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Chapter 12: Problem 63

A 1000.-kg communications satellite is released from a space shuttle to initially orbit the Earth at a radius of \(7.00 \cdot 10^{6} \mathrm{~m}\). After being deployed, the satellite's rockets are fired to put it into a higher altitude orbit of radius \(5.00 \cdot 10^{7} \mathrm{~m} .\) What is the minimum mechanical energy supplied by the rockets to effect this change in orbit?

Short Answer

Expert verified
Answer: To find the minimum mechanical energy supplied, first calculate the change in gravitational potential energy (\(\Delta U = U_2 - U_1\)) and assume the change in kinetic energy (\(\Delta K\)) is zero. Then, sum up the changes in gravitational potential energy and kinetic energy to calculate the total mechanical energy change required (\(\Delta E = \Delta U + \Delta K\)). Plug in the values from the given problem to find the minimum mechanical energy supplied by the rockets.

Step by step solution

01

Calculate Initial Gravitational Potential Energy

The gravitational potential energy (GPE) is given by the formula: $$ U = -\frac{GM_Em}{r} $$ where \(G\) is the gravitational constant (\(6.674 \times 10^{-11} N m^2 kg^{-2}\)), \(M_E\) is the mass of the Earth (\(5.972 \times 10^{24} kg\)), \(m\) is the mass of the satellite (\(1000 kg\)), and \(r\) is the distance from the center of the Earth to the satellite. Plug in the values for the initial orbit radius (\(7.00 \cdot 10^{6} m\)) and other constants to calculate the initial GPE: $$ U_1 = -\frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(1000)}{7.00 \cdot 10^{6}} $$
02

Calculate Final Gravitational Potential Energy

Using the same formula as in step 1, calculate the GPE for the final orbit with a radius of \(5.00 \cdot 10^{7} m\): $$ U_2 = -\frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(1000)}{5.00 \cdot 10^{7}} $$
03

Calculate Change in Gravitational Potential Energy

Subtract the initial GPE from the final GPE to find the change in gravitational potential energy during the orbit transition: $$ \Delta U = U_2 - U_1 $$
04

Calculate Change in Kinetic Energy

Since the minimum mechanical energy is required, we assume that the change in kinetic energy is zero. This is because the satellite's speed must be kept as low as possible to minimize the required energy input. So, the change in kinetic energy is: $$ \Delta K = 0 $$
05

Calculate the Total Mechanical Energy Change

Finally, sum up the changes in gravitational potential energy and kinetic energy to calculate the total mechanical energy change required: $$ \Delta E = \Delta U + \Delta K $$ Plug the values from steps 3 and 4 into this equation to find the minimum mechanical energy supplied by the rockets to effect the orbit change.

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Most popular questions from this chapter

A satellite of mass \(m\) is in an elliptical orbit (that satisfies Kepler's laws) about a body of mass \(M,\) with \(m\) negligible compared to \(M\) a) Find the total energy of the satellite as a function of its speed, \(v\), and distance, \(r\), from the body it is orbiting. b) At the maximum and minimum distance between the satellite and the body, and only there, the angular momentum is simply related to the speed and distance. Use this relationship and the result of part (a) to obtain a relationship between the extreme distances and the satellite's energy and angular momentum. c) Solve the result of part (b) for the maximum and minimum radii of the orbit in terms of the energy and angular momentum per unit mass of the satellite. d) Transform the results of part (c) into expressions for the semimajor axis, \(a\), and eccentricity of the orbit, \(e\), in terms of the energy and angular momentum per unit mass of the satellite.

Where the International Space Station orbits, the gravitational acceleration is just \(11.4 \%\) less than its value on the surface of the Earth. Nevertheless, astronauts in the space station float. Why is this so?

Newton's Law of Gravity specifies the magnitude of the interaction force between two point masses, \(m_{1}\) and \(m_{2}\), separated by a distance \(r\) as \(F(r)=G m_{1} m_{2} / r^{2} .\) The gravitational constant \(G\) can be determined by directly measuring the interaction force (gravitational attraction) between two sets of spheres by using the apparatus constructed in the late 18th century by the English scientist Henry Cavendish. This apparatus was a torsion balance consisting of a 6.00 -ft wooden rod suspended from a torsion wire, with a lead sphere having a diameter of 2.00 in and a weight of 1.61 lb attached to each end. Two 12.0 -in, 348 -lb lead balls were located near the smaller balls, about 9.00 in away, and held in place with a separate suspension system. Today's accepted value for \(G\) is \(6.674 \cdot 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\) Determine the force of attraction between the larger and smaller balls that had to be measured by this balance. Compare this force to the weight of the small balls.

With the usual assumption that the gravitational potential energy goes to zero at infinite distance, the gravitational potential energy due to the Earth at the center of Earth is a) positive. b) negative. c) zero. d) undetermined.

Is the orbital speed of the Earth when it is closest to the Sun greater than, less than, or equal to the orbital speed when it is farthest from the Sun? Explain.
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