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(II) \((a)\) Show that the total mechanical energy of a satellite (mass \(m\) ) orbiting at a distance \(r\) from the center of the Earth (mass \(M_{\mathrm{E}}\) ) 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 actually increase if the orbit remains a circle.

Short Answer

Expert verified
Total energy: \(E = -\frac{1}{2} \frac{GmM_E}{r}\); kinetic energy increases as \(E\) decreases with orbit shrinking.

Step by step solution

01

Understanding the problem

We need to show that the total mechanical energy \(E\) of a satellite in a circular orbit around the Earth is given by \(E = -\frac{1}{2} \frac{GmM_E}{r}\). We also need to explain why, despite friction decreasing \(E\), the kinetic energy increases if the orbit remains circular.
02

Calculate gravitational potential energy

Gravitational potential energy \(U\) of the satellite-Earth system is given by \(U = -\frac{GmM_E}{r}\), where \(G\) is the gravitational constant, \(m\) is the satellite mass, \(M_E\) is the Earth's mass, and \(r\) is the distance from the Earth's center to the satellite. The potential energy is assumed zero at infinity \(U = 0\).
03

Calculate kinetic energy

Use the formula for centripetal force for a satellite in circular motion: \(\frac{GmM_E}{r^2} = \frac{mv^2}{r}\). Solving for \(v\), we get \(v = \sqrt{\frac{GM_E}{r}}\). The kinetic energy \(K\) is then \(K = \frac{1}{2}mv^2 = \frac{1}{2}m\frac{GM_E}{r} = \frac{GmM_E}{2r}\).
04

Calculate total mechanical energy

The total mechanical energy \(E\) is the sum of kinetic and potential energy: \(E = K + U = \frac{GmM_E}{2r} - \frac{GmM_E}{r} = -\frac{GmM_E}{2r}\). This matches the given expression showing \(E\) is negative, indicating a bound system.
05

Analyze the impact of friction

Friction decreases the total mechanical energy \(E\) as the orbit starts to shrink due to energy loss. For the orbit to remain circular, as \(r\) decreases, \(v\) must increase to provide necessary centripetal force, and therefore kinetic energy \(K\) increases even as \(E\) decreases.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Potential Energy
Gravitational potential energy is an essential concept in orbital mechanics. It refers to the energy stored in a system due to the position of two masses in a gravitational field. For a satellite in orbit around the Earth, the gravitational potential energy, denoted as \(U\), is given by the formula \(U = -\frac{GmM_E}{r}\). Here, \(G\) is the gravitational constant, \(m\) is the mass of the satellite, \(M_E\) is the mass of the Earth, and \(r\) is the distance from the Earth's center to the satellite.
The negative sign reflects that work is done by gravitational force. When the satellite is far away, at infinity, potential energy is taken as zero. This point is considered because gravitational pull is virtually nonexistent at that distance. As the satellite moves closer, potential energy becomes more negative, indicating a stronger gravitational pull.
Understanding gravitational potential energy helps explain how energy shifts within an orbiting satellite system and plays a role in determining total mechanical energy, which is crucial for stable motion.
Kinetic Energy
Kinetic energy is the energy of motion, exhibited by any object in movement. For a satellite, kinetic energy \(K\) is the energy linked with its velocity as it orbits Earth. The formula for kinetic energy is given by \(K = \frac{1}{2}mv^2\), where \(v\) is the orbital velocity of the satellite.
To maintain a stable orbit, the satellite must move at a precise velocity that provides the necessary centripetal force. This can be calculated from the equation of centripetal force: \(\frac{GmM_E}{r^2} = \frac{mv^2}{r}\). Solving for \(v\) gives \(v = \sqrt{\frac{GM_E}{r}}\).
The equation above shows that the kinetic energy then becomes \(K = \frac{GmM_E}{2r}\). As friction affects movement, although the overall mechanical energy \(E\) decreases, kinetic energy may still increase if the satellite's orbit stays circular. This necessary increase is to maintain acceleration as the orbit decreases in radius.
Centripetal Force
Centripetal force is a crucial factor in circular motion, serving as the inward force that keeps an object moving in a curve. In the context of a satellite, this force keeps it in orbit around the Earth. The gravity between Earth and the satellite acts as the centripetal force necessary for the satellite's circular motion.
The formula used to represent this force in action is \(\frac{GmM_E}{r^2} = \frac{mv^2}{r}\). This equation balances gravitational force with the necessary centripetal force to keep the satellite in motion. Anything affecting this balance, like changes in velocity, will alter the orbit's circumstances.
With friction present, as total mechanical energy decreases, it's vital for centripetal force to adjust. The satellite's velocity must increase for the orbit to remain circular despite energy losses. This adjustment ensures the necessary force is sustained to counterbalance gravitational pull.
Satellite Motion
Satellite motion involves the delicate balance of forces to maintain an orbit. The interaction of gravitational potential energy and kinetic energy within total mechanical energy ensures stability in the satellite's path around Earth.
Initially, the satellite is launched at a high velocity to counter Earth's gravity and establish an orbit. The central concepts involved, such as gravitational potential and kinetic energies, centripetal force, and friction, all depict the dynamics within the satellite’s motion.
When friction plays a role, it causes the satellite to lose energy over time. Despite this, the kinetic energy may rise to maintain the orbit's circular shape, requiring an increase in speed as the orbit radius decreases.
  • The satellite’s motion remains coherent as long as these dynamic factors are managed properly.
  • Understanding these intricacies is critical for maintaining satellites within specified orbital paths for functionality.
Grasping how satellite motion operates is instrumental in predicting changes in trajectory and ensuring seamless space operations.

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Most popular questions from this chapter

(1I) Draw a potential energy diagram, \(U\) vs. \(x,\) and analyze the motion of a mass \(m\) resting on a frictionless horizontal table and connected to a horizontal spring with stiffness constant \(k\) . The mass is pulled a distance to the right so that the spring is stretched a distance \(x_{0}\) initially, and then the mass is released from rest.

(II) A 0.620 -kg wood block is firmly attached to a very light horizontal spring \((k=180 \mathrm{~N} / \mathrm{m})\) as shown in Fig. \(8-35 .\) This block-spring system, when compressed \(5.0 \mathrm{~cm}\) and released, stretches out \(2.3 \mathrm{~cm}\) beyond the equilibrium position before stopping and turning back. What is the coefficient of kinetic friction between the block and the table?

(II) A pendulum \(2.00 \mathrm{~m}\) long is released (from rest) at an angle \(\theta_{0}=30.0^{\circ}\) (Fig. \(8-14\) ). Determine the speed of the \(70.0-\mathrm{g}\) bob: \((a)\) at the lowest point \((\theta=0) ;(b)\) at \(\theta=15.0^{\circ}\) (c) at \(\theta=-15.0^{\circ}\) (i.e., on the opposite side). ( \(d\) ) Determine the tension in the cord at each of these three points. \((e)\) If the bob is given an initial speed \(v_{0}=1.20 \mathrm{~m} / \mathrm{s}\) when released at \(\theta=30.0^{\circ},\) recalculate the speeds for parts \((a),(b),\) and \((c)\)

(III) The position of a 280 -g object is given (in meters) by \(x=5.0 t^{3}-8.0 t^{2}-44 t,\) where \(t\) is in seconds. Determine the net rate of work done on this object \((a)\) at \(t=2.0 \mathrm{~s}\) and (b) at \(t=4.0 \mathrm{~s}\). (c) What is the average net power input during the interval from \(t=0 \mathrm{~s}\) to \(t=2.0 \mathrm{~s},\) and in the interval from \(t=2.0 \mathrm{~s}\) to \(4.0 \mathrm{~s} ?\)

Electric energy units are often expressed in the form of "kilowatt-hours." (a) Show that one kilowatt-hour (kWh) is equal to \(3.6 \times 10^{6} \mathrm{~J} .\) (b) If a typical family of four uses electric energy at an average rate of \(580 \mathrm{~W}\), how many \(\mathrm{kWh}\) would their electric bill show for one month, and (c) how many joules would this be? ( \(d\) ) At a cost of \(\$ 0.12\) per \(\mathrm{kWh}\), what would their monthly bill be in dollars? Does the monthly bill depend on the rate at which they use the electric energy?

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