Problem 52 - Physics for Scientists and Engineers with Modern Physics

Question

(1I) \((a)\) Show that the total mechanical energy of a satellite mass \(m\) orbiting at a distance \(r\) from the center of the Earth (mass \(M_{E} )\) is $$E=-\frac{1}{2} \frac{G m M_{\mathrm{E}}}{r}$$ if \(U=0\) at \(r=\infty .(b)\) Show that although friction causes the value of \(E\) to decrease slowly, kinetic energy must actu- ally increase if the orbit remains a circle.

Step-by-Step Solution

Answer

Mechanical energy decreases with friction, but kinetic energy increases as the radius decreases.

Step 1: Understand the Gravitational Potential Energy

The gravitational potential energy (U) of a satellite of mass \(m\) at a distance \(r\) from the center of the Earth (mass \(M_E\)) is given by \(U = -\frac{G m M_E}{r}\). Here, \(G\) is the gravitational constant. If \(U = 0\) at \(r = \infty\), this is the potential energy at a finite distance \(r\).

Step 2: Understand the Kinetic Energy of the Satellite

The kinetic energy (K) of a satellite in a circular orbit is given by \(K = \frac{1}{2} m v^2\). For a circular orbit, the centripetal force required is provided by the gravitational force: \(\frac{m v^2}{r} = \frac{G m M_E}{r^2}\). Rearranging gives \(v^2 = \frac{G M_E}{r}\), so the kinetic energy becomes \(K = \frac{1}{2} m \left(\frac{G M_E}{r}\right) = \frac{G m M_E}{2r}\).

Step 3: Derive the Total Mechanical Energy

The total mechanical energy (E) of the satellite is the sum of the kinetic and potential energy: \(E = K + U = \frac{G m M_E}{2r} - \frac{G m M_E}{r}\). Simplifying, we get \(E = -\frac{1}{2} \frac{G m M_E}{r}\). This formula shows that the total mechanical energy in a circular orbit is negative.

Step 4: Analyze Effect of Friction on Orbital Energy

If friction causes a decrease in the total mechanical energy \(E\), the orbit radius \(r\) decreases because \(E = -\frac{1}{2} \frac{G m M_E}{r}\) implies that \(E\) becomes more negative. Reducing \(r\) leads to an increase in speed \(v\) to maintain balance: \(v^2 = \frac{G M_E}{r}\). Thus, kinetic energy \(K\) increases because \(K = \frac{1}{2} m v^2\) and \(v\) increases.

Key Concepts

Gravitational Potential Energy

When we talk about gravitational potential energy of a satellite, we are considering the work done against Earth's gravitational pull. The formula used to determine this energy is \( U = -\frac{G m M_E}{r} \), with \(G\) being the gravitational constant. This equation highlights how gravitational potential energy is dependent on the distance \(r\) between the satellite and the center of Earth. Gravitational potential energy is defined as zero at an infinite distance from Earth, which helps us compare energies at various points.

• The negative sign indicates that energy is needed to bring the satellite from infinity to a point \(r\).
• As the satellite gets closer to Earth, the gravitational potential energy becomes more negative, showing that more energy is released.

Gravitational potential energy plays a key role in understanding the total energy budget of orbital mechanics. Understanding how potential energy transforms into kinetic energy and vice-versa is crucial when dealing with objects in space.

Kinetic Energy

Kinetic energy in satellite motion is all about the energy of movement. Satellites move along their orbits due to the balance between gravitational pull and their velocity. The expression for kinetic energy is \( K = \frac{1}{2} m v^2 \), where \(m\) is mass and \(v\) is velocity.For objects in orbit, the gravitational force provides the necessary centripetal force that keeps satellites moving in a circular path. Therefore:- The velocity \(v\) of a satellite can be found from \(v^2 = \frac{G M_E}{r}\).- Substituting this into the kinetic energy formula gives us: \( K = \frac{G m M_E}{2r} \).As the satellite orbits, any reduction in radius due to friction leads to an increase in velocity (to maintain circular motion), thereby increasing kinetic energy. This paradoxical increase in kinetic energy with decrease in total mechanical energy might feel counter-intuitive, but it's a well-studied principle in orbital dynamics.

Centripetal Force

The concept of centripetal force is crucial in understanding satellite mechanics. In a stable orbit, a satellite needs a force pulling it towards the center of its path, which in this case, is Earth. This force is called centripetal force, and for satellites, it is provided by Earth's gravity.The relationship is given by:- Centripetal force \( F_c = \frac{m v^2}{r} \).- Gravitational force \( F_g = \frac{G m M_E}{r^2} \) also acts as the centripetal force.By setting \( F_c = F_g \), we can solve for the orbital velocity \(v\). The delicate balance between gravitational pull and the satellite's motion (or velocity) ensures that the satellite remains in orbit and does not fall back to Earth or drift into space.Keeping the satellite in orbit requires constant adjustments to account for factors like friction and other perturbations, emphasizing how centripetal force and gravitational pull work hand in hand to maintain motion.