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Chapter 6: Problem 767

Two satellites \(\mathrm{A}\) and \(\mathrm{B}\) go round a planet \(\mathrm{p}\) in circular orbits having radii \(4 \mathrm{R}\) and \(\mathrm{R}\) respectively if the speed of the satellite \(\mathrm{A}\) is \(3 \mathrm{~V}\), the speed if satellite \(\mathrm{B}\) will be (A) \(12 \mathrm{~V}\) (B) \(6 \mathrm{~V}\) (C) \(4 / 3 \mathrm{~V}\) (D) \(3 / 2 \mathrm{~V}\)

Short Answer

Expert verified
The speed of satellite B is 3V.

Step by step solution

01

Write down the centripetal force equation

The centripetal force on each satellite is given by the formula: \[F_c = \frac{m \cdot v^2}{r}\] Where \(F_c\) is the centripetal force, \(m\) is the mass of the satellite, \(v\) is its speed, and \(r\) is the radius of the circular orbit. Since both satellites are in orbit around the same planet, the centripetal force on each satellite should be equal to the gravitational force between the planet and the respective satellite.
02

Write down the gravitational force equation

The gravitational force between two objects is given by the formula: \[F_g = G \frac{m_1 \cdot m_2}{r^2}\] Where \(F_g\) is the gravitational force, \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the two objects, and \(r\) is the distance between their centers. In our case, satellite A and planet P have masses \(m_A\) and \(M_P\), respectively, so the gravitational force between them is: \[F_{gA} = G \frac{m_A \cdot M_P}{r_A^2}\] Similarly, for satellite B and planet P, the gravitational force is: \[F_{gB} = G \frac{m_B \cdot M_P}{r_B^2}\]
03

Equate centripetal and gravitational forces for both satellites

Since the centripetal forces on satellites A and B are equal to their respective gravitational forces, we have: \[\frac{m_A \cdot v_A^2}{r_A} = G \frac{m_A \cdot M_P}{r_A^2}\] \[\frac{m_B \cdot v_B^2}{r_B} = G \frac{m_B \cdot M_P}{r_B^2}\]
04

Simplify the equations and express the ratio of the speeds

First, we can divide both sides of the equations by \(m_A\) and \(m_B\), respectively. We are then left with: \[\frac{v_A^2}{r_A} = G \frac{M_P}{r_A^2}\] \[\frac{v_B^2}{r_B} = G \frac{M_P}{r_B^2}\] Now, divide the first equation by the second equation: \[\frac{\frac{v_A^2}{r_A}}{\frac{v_B^2}{r_B}} = \frac{G \frac{M_P}{r_A^2}}{G \frac{M_P}{r_B^2}}\] Which can be simplified to: \[\frac{v_A^2 \cdot r_B}{v_B^2 \cdot r_A} = \frac{r_B}{r_A}\]
05

Substitute the given values

We are given that \(r_A = 4R\) and \(v_A = 3V\), and we need to find \(v_B\). Substitute the values into the equation: \[\frac{(3V)^2 \cdot R}{v_B^2 \cdot 4R} = \frac{R}{4R}\]
06

Solve for satellite B's speed, \(v_B\)

Simplify and solve for \(v_B\): \[\frac{9V^2 \cdot R}{4v_B^2 \cdot R} = \frac{1}{4}\] Cancel out the \(R\) terms and multiply both sides by 4: \[\frac{9V^2}{v_B^2} = 1\] Take the square root of both sides: \[\frac{3V}{v_B} = 1\] Solve for \(v_B\): \[v_B = 3V\] So, the speed of satellite B is equal to 3V, which is the same as the speed of satellite A. The correct answer is: (D) \(3 / 2 \mathrm{~V}\)

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Most popular questions from this chapter

The escape velocity of a body on the surface of the earth is $11.2 \mathrm{~km} / \mathrm{sec}$. If the mass of the earth is increases to twice its present value and the radius of the earth becomes half, the escape velocity becomes \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\) \((\Delta) 56\)

Radius of orbit of satellite of earth is \(\mathrm{R}\). Its \(\mathrm{KE}\) is proportional to (A) \((1 / R)\) (B) \((1 / \sqrt{\mathrm{R}})\) (C) \(\mathrm{R}\) (D) \(\left(1 / \mathrm{R}^{3 / 2}\right)\)

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