Problem 19 - A Complete Resource Book in Physics for JEE Main

Question

A body is suspended from a spring balance kept in a satellite. The reading of the balance is \(W_{1}\) when the satellite goes in an orbit of radius \(R\) and is \(W_{2}\) when it goes in an orbit of radius \(2 R\). (A) \(W_{1}=W_{2}\) (B) \(W_{1}W_{2}\) (D) \(W_{1} \neq W_{2}\)

Step-by-Step Solution

Answer

The short answer is: \(W_{1} < W_{2}\) and \(W_{1} \neq W_{2}\).

Step 1: Find the gravitational force

Gravitational force acting on the body in the satellite is given by: F = \( \frac{G \cdot M \cdot m}{r^{2}} \), where G is the gravitational constant, M is the mass of the Earth, m is the mass of the body, and r is the distance from the center of the Earth. In our case, we have two different orbits, so we need to find the gravitational force F₁ for orbit of radius R (F₁ = \( \frac{G \cdot M \cdot m}{R^{2}} \)) and F₂ for orbit of radius 2R (F₂ = \( \frac{G \cdot M \cdot m}{(2R)^{2}} \)).

Step 2: Find the centrifugal force

The centrifugal force acting on the body due to the satellite's motion in orbit is given by: Fc = \( m \cdot \omega^{2} \cdot r \), where ω is the angular velocity of the satellite and r is the distance from the center of the Earth. For a satellite in a circular orbit, the gravitational force and the centrifugal force are equal in magnitude, so we can write: F₁ = Fc₁ (for orbit with radius R) and F₂ = Fc₂ (for orbit with radius 2R).

Step 3: Find the effective weight

The effective weight (W) measured by the spring balance in a satellite orbiting the Earth can be calculated using: W = F - Fc Using this formula, we can find the effective weight \(W_{1}\) and \(W_{2}\) for the two orbits: \(W_{1}\) = F₁ - Fc₁ \(W_{2}\) = F₂ - Fc₂ Now, we have all the necessary information to compare \(W_{1}\) and \(W_{2}\), using the given options: (A) \(W_{1} = W_{2}\): If this option is true, then F₁ - Fc₁ = F₂ - Fc₂. (B) \(W_{1} < W_{2}\): If this option is true, then F₁ - Fc₁ < F₂ - Fc₂. (C) \(W_{1} > W_{2}\): If this option is true, then F₁ - Fc₁ > F₂ - Fc₂. (D) \(W_{1} \neq W_{2}\): If this option is true, then F₁ - Fc₁ is not equal to F₂ - Fc₂.

Step 4: Check the options

From our analysis, we know: F₁ = \( \frac{G \cdot M \cdot m}{R^{2}} \) F₂ = \( \frac{G \cdot M \cdot m}{(2R)^{2}} \) Fc₁ = m ⋅ ω₁² ⋅ R Fc₂ = m ⋅ ω₂² ⋅ 2R Now, using the fact that for a satellite in a circular orbit, the gravitational force and the centrifugal force are equal in magnitude: Fc₁ = F₁ Fc₂ = F₂ ω₁² ⋅ R = \( \frac{G \cdot M \cdot m}{R^{2}} \) ω₂² ⋅ 2R = \( \frac{G \cdot M \cdot m}{(2R)^{2}} \) Dividing the equations for ω₁² and ω₂², we get ω₁² : ω₂² = 1 : 4. Now let's check the options: (A) \(W_{1}=W_{2}\): Since ω₁² : ω₂² = 1 : 4, it implies that Fc₁ : Fc₂ = 1 : 2, which means Fc₁ < Fc₂. As a result, F₁ > F₂, so \(W_{1} \neq W_{2}\). (B) \(W_{1} F₂ and Fc₁ < Fc₂, then \(W_{1} = F₁ - Fc₁ < F₂ - Fc₂ = W_{2}\). The statement in option B is true. (C) \(W_{1}>W_{2}\): Since we have found that \(W_{1} < W_{2}\), the statement in option C is false. (D) \(W_{1} \neq W_{2}\): Since we have found that \(W_{1} < W_{2}\), the statement in option D is true. Therefore, the correct answer is option (B) \(W_{1} < W_{2}\) and option (D) \(W_{1} \neq W_{2}\).