Problem 729 - JEE Physics

Question

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) \(\left(1 / \mathrm{R}^{3 / 2}\right)\)

Step-by-Step Solution

Answer

The kinetic energy (KE) of a satellite orbiting Earth is given by the expression \(KE = \frac{1}{2} m_s * v^2\), where \(m_s\) is the mass of the satellite and \(v\) is the orbital speed. Using the relationship between gravitational force and centripetal force, we find that \(v^2 = \frac{G * m_e}{R}\), where \(G\) is the gravitational constant, \(m_e\) is the mass of Earth, and \(R\) is the radius of the orbit. Substituting this expression into the kinetic energy formula, we get \(KE = \frac{1}{2} m_s * \frac{G * m_e}{R}\), which simplifies to \(KE \propto \frac{1}{R}\). Therefore, the correct answer is (A) \(\frac{1}{R}\).

Step 1: Write the expression for gravitational force

The gravitational force between Earth and the satellite can be expressed as: \[F_G = \frac{G * m_e * m_s}{R^2}\] where \(F_G\) is the gravitational force, \(G\) is the gravitational constant, \(m_e\) is the mass of Earth, and \(m_s\) is the mass of the satellite.

Step 2: Write the expression for the centripetal force

The centripetal force required for the satellite's circular motion can be written as: \[F_c = \frac{m_s * v^2}{R}\] where \(F_c\) is the centripetal force and \(v\) is the satellite's orbital speed.

Step 3: Equate gravitational force and centripetal force

Since the gravitational force between Earth and the satellite provides the centripetal force required for the satellite's circular motion, we can set the two expressions equal: \[\frac{G * m_e * m_s}{R^2} = \frac{m_s * v^2}{R}\]

Step 4: Solve for the satellite's orbital speed

To find the satellite's orbital speed, solve the equation above for \(v^2\): \[v^2 = \frac{G * m_e}{R}\]

Step 5: Write the expression for kinetic energy

The kinetic energy (KE) of the satellite orbiting Earth is given by the expression: \[KE = \frac{1}{2} m_s * v^2\]

Step 6: Substitute the expression for orbital speed into the kinetic energy expression

By substituting the expression from step 4 into the kinetic energy expression, we get: \[KE = \frac{1}{2} m_s * \frac{G * m_e}{R}\]

Step 7: Determine how kinetic energy is proportional to \(R\)

Simplifying the expression from step 6, we have: \[KE \propto \frac{1}{R}\] which means that the kinetic energy of the satellite is proportional to \(\frac{1}{R}\).

Step 8: Select the correct answer

Based on the above derivation, the correct answer is: (A) \(\frac{1}{R}\)

Key Concepts

Gravitational Force

Gravitational force is the glue that holds the universe together. It is an attractive force that acts between any two masses in the universe. The formula to calculate the gravitational force between two objects is given by Newton's law of gravitation:

• \( F_G = \frac{G \cdot m_1 \cdot m_2}{r^2} \)

where:

• \( F_G \) is the gravitational force.
• \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, m^3 \, kg^{-1} \, s^{-2}) \).
• \( m_1 \) and \( m_2 \) are the masses of the objects.
• \( r \) is the distance between the centers of the two masses.

Gravitational force is what keeps satellites orbiting the Earth. When a satellite is in orbit, it is constantly being pulled towards the Earth by gravity. However, its forward motion means it keeps missing, resulting in a stable orbit. This balance of forces is crucial in understanding the dynamics of orbital mechanics.

Kinetic Energy

Kinetic energy is the energy of motion. Any object in motion possesses kinetic energy. The formula for kinetic energy is

• \( KE = \frac{1}{2} m v^2 \)

where:

• \( KE \) is the kinetic energy,
• \( m \) is the mass of the object,
• \( v \) is the velocity of the object.

In the context of a satellite, kinetic energy helps us understand how fast the satellite travels through its orbit. More kinetic energy means more speed. The balance between a satellite's kinetic energy and gravitational potential energy determines its orbit. For a satellite in circular orbit, its kinetic energy is derived from the gravitational forces acting on it.
Kinetic energy's dependence on velocity squared is critical in answering exercises like determining proportional relationships. For example, in the original exercise, we see that kinetic energy varies inversely with the radius \( (R) \) of the orbit as noticed from the substitution:

• \( KE = \frac{1}{2} m_s \cdot \frac{G \cdot m_e}{R} \)

Thus, \( KE \propto \frac{1}{R} \). This inverse relationship implies that as the satellite's orbit radius increases, its speed decreases, hence reducing its kinetic energy.

Orbital Mechanics

Orbital mechanics is the study of the motions of objects in space, influenced primarily by gravitational forces. It applies physical laws of motion to predict and understand paths of satellites and other celestial bodies. A key concept in orbital mechanics is the balance between gravitational force and an object's velocity, which together dictate the shape and size of an orbit.
One must consider centripetal force when diving into orbital mechanics. In a stable orbit, the gravitational force provides the centripetal force necessary to maintain the satellite's path. This means:

• \( \frac{G \cdot m_e \cdot m_s}{R^2} = \frac{m_s \cdot v^2}{R} \)

By solving for orbital speed \( v \), we find it has a dependency on the distance to the center of the mass it orbits:

• \( v^2 = \frac{G \cdot m_e}{R} \)

Orbital mechanics explains how changing a satellite's orbit radius affects its speed and energy. For instance, a satellite closer to Earth must move faster to counterbalance the stronger gravitational pull. Understanding these relationships helps in designing stable orbits for satellites, ensuring they provide needed services like GPS or weather monitoring without falling back to Earth or drifting aimlessly into space.