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(II) Planet \(A\) and planet \(B\) are in circular orbits around a distant star. Planet \(A\) is 9.0 times farther from the star than is planet B. What is the ratio of their speeds \(v_{A} / v_{B} ?\)

Short Answer

Expert verified
The ratio of their speeds is \( \frac{1}{3} \).

Step by step solution

01

Understand the Problem

We have two planets orbiting a star. Planet A is 9 times farther away from the star compared to planet B. We need to find the ratio of the speeds of these two planets, i.e., \( \frac{v_A}{v_B} \).
02

Apply Kepler's Third Law

Kepler's Third Law states that the square of the orbital period \( T \) is proportional to the cube of the semi-major axis \( r \): \( T^2 \propto r^3 \). Since the planets are in circular orbits, we can assume \( T_A^2 / T_B^2 = (r_A / r_B)^3 = 9^3 \).
03

Relate Orbital Period to Orbital Speed

The orbital speed \( v \) is related to the orbital period \( T \) by the equation \( v = \frac{2\pi r}{T} \). Thus, \( T \propto \frac{r}{v} \), which implies \( v \propto \frac{1}{\sqrt{r}} \).
04

Calculate the Speed Ratio

Since \( v \propto \frac{1}{\sqrt{r}} \), we have \( \frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A}} \). Given \( r_A = 9r_B \), \( \frac{v_A}{v_B} = \sqrt{\frac{r_B}{9r_B}} = \frac{1}{3} \). Thus, the speed of Planet A is \( \frac{1}{3} \) the speed of Planet B.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Orbital Speed
The concept of orbital speed is a fundamental part of understanding how celestial bodies, like planets and moons, move through space. Orbital speed refers to the constant speed at which an object travels along its circular orbit around another larger body, such as a star or planet. For a body to maintain a stable orbit, its speed must balance the gravitational pull of the object it is orbiting.

To calculate the orbital speed (\(v\), the formula used is:\[v = \frac{2\pi r}{T}\]where \(r\) is the radius of the orbit (distance from the center of the object being orbited to the orbiting body), and \(T\) is the orbital period (time taken to complete one full orbit).

From Kepler's third law, we also understand that orbital speed is inversely proportional to the square root of the orbital radius when comparing different bodies orbiting the same central object. This relationship highlights that the further away a planet is from the star it orbits, the slower its orbital speed. This inverse relationship is crucial in solving problems that involve multiple planetary bodies as their respective distances from the central star determine their speeds.
Circular Orbits
When we talk about circular orbits, we refer to a special type of orbit where an object travels around another body in a perfect circle. In reality, most orbits are elliptical, but for simplicity, and many educational problems, we consider circular orbits. This assumption simplifies the calculations and understanding of basic orbital mechanics.

In a circular orbit:
  • The distance between the orbiting object and the center of the object it orbits remains constant.
  • The gravitational force acts as the centripetal force keeping the object in orbit.
  • The speed of the orbiting object remains constant.
For example, planets in nearly circular orbits revolve around their stars due to the gravitational forces and their initial tangential speeds. This balance explains why they don't spiral into their star or drift out into space.

Analyzing circular orbits helps to build foundational knowledge in astronomy and physics by offering a simplified model of more complex, real-world phenomena. It's a stepping stone towards understanding the dynamics of elliptical orbits and the entire celestial mechanics.
Ratio of Speeds
In our example, the ratio of speeds between two planets, \(A\) and \(B\), is determined by their distances from a central star. Given that Planet \(A\) is nine times further from the star than Planet \(B\), we can calculate the speed ratio using the relationship derived from Kepler's Third Law and the proportional relationship of orbital speed to radius.Using the proportional relationship \(v \propto \frac{1}{\sqrt{r}}\), we calculate the ratio of their speeds as follows:
  • Let the distance of Planet \(B\) from the star be denoted by \(r_B\).
  • Then, the distance of Planet \(A\) is \(r_A = 9 \times r_B\).
  • The speed ratio \(\frac{v_A}{v_B}\) is:
\[\frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A}} = \sqrt{\frac{r_B}{9r_B}} = \frac{1}{3}\]This calculation shows that Planet A travels at one-third the speed of Planet B. Such calculations and understanding of speed ratios are essential in astrophysics and help predict and compare movements of different celestial bodies moving around a common center, such as multiple planets orbiting a single star.

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Most popular questions from this chapter

A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the size of our Moon but has the mass of our Sun. ( \(a\) ) Estimate gravity on the surface on this star. \((b)\) How much would a \(65-\mathrm{kg}\) person weigh on this star? (c) What would be the speed of a baseball dropped from a height of \(1.0 \mathrm{~m}\) when it hit the surface?

(II) A mass \(M\) is ring shaped with radius \(r .\) A small mass \(m\) is placed at a distance \(x\) along the ring's axis as shown in Fig. \(27 .\) Show that the gravitational force on the mass \(m\) due to the ring is directed inward along the axis and has magnitude $$F=\frac{G M m x}{\left(x^{2}+r^{2}\right)^{\frac{3}{2}}}$$ [Hint: Think of the ring as made up of many small point masses \(d M ;\) sum over the forces due to each \(d M,\) and use symmetry.]

(III) \((a)\) Use Kepler's second law to show that the ratio of the speeds of a planet at its nearest and farthest points from the Sun is equal to the inverse ratio of the near and far distances: \(v_{\mathrm{N}} / v_{\mathrm{F}}=d_{\mathrm{F}} / d_{\mathrm{N}}\). ( \(b\) ) Given that the Earth's distance from the Sun varies from 1.47 to \(1.52 \times 10^{11} \mathrm{~m},\) determine the minimum and maximum velocities of the Earth in its orbit around the Sun.

(III) Two identical particles, each of mass \(m,\) are located on the \(x\) axis at \(x=+x_{0}\) and \(x=-x_{0}\). (a) Determine a formula for the gravitational field due to these two particles for points on the \(y\) axis; that is, write \(\vec{g}\) as a function of \(y, m,\) \(x_{0},\) and so on. \((b)\) At what point (or points) on the \(y\) axis is the magnitude of \(\vec{g}\) a maximum value, and what is its value there? [Hint: Take the derivative \(d \overrightarrow{\mathrm{g}} / d y .\)

(I) Neptune is an average distance of \(4.5 \times 10^{9} \mathrm{~km}\) from the Sun. Estimate the length of the Neptunian year using the fact that the Earth is \(1.50 \times 10^{8} \mathrm{~km}\) from the Sun on the average.

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