Problem 45 - Physics for Scientists and Engineers with Modern Physics

Question

(I) For a satellite of mass \(m_{\mathrm{s}}\) in a circular orbit of radius \(r_{\mathrm{S}}\) around the Earth, determine \((a)\) its kinetic energy \(K,\) (b) its potential energy \(U(U=0\) at infinity \(),\) and \((c)\) the ratio \(K / U\)

Step-by-Step Solution

Answer

(a) \( K = \frac{G M_{ ext{E}} m_{ ext{s}}}{2 r_{ ext{s}}} \), (b) \( U = -\frac{G M_{ ext{E}} m_{ ext{s}}}{r_{ ext{s}}} \), (c) \( \frac{K}{U} = -\frac{1}{2} \).

Step 1: Understand the Basics

A satellite in a circular orbit around the Earth is under the influence of gravitational force providing the necessary centripetal force. The gravitational force acting on the satellite is given by Newton's law of universal gravitation.

Step 2: Apply Gravitational Force Formula

The gravitational force between the Earth and the satellite is given by\[ F_{ ext{gravity}} = \frac{G M_{ ext{E}} m_{ ext{s}}}{r_{ ext{s}}^2} \]where \( G \) is the gravitational constant and \( M_{ ext{E}} \) is the mass of the Earth. This force provides the necessary centripetal force to maintain the satellite's circular motion.

Step 3: Equate Gravitational and Centripetal Forces

For a stable circular orbit, the gravitational force is equal to the centripetal force, which is given by\[ F_{ ext{centripetal}} = \frac{m_{ ext{s}} v^2}{r_{ ext{s}}} \]Setting these two forces equal gives us:\[ \frac{G M_{ ext{E}} m_{ ext{s}}}{r_{ ext{s}}^2} = \frac{m_{ ext{s}} v^2}{r_{ ext{s}}} \]which simplifies to:\[ v^2 = \frac{G M_{ ext{E}}}{r_{ ext{s}}} \]

Step 4: Calculate Kinetic Energy

The kinetic energy \( K \) of the satellite is given by\[ K = \frac{1}{2} m_{ ext{s}} v^2 \]Substituting the expression for \( v^2 \), we have\[ K = \frac{1}{2} m_{ ext{s}} \left( \frac{G M_{ ext{E}}}{r_{ ext{s}}} \right) = \frac{G M_{ ext{E}} m_{ ext{s}}}{2 r_{ ext{s}}} \]

Step 5: Calculate Potential Energy

The potential energy \( U \) of the satellite at a distance \( r_{ ext{s}} \) from the center of the Earth is given by\[ U = -\frac{G M_{ ext{E}} m_{ ext{s}}}{r_{ ext{s}}} \]

Step 6: Determine the Ratio of Kinetic to Potential Energy

The ratio \( \frac{K}{U} \) is given by:\[ \frac{K}{U} = \frac{\frac{G M_{ ext{E}} m_{ ext{s}}}{2 r_{ ext{s}}}}{-\frac{G M_{ ext{E}} m_{ ext{s}}}{r_{ ext{s}}}} = -\frac{1}{2} \]

Key Concepts

Kinetic Energy

Kinetic energy is a measure of an object's motion energy. For any object in motion, its kinetic energy can be calculated using the formula: \( K = \frac{1}{2} m v^2 \), where:

• \( m \) is the mass of the object.
• \( v \) is the speed or velocity of the object.

In the context of a satellite orbiting Earth, kinetic energy represents how fast the satellite is moving along its circular path. The speed is determined by the gravitational force that keeps the satellite in orbit. By setting the gravitational force equal to the centripetal force needed to maintain the satellite's circular motion, we derive the satellite's velocity: \( v^2 = \frac{G M_{\mathrm{E}}}{r_{\mathrm{s}}} \). Substituting this into the kinetic energy equation, we find: \( K = \frac{1}{2} m_{\mathrm{s}} \left( \frac{G M_{\mathrm{E}}}{r_{\mathrm{s}}} \right) = \frac{G M_{\mathrm{E}} m_{\mathrm{s}}}{2 r_{\mathrm{s}}} \). This formula shows that the kinetic energy is proportional to the mass of the satellite and inversely proportional to the orbital radius.

Gravitational Force

Gravitational force is the attractive force between two masses. According to Newton's law of universal gravitation, the gravitational force \( F_{\text{gravity}} \) between two masses is described by: \[ F_{\text{gravity}} = \frac{G M_1 M_2}{r^2} \] The variables in the formula are:

• \( G \) is the gravitational constant.
• \( M_1 \) and \( M_2 \) are the masses of the two objects.
• \( r \) is the distance between the centers of the two masses.

When considering a satellite orbiting around Earth, the mass \( M_{\mathrm{E}} \) is the Earth's mass, and \( m_{\mathrm{s}} \) is the satellite's mass. This gravitational force acts as the centripetal force necessary to keep the satellite moving in a circular path around the Earth. The balance between these forces ensures that the satellite remains in stable orbit.

Centripetal Force

Centripetal force is required to keep an object moving in a circular path. This inward force is directed towards the center of the circular path and is crucial for circular motion. For an object of mass \( m \), moving at a speed \( v \), and at a radius \( r \), the centripetal force \( F_{\text{centripetal}} \) is calculated by: \[ F_{\text{centripetal}} = \frac{m v^2}{r} \] Key points to consider about centripetal force:

• It does not do work on the moving object since the force is perpendicular to the direction of motion.
• Without this force, objects would move in a straight line instead of a circular path.

In the case of a satellite, the gravitational force from the Earth acts as the required centripetal force. This ensures that the satellite stays in orbit. By equating gravitational force with centripetal force, we can solve for the satellite’s velocity to maintain orbital stability. This fundamental relationship underpins many aspects of orbital mechanics.