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Chapter 12: Problem 72

The distances from the Sun at perihelion and aphelion for Pluto are \(4410 \cdot 10^{6} \mathrm{~km}\) and \(7360 \cdot 10^{6} \mathrm{~km},\) respectively. What is the ratio of Pluto's orbital speed around the Sun at perihelion to that at aphelion?

Short Answer

Expert verified
Answer: The ratio of Pluto's orbital speed around the Sun at perihelion to that at aphelion is approximately 1.667.

Step by step solution

01

Determine the distances at perihelion and aphelion

Given that the distances from the Sun at perihelion and aphelion for Pluto are \(4410 \cdot 10^{6} \mathrm{~km}\) and \(7360 \cdot 10^{6} \mathrm{~km},\) respectively. Convert these distances to meters by multiplying by 1000. \(r_{peri} = 4410 \cdot 10^{6} \mathrm{~km} \times 1000 = 4410 \cdot 10^{9} \mathrm{~m}\) \(r_{aph} = 7360 \cdot 10^{6} \mathrm{~km} \times 1000 = 7360 \cdot 10^{9} \mathrm{~m}\)
02

Apply Kepler's second law of planetary motion

Kepler's second law states that a line connecting a planet and the Sun sweeps out equal areas during equal intervals of time. Mathematically, we can express this as: \(r_{peri} \cdot v_{peri} = r_{aph} \cdot v_{aph}\) Where \(r_{peri}\) and \(r_{aph}\) are the distances from the Sun at perihelion and aphelion, and \(v_{peri}\) and \(v_{aph}\) are the orbital speeds at perihelion and aphelion respectively.
03

Solve for the ratio of the orbital speeds

We need to find the ratio of \(v_{peri}\) to \(v_{aph}\): \(\frac{v_{peri}}{v_{aph}} = \frac{r_{aph} \cdot v_{aph}}{r_{peri} \cdot v_{aph}}\) Since we want the ratio of the orbital speeds, the \(v_{aph}\) in the numerator and denominator cancels out: \(\frac{v_{peri}}{v_{aph}} = \frac{r_{aph}}{r_{peri}}\) Now, substitute the values of \(r_{peri}\) and \(r_{aph}\): \(\frac{v_{peri}}{v_{aph}} = \frac{7360 \cdot 10^{9} \mathrm{~m}}{4410 \cdot 10^{9} \mathrm{~m}}\) Finally, calculate the ratio: \(\frac{v_{peri}}{v_{aph}} = 1.667\) Therefore, the ratio of Pluto's orbital speed around the Sun at perihelion to that at aphelion is approximately \(1.667\).

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Most popular questions from this chapter

A planet with a mass of \(7.00 \cdot 10^{21} \mathrm{~kg}\) is in a circular orbit around a star with a mass of \(2.00 \cdot 10^{30} \mathrm{~kg} .\) The planet has an orbital radius of \(3.00 \cdot 10^{10} \mathrm{~m}\). a) What is the linear orbital velocity of the planet? b) What is the period of the planet's orbit? c) What is the total mechanical energy of the planet?

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