Problem 29 - Master Resource Book in JEE Main Physics

Question

Two identical satellites are at \(R\) and \(7 R\) away from earth surface, the wrong statement is ( \(R=\) Radius of earth) (a) ratio of total energy will be \(y\) (b) ratio of kinetic energies will be \(y\) (c) ratio of potential energies will be \(y\) (d) ratio of total energy will be 4 but ratio of potential and kinetic energy will be 1

Step-by-Step Solution

Answer

(d) is the incorrect statement.

Step 1: Understanding the Problem

We have two satellites at distances of \(R\) and \(7R\) from the Earth's surface, which means they are actually \(2R\) and \(8R\) from the Earth's center because \(R\) is the radius of the Earth. The task is to identify the incorrect statement about the ratios of energies (total, kinetic, and potential) for these satellites.

Step 2: Calculating the Gravitational Potential Energy

The gravitational potential energy for a satellite at a distance \(r\) from the center of the Earth is given by \(U = -\frac{GMm}{r}\), where \(M\) is the mass of the Earth and \(m\) is the mass of the satellite. Calculate the potential energies at distances \(2R\) and \(8R\):\[ U_1 = -\frac{GMm}{2R}, \quad U_2 = -\frac{GMm}{8R} \]

Step 3: Finding the Ratio of Potential Energies

The ratio of potential energies is:\[ \frac{U_1}{U_2} = \frac{-\frac{GMm}{2R}}{-\frac{GMm}{8R}} = 4 \]

Step 4: Calculating Kinetic Energy in Orbit

The kinetic energy for a satellite in orbit is half the magnitude of the potential energy and given by \(K = \frac{GMm}{2r}\). Calculate the kinetic energies at \(2R\) and \(8R\):\[ K_1 = \frac{GMm}{4R}, \quad K_2 = \frac{GMm}{16R} \]

Step 5: Finding the Ratio of Kinetic Energies

The ratio of kinetic energies is:\[ \frac{K_1}{K_2} = \frac{\frac{GMm}{4R}}{\frac{GMm}{16R}} = 4 \]

Step 6: Calculating Total Energy

Total energy is the sum of kinetic and potential energy, \( E = U + K = -\frac{GMm}{2r}\). Find the total energies for \(2R\) and \(8R\):\[ E_1 = -\frac{GMm}{4R}, \quad E_2 = -\frac{GMm}{16R} \]

Step 7: Finding the Ratio of Total Energies

The ratio of total energies is:\[ \frac{E_1}{E_2} = \frac{-\frac{GMm}{4R}}{-\frac{GMm}{16R}} = 4 \]

Step 8: Determining the Incorrect Statement

All results show that the ratio of potential, kinetic, and total energies is \(4\). Therefore, statement (d) claiming that the ratio of total energy is 4 but the ratios of potential and kinetic energies are 1 is incorrect.

Key Concepts

Gravitational Potential Energy

Gravitational potential energy, often denoted as "U," is the energy held by an object because of its position relative to other objects, due to gravitation. For satellites orbiting Earth, this energy determines how strongly the Earth attracts them. The formula for calculating the gravitational potential energy of a satellite at a distance "r" from the Earth's center is:\[ U = -\frac{GMm}{r} \]where "G" is the gravitational constant, "M" is the mass of the Earth, and "m" is the mass of the satellite. The negative sign indicates that energy is lower when two bodies are closer, i.e., when "r" is small.

• When the satellite is closer to Earth (e.g., at a distance of \(2R\) from Earth's center), the potential energy is more negative compared to when it is further away (e.g., at \(8R\)).
• This indicates a stronger gravitational attraction when the satellite is nearer.
• Potential energy differences are crucial for understanding energy conservation in celestial mechanics.

For our problem, satellites at distances \(2R\) and \(8R\) have potential energies \(U_1 = -\frac{GMm}{2R}\) and \(U_2 = -\frac{GMm}{8R}\), respectively. The ratio \(\frac{U_1}{U_2} = 4\) confirms the practical effect of distance on gravitational potential energy.

Kinetic Energy in Orbit

Kinetic energy, represented as "K," describes the energy that an object possesses due to its motion. For satellites, it plays a crucial role in maintaining their stable orbits around Earth. The kinetic energy of a satellite in orbit depends on its velocity and can be derived from its gravitational motion as:\[ K = \frac{GMm}{2r} \]

• Kinetic energy is proportional to the inverse of the radius; therefore, as distance from Earth decreases, kinetic energy increases.
• Satellites in closer orbits (e.g., at \(2R\)) have greater kinetic energies than those farther away at \(8R\).
• This explains why closer satellites must travel faster to maintain their orbit.

For our satellites, the kinetic energies are \(K_1 = \frac{GMm}{4R}\) and \(K_2 = \frac{GMm}{16R}\), showing that they also have a ratio of \(4\), much like the potential energy.

Total Orbital Energy

Total orbital energy "E" is a key concept in understanding satellite dynamics. It is the sum of gravitational potential energy and kinetic energy:\[ E = U + K = -\frac{GMm}{2r} \]

• This total energy remains constant if a satellite's orbit remains stable, fulfilling the conservation of energy law.
• The formula indicates that total energy is negative, signifying a bound system where the satellite is gravitationally tied to Earth.
• Both components, kinetic and potential energies, shift with satellite altitude changes, but their sum is consistent with stable orbits.

In our scenario, at distances \(2R\) and \(8R\), the total energies are \(E_1 = -\frac{GMm}{4R}\) and \(E_2 = -\frac{GMm}{16R}\), respectively, sharing the same ratio of \(4\). Understanding this ratio simplifies analyzing satellite behavior and predicting orbital changes.