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Chapter 3: Problem 28

The radius of the earth's orbit around the sun (assumed to be circular) is 1.50 \(\times\) 10\(^8\) km, and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in m/s? (b) What is the radial acceleration of the earth toward the sun, in m/s\(^2\) ? (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius = 5.79 \(\times\) 10\(^7\) km, orbital period = 88.0 days).

Short Answer

Expert verified
Earth's velocity is 29,860 m/s, radial acceleration is 5.93 脳 10鈦宦� m/s虏; Mercury's velocity is 47,900 m/s, radial acceleration is 3.97 脳 10鈦宦� m/s虏.

Step by step solution

01

Convert Units for Earth's Orbit

First, convert the radius from kilometers to meters by multiplying by 1,000 (since 1 km = 1,000 m). So the radius becomes:\[ r = 1.50 \times 10^8 \times 1000 = 1.50 \times 10^{11} \text{ m} \]Convert the period from days to seconds knowing that there are 86400 seconds in a day:\[ T = 365 \times 86400 = 3.1536 \times 10^7 \text{ seconds} \]
02

Calculate Earth's Orbital Velocity

Use the formula for the circumference of a circle \( C = 2 \pi r \) to find the distance traveled in one orbit:\[ C = 2 \pi \times 1.50 \times 10^{11} = 9.42 \times 10^{11} \text{ meters} \]The orbital velocity \( v \) is then given by:\[ v = \frac{C}{T} = \frac{9.42 \times 10^{11}}{3.1536 \times 10^7} = 2.986 \times 10^4 \text{ m/s} \]
03

Find Earth's Radial Acceleration

The radial acceleration \( a_r \) is given by:\[ a_r = \frac{v^2}{r} = \frac{(2.986 \times 10^4)^2}{1.50 \times 10^{11}} \approx 5.93 \times 10^{-3} \text{ m/s}^2 \]
04

Convert Units for Mercury's Orbit

Similarly, convert Mercury's orbit radius from kilometers to meters:\[ r = 5.79 \times 10^7 \times 1000 = 5.79 \times 10^{10} \text{ m} \]Convert its period from days to seconds:\[ T = 88 \times 86400 = 7.6032 \times 10^6 \text{ seconds} \]
05

Calculate Mercury's Orbital Velocity

Find the circumference of Mercury's orbit:\[ C = 2 \pi \times 5.79 \times 10^{10} = 3.64 \times 10^{11} \text{ meters} \]Calculate the orbital velocity \( v \):\[ v = \frac{C}{T} = \frac{3.64 \times 10^{11}}{7.6032 \times 10^6} = 4.79 \times 10^4 \text{ m/s} \]
06

Find Mercury's Radial Acceleration

Using the radial acceleration formula:\[ a_r = \frac{v^2}{r} = \frac{(4.79 \times 10^4)^2}{5.79 \times 10^{10}} \approx 3.97 \times 10^{-2} \text{ m/s}^2 \]

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Key Concepts

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

Orbital Velocity
Orbital velocity is the speed at which a planet revolves around a star, such as Earth around the Sun. To calculate it, you need to find the circumference of the planet's orbit, which is the circle that the Earth traces as it moves around the Sun.
The formula to find the circumference is:
  • \( C = 2 \pi r \)
The orbital velocity, \( v \), is determined by how quickly this distance is covered, found using the formula:
  • \( v = \frac{C}{T} \)
Here, \( T \) is the period of one complete orbit.
For Earth, the orbital velocity is calculated to be approximately \( 2.986 \times 10^4 \text{ m/s} \).
This velocity ensures Earth鈥檚 stable path around the Sun, balancing gravitational pull and the inertia of Earth moving straight forward.
Radial Acceleration
Radial acceleration refers to the acceleration that is directed towards the center of a circular path. It is essential to maintain the circular motion of an object, countering the inertia that would otherwise cause it to fly off in a straight line.
The radial acceleration \( a_r \) can be calculated using the formula:
  • \( a_r = \frac{v^2}{r} \)
where:
  • \( v \) is the orbital velocity
  • \( r \) is the radius of the orbit
For Earth, the radial acceleration comes out to be approximately \( 5.93 \times 10^{-3} \text{ m/s}^2 \).
This acceleration is crucial as it continuously changes the direction of Earth's velocity, keeping it in orbit around the Sun.
Earth's Orbit
The orbit of Earth around the Sun is often simplified into a circular path for easier calculation, although it is actually somewhat elliptical.
The average radius of this orbit is \( 1.50 \times 10^{11} \text{ m} \).
This massive radius means that it takes 365 days for Earth to complete one full orbit, which we commonly refer to as a year.
Calculating with a circular model also allows for straightforward computations of Earth's orbital velocity and radial acceleration, parameters that help in understanding not just Earth's motion, but those of other planets as well.
By studying Earth鈥檚 orbit, scientists learn how our planet's movement affects seasons, climate, and ultimately life on Earth.

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Most popular questions from this chapter

You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determine the width of the river and the current speed (the speed of the water relative to the earth). You have a small boat with an outboard motor. By measuring the time it takes to cross a pond where the water isn't flowing, you have calibrated the throttle settings to the speed of the boat in still water. You set the throttle so that the speed of the boat relative to the river is a constant 6.00 m/s. Traveling due north across the river, you reach the opposite bank in 20.1 s. For the return trip, you change the throttle setting so that the speed of the boat relative to the water is 9.00 m/s. You travel due south from one bank to the other and cross the river in 11.2 s. (a) How wide is the river, and what is the current speed? (b) With the throttle set so that the speed of the boat relative to the water is 6.00 m/s, what is the shortest time in which you could cross the river, and where on the far bank would you land?

Two students are canoeing on a river. While heading upstream, they accidentally drop an empty bottle overboard. They then continue paddling for 60 minutes, reaching a point 2.0 km farther upstream. At this point they realize that the bottle is missing and, driven by ecological awareness, they turn around and head downstream. They catch up with and retrieve the bottle (which has been moving along with the current) 5.0 km downstream from the turnaround point. (a) Assuming a constant paddling effort throughout, how fast is the river flowing? (b) What would the canoe speed in a still lake be for the same paddling effort?

CALC The coordinates of a bird flying in the \(xy\)-plane are given by \(x(t)\) = \(at\) and \(y(t)\) = 3.0 m - \(\beta t^2\), where \(\alpha\) = 2.4 m/s and \(\beta\) = 1.2 m/s\(^2\). (a) Sketch the path of the bird between \(t\) = 0 and \(t\) = 2.0 s. (b) Calculate the velocity and acceleration vectors of the bird as functions of time. (c) Calculate the magnitude and direction of the bird's velocity and acceleration at \(t\) = 2.0 s. (d) Sketch the velocity and acceleration vectors at \(t\) = 2.0 s. At this instant, is the bird's speed increasing, decreasing, or not changing? Is the bird turning? If so, in what direction?

Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in 2.70 s, while Milada jumps horizontally with an initial speed of 95.0 cm/s. How far from the base of the cliff will Milada hit the ground? Ignore air resistance.

A toy rocket is launched with an initial velocity of 12.0 m/s in the horizontal direction from the roof of a 30.0-m-tall building. The rocket's engine produces a horizontal acceleration of \((1.60 m/s^3)t\), in the same direction as the initial velocity, but in the vertical direction the acceleration is \(g\), downward. Ignore air resistance. What horizontal distance does the rocket travel before reaching the ground?
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