(III) Show that to slow down a spacecraft (perhaps to put it into orbit around a planet or moon) must be in the same direction as and have magnitude greater than u. Show that this means the spacecraft will then pass in front of Jupiter. Draw a new diagram to replace Fig. 8–25b for this situation.
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Energy of Circular Orbits
Problem 51
Textbook Question
The 75,000 kg space shuttle used to fly in a 250-km-high circular orbit. It needed to reach a 610-km-high circular orbit to service the Hubble Space Telescope. How much energy was required to boost it to the new orbit?

1
Step 1: Understand the problem. The space shuttle is moving from a lower circular orbit (250 km above Earth's surface) to a higher circular orbit (610 km above Earth's surface). The energy required is the difference in total mechanical energy (kinetic + potential) between the two orbits.
Step 2: Write the formula for the total mechanical energy of an orbiting object: \( E = -\frac{GMm}{2r} \), where \( G \) is the gravitational constant, \( M \) is the mass of Earth, \( m \) is the mass of the shuttle, and \( r \) is the orbital radius (distance from the center of Earth).
Step 3: Calculate the orbital radius for each orbit. Add Earth's radius (\( R_E = 6,371 \; \text{km} \)) to the altitude of each orbit. For the lower orbit, \( r_1 = R_E + 250 \; \text{km} \). For the higher orbit, \( r_2 = R_E + 610 \; \text{km} \).
Step 4: Compute the total mechanical energy for each orbit using the formula \( E = -\frac{GMm}{2r} \). Substitute \( r_1 \) for the lower orbit and \( r_2 \) for the higher orbit. Note that \( G = 6.674 \times 10^{-11} \; \text{N·m}^2/\text{kg}^2 \) and \( M = 5.972 \times 10^{24} \; \text{kg} \).
Step 5: Find the energy required to boost the shuttle by calculating the difference in total mechanical energy: \( \Delta E = E_2 - E_1 \), where \( E_2 \) is the energy in the higher orbit and \( E_1 \) is the energy in the lower orbit. This \( \Delta E \) represents the energy needed to move the shuttle to the new orbit.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Gravitational Potential Energy
Gravitational potential energy (U) is the energy an object possesses due to its position in a gravitational field. It is calculated using the formula U = -G(m1*m2)/r, where G is the gravitational constant, m1 and m2 are the masses involved, and r is the distance from the center of the mass creating the gravitational field. In the context of orbits, changes in altitude directly affect the gravitational potential energy of the spacecraft.
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Kinetic Energy in Orbital Motion
Kinetic energy (KE) in orbital motion is the energy an object has due to its velocity. For an object in a circular orbit, KE can be expressed as KE = 0.5mv^2, where m is the mass and v is the orbital speed. As the shuttle moves to a higher orbit, its speed changes, which affects its kinetic energy and must be accounted for when calculating the total energy required for the orbital transition.
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Energy of Circular Orbits
Energy Conservation in Orbital Mechanics
Energy conservation in orbital mechanics states that the total mechanical energy (the sum of kinetic and potential energy) of an object in orbit remains constant if only conservative forces are acting. When changing orbits, the shuttle must perform work to increase its potential energy while also adjusting its kinetic energy. The energy required to boost the shuttle to a higher orbit can be determined by calculating the difference in total mechanical energy between the two orbits.
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Energy Conservation in Changing Elliptical Orbits
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