/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 The Sun's center is at one focus... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 52

The Sun's center is at one focus of Earth's orbit. How far from this focus is the other focus, (a) in meters and (b) in terms of the solar radius, \(6.96 \times 10^{8} \mathrm{~m}\) ? The eccentricity is \(0.0167\), and the semimajor axis is \(1.50 \times 10^{11} \mathrm{~m}\).

Short Answer

Expert verified
(a) The other focus is approximately 2.505 脳 10^9 m from the Sun. (b) This is about 3.60 solar radii.

Step by step solution

01

Understanding the Ellipse

In an ellipse, there are two focal points (foci). The distance between these foci defines the ellipse's shape and can be determined by the formula for the distance between foci, which involves the eccentricity (蔚) and the semi-major axis (a). The formula is:\[ 2c = 2a \times e \] where \( e \) is the eccentricity, and \( a \) is the semi-major axis.
02

Calculate the Distance Between Foci

Use the formula \( 2c = 2a \times e \) to calculate the distance between the two foci:- Given: \( a = 1.50 \times 10^{11} \, \mathrm{m} \), \( e = 0.0167 \).- Substitute these values into the equation:\[ 2c = 2 \times 1.50 \times 10^{11} \times 0.0167 \]- Calculate \( 2c \):\[ 2c = 5.01 \times 10^9 \, \mathrm{m} \]
03

Determine the Distance from the Sun to the Other Focus

The distance from one focus (Sun) to the other focus is \( c \). Since \( 2c \) is the total distance between the foci, we have:\[ c = \frac{2c}{2} = \frac{5.01 \times 10^9}{2} \, \mathrm{m} \]\[ c = 2.505 \times 10^9 \, \mathrm{m} \]
04

Convert the Distance to Solar Radii

To convert the distance from the other focus to the solar radius units, divide the distance \( c \) by the solar radius \( 6.96 \times 10^8 \, \mathrm{m} \):\[ \text{Distance in solar radii} = \frac{2.505 \times 10^9}{6.96 \times 10^8} \]\[ \text{Distance in solar radii} \approx 3.60 \]

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Key Concepts

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

Eccentricity
Eccentricity is a fundamental characteristic that describes the shape of an elliptical orbit. It determines how much an orbit deviates from being perfectly circular.
Compared to a circle鈥攚here the eccentricity (\( e \)) is zero鈥攁n ellipse has an eccentricity greater than zero but less than one. The closer the eccentricity is to one, the more elongated the ellipse.
For Earth's orbit around the Sun, the eccentricity is about 0.0167, indicating that the orbit is nearly circular, given the value is close to zero.
  • High eccentricity: More stretched and oval-like ellipse.
  • Low eccentricity: Shape approaches a circle.
  • Zero eccentricity: Perfect circle.
Understanding eccentricity helps in studying the orbital dynamics of planets and calculating potential distances between locations within the orbit, such as the two foci in Earth's orbit.
Semi-major Axis
The semi-major axis is one of the key parameters defining the size of an elliptical orbit. It is the longest diameter of the ellipse, running through its center and both foci.
For the Earth's orbit around the Sun, the semi-major axis (\( a \)) measures approximately 1.50 x 10^{11} meters. This distance is crucial for calculating the overall shape and scale of the orbit.
  • The length of the semi-major axis affects the orbital period鈥攈ow long it takes for one complete orbit.
  • Using the semi-major axis and eccentricity, we can calculate other orbital elements such as the distance between foci.
In our exercise, the given semi-major axis helps specifically to determine the exact position of the two focal points for Earth's nearly circular path around the Sun.
Solar Radius
The solar radius is a unit of distance used to express astronomical distances relative to the size of the Sun. This unit simplifies comparisons between celestial scales and planetary measurements.
The Sun's radius is roughly 6.96 x 10^{8} meters. This value is used when making calculations for distances in terms of solar radii, offering a direct metric.
  • Helpful in comparing distances in solar systems without dealing with large numbers.
  • Provides insight into the relative scale of celestial distances.
For instance, converting the distance between the foci of Earth's orbit into solar radii allows us to contextualize and understand its scale compared to the massive size of the Sun. This concept is particularly useful in the final step of our exercise.

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Most popular questions from this chapter

In a shuttle craft of mass \(m=3000 \mathrm{~kg}\), Captain Janeway orbits a planet of mass \(M=9.50 \times 10^{25} \mathrm{~kg}\), in a circular orbit of radius \(r=4.20 \times 10^{7} \mathrm{~m} .\) What are (a) the period of the orbit and (b) the speed of the shuttle craft? Janeway briefly fires a forwardpointing thruster, reducing her speed by \(2.00 \%\). Just then, what are (c) the speed, (d) the kinetic energy, (e) the gravitational potential energy, and (f) the mechanical energy of the shuttle craft? (g) What is the semimajor axis of the elliptical orbit now taken by the craft? (h) What is the difference between the period of the original circular orbit and that of the new elliptical orbit? (i) Which orbit has the smaller period?

Mile-high building. In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constructed. Ignoring Earth's rotation, find the change in your weight if you were to ride an elevator from the street level, where you weigh \(600 \mathrm{~N}\), to the top of the building.

The mean distance of Mars from the Sun is \(1.52\) times that of Earth from the Sun. From Kepler's law of periods, calculate the number of years required for Mars to make one revolution around the Sun; compare your answer with the value given in Appendix \(\mathrm{C}\).

The orbit of Earth around the Sun is almost circular: The closest and farthest distances are \(1.47 \times 10^{8} \mathrm{~km}\) and \(1.52 \times 10^{8} \mathrm{~km}\) respectively. Determine the corresponding variations in (a) total energy, (b) gravitational potential energy, (c) kinetic energy, and (d) orbital speed. (Hint: Use conservation of energy and conservation of angular momentum.)

A typical neutron star may have a mass equal to that of the Sun but a radius of only \(10 \mathrm{~km}\). (a) What is the gravitational acceleration at the surface of such a star? (b) How fast would an object be moving if it fell from rest through a distance of \(1.0 \mathrm{~m}\) on such a star? (Assume the star does not rotate.)
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