm{R}^{[(\mathrm{n}-1) / 2]}\)

Short Answer

Expert verified
The time period (T) of a planet in a circular orbit of radius R around the sun, when the gravitational force varies inversely as the nth power of distance, is proportional to \(R^{[(n+1) / 2]}\). Therefore, the correct answer is (A) \(R^{[(n+1) / 2]}\).

Step by step solution

01

Recall Kepler's Third Law

Kepler's third law states that the square of the time period of a planet's orbit around the sun is proportional to the cube of the semi-major axis of the orbit. For a circular orbit, the semi-major axis is equal to the radius of the orbit. Mathematically, this can be written as: \(T^2 = k R^3\), where T is the time period, R is the radius of the orbit, and k is the constant of proportionality.
02

Modify Kepler's Third Law for the given force variation

According to the problem, the gravitational force varies inversely as the nth power of distance, which can be written as: \(F = \frac{G M m}{R^n}\), where F is the gravitational force, G is the gravitational constant, M is the mass of the sun, m is the mass of the planet, and n is the power of distance dependence.
03

Express centripetal force in terms of radius

Centripetal force is required for the planet to stay in a circular orbit around the sun. This force is equal to the gravitational force. Therefore, we can write: \(\frac{G M m}{R^n} = \frac{m v^2}{R}\), where v is the tangential velocity of the planet. Now, we can rearrange this equation to find the relationship between v and R: \(v^2 = \frac{G M R^{(n-2)}}{R^n} = G M R^{(2-n)}\).
04

Relate time period, velocity and radius

The time period can also be defined as the time taken for the planet to complete one full orbit around the sun, which can be defined as: \(T = \frac{2 \pi R}{v}\), Now, square T to find the relation between T and R: \(T^2 = (\frac{2 \pi R}{v})^2\).
05

Derive the relationship between time period and radius

Now, substitute the expression for v^2 from step 3 into the equation for T^2 from step 4: \(T^2 = (\frac{2 \pi R}{\sqrt{G M R^{(2-n)}}})^2\) \(T^2 = \frac{4 \pi^2 R^2}{G M R^{(2-n)}}\), Then, simplify to find the relationship between T and R: \(T^2 = \frac{4 \pi^2}{G M} R^{(n+1)}\) Hence, \(T \propto R^{[(n+1)/2]}\).
06

Find the correct answer

Comparing the expression found in step 5 with the given options, we can conclude that the correct answer is: (A) \(R^{[(n+1) / 2]}\).

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