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Chapter 3: Problem 11

The earth moves around the sun in a nearly circular orbit of radius \(1.50 \times 10^{11} \mathrm{m} .\) During the three summer months (an elapsed time of \(\left.7.89 \times 10^{6} \mathrm{s}\right),\) the earth moves one-fourth of the distance around the sun. (a) What is the average speed of the earth? (b) What is the magnitude of the average velocity of the earth during this period?

Short Answer

Expert verified
(a) \(2.99 \times 10^{4} \text{m/s}\); (b) \(3.80 \times 10^{4} \text{m/s}\).

Step by step solution

01

Understand the Question

We need to determine two things: (a) the average speed of the Earth as it moves one-fourth around its orbit during the summer months, and (b) the magnitude of the average velocity of the Earth over this period.
02

Calculate the Circumference of Earth's Orbit

The Earth's orbit is nearly circular with a radius of \(1.50 \times 10^{11} \text{m}\). The circumference of a circle is calculated using the formula \(C = 2 \pi r\). Thus, we find:\[ C = 2 \pi (1.50 \times 10^{11}) \approx 9.42 \times 10^{11} \text{m} \]
03

Determine the Distance Traveled in 3 Months

In three months, the Earth moves one-fourth of its orbit. Therefore, the distance traveled \(d\) is one-fourth of the circumference:\[ d = \frac{1}{4} \times 9.42 \times 10^{11} = 2.36 \times 10^{11} \text{m} \]
04

Calculate the Average Speed

Average speed is given by the formula \(\text{average speed} = \frac{\text{total distance}}{\text{total time}}\). Using the distance from Step 3 and the given time:\[ \text{average speed} = \frac{2.36 \times 10^{11}}{7.89 \times 10^{6}} \approx 2.99 \times 10^{4} \text{m/s} \]
05

Calculate Average Velocity's Magnitude

Average velocity's magnitude is calculated as the displacement divided by time. Since the Earth completes a quarter of a circle, the displacement forms a straight line across the orbit's quarter, equal to the diameter, \(2r\):\[ \text{displacement} = 2 \times 1.50 \times 10^{11} = 3.00 \times 10^{11} \text{m} \] Thus, \[ \text{average velocity magnitude} = \frac{3.00 \times 10^{11}}{7.89 \times 10^{6}} \approx 3.80 \times 10^{4} \text{m/s} \]

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Key Concepts

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

Orbital Mechanics
Understanding orbital mechanics is key for analyzing how celestial bodies such as the Earth move in space. Orbital mechanics deals with the principles governing the motion of objects orbiting a larger body due to gravitational forces. In our example, Earth revolves around the Sun in a circular-like orbit due to the gravitational pull from the Sun.
- Key considerations in orbital mechanics include:
  • Gravitational force: This is the attractive force that causes planets to orbit the Sun. It provides the centripetal force necessary for maintaining the Earth's nearly circular orbit.
  • Orbit shape: While the Earth's orbit is nearly circular, most orbits are elliptical, described by Kepler鈥檚 laws of planetary motion.
  • Orbital elements: These include properties such as semi-major axis (half of the orbit's longest diameter) and eccentricity (a measure of how an orbit deviates from being circular).
Understanding these fundamentals of orbital mechanics allows us to calculate significant properties of the orbital paths, such as circumference, radius, and distance covered during a portion of the orbit.
Average Speed
Average speed is a scalar quantity that refers to the total distance traveled divided by the total time taken. In the case of the Earth moving around the Sun, we analyze the average speed over a quarter of its orbit.
To determine the Earth鈥檚 average speed:
  • Calculate the orbit circumference: As done above, the circumference of Earth's orbit is gone using the formula \(C = 2 \pi r\), yielding approximately \(9.42 \times 10^{11} \text{m}\).
  • Determine distance traveled: For one-fourth of its orbit, Earth travels \(2.36 \times 10^{11} \text{m}\).
  • Calculate average speed: Using the formula \(\text{average speed} = \frac{\text{total distance}}{\text{total time}}\), we find Earth's average speed to be approximately \(2.99 \times 10^{4} \text{m/s}\).
This measurement of speed helps astronomers understand the dynamic nature of Earth's motion around the Sun in real-time terms.
Average Velocity
While average speed focuses on the total distance traveled, average velocity considers the overall displacement divided by the total time, making it a vector quantity that has both magnitude and direction.
During a quarter orbit of the Earth:
  • Determine displacement: Since only a quarter of the orbit is completed, the straight-line displacement is the diameter of the orbit, equal to \(3.00 \times 10^{11} \text{m}\).
  • Calculate total time: This is consistent with the time period in question, \(7.89 \times 10^{6}\) seconds.
  • Find average velocity's magnitude: Using the formula \(\text{average velocity magnitude} = \frac{\text{displacement}}{\text{time}}\), it results in \(3.80 \times 10^{4} \text{m/s}\).
The vector nature of average velocity highlights how it considers the direct path between the start and end points, providing deeper insights into the straight-line movement relative to total orbit completion.

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Most popular questions from this chapter

An airplane with a speed of \(97.5 \mathrm{m} / \mathrm{s}\) is climbing upward at an angle of \(50.0^{\circ}\) with respect to the horizontal. When the plane's altitude is \(732 \mathrm{m},\) the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

On a spacecraft, two engines are turned on for 684 s at a moment when the velocity of the craft has \(x\) and \(y\) components of \(v_{0 x}=4370 \mathrm{m} / \mathrm{s}\) and \(v_{0 y}=6280 \mathrm{m} / \mathrm{s} .\) While the engines are firing, the craft undergoes a displacement that has components of \(x=4.11 \times 10^{6} \mathrm{m}\) and \(y=6.07 \times 10^{6} \mathrm{m} .\) Find the \(x\) and \(y\) components of the craft's acceleration.

A baseball player hits a home run, and the ball lands in the left-field seats, \(7.5 \mathrm{m}\) above the point at which it was hit. It lands with a velocity of \(36 \mathrm{m} / \mathrm{s}\) at an angle of \(28^{\circ}\) below the horizontal (see the drawing). The positive directions are upward and to the right in the drawing. Ignoring air resistance, find the magnitude and direction of the initial velocity with which the ball leaves the bat.

A quarterback claims that he can throw the football a horizontal distance of \(183 \mathrm{m}(200 \mathrm{yd}) .\) Furthermore, he claims that he can do this by launching the ball at the relatively low angle of \(30.0^{\circ}\) above the horizontal. To evaluate this claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison, a baseball pitcher who can accurately throw a fastball at \(45 \mathrm{m} / \mathrm{s}(100 \mathrm{mph})\) would be considered exceptional.

A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of \(5.3 \mathrm{m} / \mathrm{s},\) hoping to land on the roof of an adjacent building. Air resistance is negligible. The horizontal distance between the two buildings is \(D,\) and the roof of the adjacent building is \(2.0 \mathrm{m}\) below the jumping-off point. Find the maximum value for \(D\).
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