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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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