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Chapter 8: Problem 83

The Moon's distance from Earth varies between \(3.56 \times 10^{5} \mathrm{km}\) at perigee and \(4.07 \times 10^{5} \mathrm{km}\) at apogee. What is the ratio of its orbital speed around Earth at perigee to that at apogee?

Short Answer

Expert verified
Answer: The Moon's orbital speed at perigee is about 1.143 times that at apogee.

Step by step solution

01

Express the conservation of angular momentum equation

Kepler's Second Law states that the angular momentum of a celestial body is conserved. For the Moon orbiting the Earth, the angular momentum can be expressed as: $$ L = m v r $$ Where L is the angular momentum, m is the moon's mass, v is the orbital speed, and r is the distance from the Earth. Since the angular momentum is conserved, we can write the equation for perigee and apogee as: $$ L_{perigee} = L_{apogee} $$ Since the mass of the Moon is constant, we can write: $$ v_{perigee} r_{perigee} = v_{apogee} r_{apogee} $$
02

Solve the equation for the ratio of orbital speed at perigee to that at apogee

Now we need to solve the equation for the ratio of orbital speed at perigee (shorter distance) to that at apogee (longer distance). We need to find: $$ \frac{v_{perigee}}{v_{apogee}} $$ Dividing both sides of the equation by \(v_{apogee} r_{apogee}\), we find: $$ \frac{v_{perigee}}{v_{apogee}} = \frac{r_{apogee}}{r_{perigee}} $$ Now substituting the given distances: $$ \frac{v_{perigee}}{v_{apogee}} = \frac{4.07 \times 10^{5} \mathrm{km}}{3.56 \times 10^{5} \mathrm{km}} $$ Finally, calculate the ratio: $$ \frac{v_{perigee}}{v_{apogee}} \approx 1.143 $$ So, the Moon's orbital speed at perigee is about 1.143 times that at apogee.

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Most popular questions from this chapter

A large clock has a second hand with a mass of \(0.10 \mathrm{kg}\) concentrated at the tip of the pointer. (a) If the length of the second hand is $30.0 \mathrm{cm},$ what is its angular momentum? (b) The same clock has an hour hand with a mass of \(0.20 \mathrm{kg}\) concentrated at the tip of the pointer. If the hour hand has a length of \(20.0 \mathrm{cm},\) what is its angular momentum?
A 68 -kg woman stands straight with both feet flat on the floor. Her center of gravity is a horizontal distance of \(3.0 \mathrm{cm}\) in front of a line that connects her two ankle joints. The Achilles tendon attaches the calf muscle to the foot a distance of \(4.4 \mathrm{cm}\) behind the ankle joint. If the Achilles tendon is inclined at an angle of \(81^{\circ}\) with respect to the horizontal, find the force that each calf muscle needs to exert while she is standing. [Hint: Consider the equilibrium of the part of the body above the ankle joint.]
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A solid sphere is released from rest and allowed to roll down a board that has one end resting on the floor and is tilted at \(30^{\circ}\) with respect to the horizontal. If the sphere is released from a height of \(60 \mathrm{cm}\) above the floor, what is the sphere's speed when it reaches the lowest end of the board?
A mountain climber is rappelling down a vertical wall. The rope attaches to a buckle strapped to the climber's waist \(15 \mathrm{cm}\) to the right of his center of gravity. If the climber weighs \(770 \mathrm{N},\) find (a) the tension in the rope and (b) the magnitude and direction of the contact force exerted by the wall on the climber's feet.
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