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Chapter 6: Problem 39

(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.

Short Answer

Expert verified
The Neptunian year is approximately 164 Earth years.

Step by step solution

01

Understanding Kepler's Third Law

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, it can be given as \( T^2 \propto r^3 \), where \( T \) is the orbital period and \( r \) is the distance from the sun.
02

Setting Up the Ratio for Earth and Neptune

Let's set the relationship for both Earth and Neptune using Kepler's Law as \( \frac{T_{\text{Neptune}}^2}{T_{\text{Earth}}^2} = \frac{r_{\text{Neptune}}^3}{r_{\text{Earth}}^3} \). We know that \( T_{\text{Earth}} = 1 \) year and \( r_{\text{Earth}} = 1.50 \times 10^{8} \, \text{km} \), \( r_{\text{Neptune}} = 4.5 \times 10^{9} \, \text{km} \).
03

Calculating the Distance Ratio

Calculate the cube of the distance ratio: \( \frac{r_{\text{Neptune}}^3}{r_{\text{Earth}}^3} = \left( \frac{4.5 \times 10^{9}}{1.50 \times 10^{8}} \right)^3 = \left( 30 \right)^3 = 27000 \).
04

Solving for the Neptunian Year

From \( \frac{T_{\text{Neptune}}^2}{1^2} = 27000 \), we get \( T_{\text{Neptune}}^2 = 27000 \). Taking the square root gives \( T_{\text{Neptune}} = \sqrt{27000} \approx 164.32 \) Earth years.

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Key Concepts

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

Orbital Period
The orbital period of a planet is the time it takes to complete one full orbit around the Sun. In planetary motion, this is an essential aspect, as it determines how long a year on that planet is. For Earth, this period is one year, about 365.25 days. However, for other planets, especially those farther from the Sun, the orbital period is often much longer.

This period is not constant for all planets due to varying distances from the Sun, known as the semi-major axis. Kepler's Third Law provides a relationship between the orbital period and the semi-major axis, allowing us to calculate the period of distant planets, like Neptune, by comparing it to Earth's own known values.
Semi-Major Axis
The semi-major axis refers to the longest diameter of an elliptical orbit. In simpler terms, it's half of the longest length across the ellipse. This axis plays a crucial role in determining the characteristics of a planet's orbit around a star like the Sun.

For planets, the semi-major axis is an average distance from the Sun over one complete orbit. Since orbits are elliptical, this axis provides a standard measure that is consistent over time. In Kepler's Third Law, the cube of the semi-major axis is directly proportional to the square of the orbital period, indicating the powerful influence of distance on the speed and duration of a planet's orbit. Understanding this concept is key to calculating differences in orbital behavior between planets at varying distances.
Planetary Motion
Planetary motion describes how planets move around stars, such as the Sun. This motion is dictated by the gravitational forces between the planet and the Sun. Kepler's laws of planetary motion give us insights into how these celestial movements work. Of particular interest here is Kepler's Third Law, which helps explain the motion of planets in our solar system.

The fundamental idea behind planetary motion is that a planet's speed and path depend on its distance from its star. The further a planet is, the longer it takes to orbit the star, which is why Neptune's year is much longer than Earth's. The complexity of planetary motion is simplified into general laws that predict how planets behave in an elliptical orbit, emphasizing the perfect blend of science and mathematical elegance.

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Most popular questions from this chapter

(II) Determine the time it takes for a satellite to orbit the Earth in a circular "near-Earth" orbit. A "near-Earth" orbit is at a height above the surface of the Earth that is very small compared to the radius of the Earth. [Hint: You may take the acceleration due to gravity as essentially the same as that on the surface.] Does your result depend on the mass of the satellite?

Jupiter is about 320 times as massive as the Earth. Thus, it has been claimed that a person would be crushed by the force of gravity on a planet the size of Jupiter since people can't survive more than a few \(g\) 's. Calculate the number of \(g\) 's a person would experience at the equator of such a planet. Use the following data for Jupiter: mass \(=1.9 \times 10^{27} \mathrm{~kg}, \quad\) equatorial radius \(=7.1 \times 10^{4} \mathrm{~km}\) rotation period \(=9 \mathrm{hr} 55 \mathrm{~min} .\) Take the centripetal acceleration into account.

(III) The center of a \(1.00 \mathrm{~km}\) diameter spherical pocket of oil is \(1.00 \mathrm{~km}\) beneath the Earth's surface. Estimate by what percentage \(g\) directly above the pocket of oil would differ from the expected value of \(g\) for a uniform Earth? Assume the density of oil is \(8.0 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}\)

An asteroid of mass \(m\) is in a circular orbit of radius \(r\) around the Sun with a speed \(v\). It has an impact with another asteroid of mass \(M\) and is kicked into a new circular orbit with a speed of \(1.5 v .\) What is the radius of the new orbit in terms of \(r ?\)

(II) Four \(8.5-\mathrm{kg}\) spheres are located at the corners of a square of side \(0.80 \mathrm{~m}\). Calculate the magnitude and direction of the gravitational force exerted on one sphere by the other three.
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