Two bodies of equal masses revolve in circular orbits of radii \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) with the same period Their centripetal forces are in the ratio. (A) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)^{2}\) (B) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\) (C) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{2}\) (D) \(\left.\sqrt{(}_{1} R_{2}\right)\)

Short Answer

Expert verified
The ratio of the centripetal forces is \(\frac{R_1}{R_2}\), which is the answer (B).

Step by step solution

01

Recall the centripetal force formula

The formula for centripetal force is given by: \(F = m \frac{v^2}{r}\) Where \(F\) is the centripetal force, \(m\) is the mass of the body, \(v\) is the tangential velocity, and \(r\) is the radius of the circular orbit.
02

Express velocity in terms of period

Given that the two bodies have the same period (\(T\)), we can express their velocities in terms of the period. The relation between tangential velocity, radius, and period is given by: \(v = \frac{2 \pi r}{T}\)
03

Substitute velocity in the centripetal force formula

We can now substitute the expression of tangential velocity from Step 2 into the centripetal force formula from Step 1 for both orbits. Body 1: \(F_1 = m \left(\frac{2\pi R_1}{T}\right)^2 \cdot \frac{1}{R_1}\) Body 2: \(F_2 = m \left(\frac{2\pi R_2}{T}\right)^2 \cdot \frac{1}{R_2}\)
04

Simplify and find the ratio of the centripetal forces

We can simplify each equation and then find the ratio of the centripetal forces. \(F_1 = \frac{4\pi^2 m R_1}{T^2}\) \(F_2 = \frac{4\pi^2 m R_2}{T^2}\) Now find the ratio \(\frac{F_1}{F_2}\): \(\frac{F_1}{F_2} = \frac{\frac{4\pi^2 m R_1}{T^2}}{\frac{4\pi^2 m R_2}{T^2}}\) Cancel out the common terms: \(\frac{F_1}{F_2} = \frac{R_1}{R_2}\) So, the ratio of the centripetal forces is \(\frac{R_1}{R_2}\), which is the answer (B).

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