Problem 767 - JEE Physics

Question

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) \(3 / 2 \mathrm{~V}\)

Step-by-Step Solution

Answer

The speed of satellite B is 3V.

Step 1: 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.

Step 2: 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: \[F_{gB} = G \frac{m_B \cdot M_P}{r_B^2}\]

Step 3: 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: \[\frac{m_A \cdot v_A^2}{r_A} = G \frac{m_A \cdot M_P}{r_A^2}\] \[\frac{m_B \cdot v_B^2}{r_B} = G \frac{m_B \cdot M_P}{r_B^2}\]

Step 4: 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: \[\frac{v_A^2 \cdot r_B}{v_B^2 \cdot r_A} = \frac{r_B}{r_A}\]

Step 5: 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: \[\frac{(3V)^2 \cdot R}{v_B^2 \cdot 4R} = \frac{R}{4R}\]

Step 6: 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) \(3 / 2 \mathrm{~V}\)

Key Concepts

Gravitational Force

Gravitational Force is a fundamental concept in physics that describes the attraction between two masses. Every object in the universe exerts a gravitational force on every other object. The strength of this force depends on two factors:

• The masses of the objects.
• The distance between their centers.

The formula for gravitational force is given by:\[F_g = G \frac{m_1 \cdot m_2}{r^2},\]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.
The gravitational force is always attractive, pulling objects towards each other. When considering masses like satellites and planets, the gravitational force plays a crucial role in keeping the satellites in orbit. It acts as the centripetal force required to maintain circular motion. This relationship is the core concept behind problems involving satellite motion and circular orbits.

Satellite Motion

Satellite Motion refers to how satellites orbit planets due to the interplay of gravitational and centripetal force. Satellites travel in an elliptical or circular path around a planet, which requires a delicate balance of forces.

• Gravitational force pulls the satellite towards the planet.
• Centripetal force is the "inward force" needed for circular motion.
• The satellite's velocity keeps it moving forward, counteracting the inward pull.

If the gravitational force is too strong or too weak compared to the satellite's speed, the path of the satellite changes. Too much speed and the satellite could escape into space; too little and it might crash into the planet. The constant speed and radius of a satellite in a perfect circular orbit are determined by the balance of gravitational pull acting as the centripetal force.

Circular Orbits

Circular Orbits occur when a satellite maintains a constant distance from the center of the planet as it travels around it. This can be achieved when:

• The satellite's speed is just right for the gravitational pull it experiences.
• The centripetal force needed for circular motion is exactly provided by gravity.

To find the speed for a circular orbit, you equate the gravitational force and the centripetal force:\[ \frac{m \cdot v^2}{r} = G \frac{m \cdot M}{r^2}, \]where \( m \) is the satellite's mass, \( v \) is its orbital speed, \( r \) is the orbit's radius, and \( M \) is the planet's mass.
This equation shows that for a given radius, only certain speeds will keep the satellite in a stable, circular path.
The interesting thing about circular orbits is that they simplify calculations and are often used as a first approximation when studying more complex orbital mechanics.