Problem 815 - JEE Physics

Question

An artificial satellite moving in a circular orbit around earth has a total (kinetic + potential energy) \(E_{0}\), its potential energy is (A) \(-\mathrm{E}_{0}\) (B) \(1.5 \mathrm{E}_{0}\) (C) \(2 \mathrm{E}_{0}\) (D) \(\mathrm{E}_{0}\)

Step-by-Step Solution

Answer

The potential energy (U) of an artificial satellite moving in a circular orbit around Earth is twice the total energy (E): \(U = -2E_{0}\). The correct option is (C) \(2 \mathrm{E}_{0}\).

Step 1: Recall the formulas for gravitational potential energy and kinetic energy

Gravitational potential energy (U) is given by the formula: \(U = -\dfrac{GMm}{r}\), where G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance between the centers of the Earth and the satellite. Kinetic energy (K) is given by the formula: \(K = \dfrac{1}{2}mv^2\), where m is the mass of the satellite, and v is its orbital velocity.

Step 2: Express the orbital velocity in terms of the gravitational constant, Earth's mass, and distance

For a satellite in a circular orbit, the centripetal force is balanced by the gravitational force. That is, \(m\dfrac{v^2}{r} = \dfrac{GMm}{r^2}\). We can rearrange this equation for the orbital velocity: \(v^2 = \dfrac{GM}{r}\).

Step 3: Substitute the orbital velocity into the kinetic energy formula

Now, use the expression for orbital velocity found in Step 2 to substitute into the kinetic energy formula: \(K = \dfrac{1}{2}m\left(\dfrac{GM}{r}\right)\).

Step 4: Find the total energy (kinetic and potential energy) of the satellite

Add the kinetic and potential energy formulas to find the total energy of the satellite (E): \(E_{0} = U + K = -\dfrac{GMm}{r} + \dfrac{1}{2}m\left(\dfrac{GM}{r}\right)\).

Step 5: Solve for potential energy in terms of total energy

We have \(E_{0} = -\dfrac{GMm}{r} + \dfrac{1}{2}m\left(\dfrac{GM}{r}\right)\). Simplifying this equation, we get: \(E_{0} = - \dfrac{1}{2}\dfrac{GMm}{r}\). Thus, potential energy (U) in terms of total energy (E) is twice of the total energy: \(U = -2E_{0}\).

Step 6: Choose the correct answer from the given options

From the result obtained in Step 5, the potential energy (U) is -2 times the total energy (E). Hence, the correct option is (C) \(2 \mathrm{E}_{0}\).

Key Concepts

Gravitational Potential Energy

Gravitational potential energy is a fundamental concept in orbital mechanics that measures the energy due to position in a gravitational field. It is especially relevant when discussing satellites such as those orbiting Earth.
Gravitational potential energy (\( U \)) is expressed as \( U = -\dfrac{GMm}{r} \). Here's a breakdown of what each symbol represents:

• \( G \) is the gravitational constant, a value that defines the intensity of the gravitational force.
• \( M \) stands for the mass of the Earth.
• \( m \) is the mass of the satellite.
• \( r \) is the distance from the center of the Earth to the satellite.

The negative sign in the equation indicates that gravitational forces are attractive. As the satellite's altitude increases, its potential energy becomes less negative, implying an increase in the satellite's energy state.
In a circular orbit, gravitational potential energy is a key factor in determining how much energy the satellite possesses due to its position relative to the Earth.

Circular Orbit

A circular orbit occurs when a satellite travels around a planet, like Earth, with a constant distance from the center of the planet. This type of orbit is characterized by a balance between two forces: gravitational force and centripetal force.
In a circular orbit, the gravitational force acting on the satellite provides the necessary centripetal force to keep it moving in a circle. This balance can be expressed as:\[ m\dfrac{v^2}{r} = \dfrac{GMm}{r^2} \]Rearranging this equation, we find the orbital velocity:\[ v^2 = \dfrac{GM}{r} \]This equation shows that in a circular orbit:

• The orbital velocity is constant.
• The orbital radius \( r \) directly affects velocity.

Understanding these forces helps us predict how changes in velocity or altitude will affect the orbit of a satellite. This knowledge is essential for space missions and the placement of satellites.

Kinetic Energy

Kinetic energy refers to the energy an object possesses due to its motion. In the context of satellite orbits, it is crucial to balance kinetic and potential energy to maintain a stable orbit.
The kinetic energy (\( K \)) of a satellite is given by:\[ K = \dfrac{1}{2}mv^2 \]Substituting the expression for orbital velocity from a circular orbit, it becomes:\[ K = \dfrac{1}{2}m\left(\dfrac{GM}{r}\right) \]Key points about kinetic energy in orbits include:

• It contributes to the total mechanical energy of the satellite.
• A balance between kinetic and potential energy, summed as total energy \( E_0 \), is vital for orbit maintenance.

As shown in the original solution, the sum of kinetic and potential energy leads to the total energy of the satellite. This understanding aids in calculating the required speed and energy for launching and stabilizing satellites in orbit.