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

Short Answer

Expert verified
Earth has an orbital velocity of approximately 29,784 m/s and a radial acceleration of about 0.006 m/s². Mercury has an orbital velocity of approximately 47,872 m/s and a radial acceleration of about 0.025 m/s².

Step by step solution

01

Calculation of Earth's orbital velocity

We calculate the orbital velocity using the formula \(v = \frac{2 \pi r}{t}\) , where r is the radius of the orbit and t is the period of revolution. The radius r is \(1.50 \times 10^{8} \mathrm{~km}\) or \(1.50 \times 10^{11} \mathrm{~m}\). The period t is \(365 \times 24 \times 3600 \mathrm{~s}\). Substituting the values we get, \(v = \frac{2 \pi (1.50 \times 10^{11})}{365 \times 24 \times 3600}\)
02

Calculation of Earth's radial acceleration

We calculate the radial acceleration using the formula \(a = \frac{v^{2}}{r}\), where v is the orbital velocity as calculated previously and r is the radius of the orbit. Substituting the values we get, \(a = \frac{v^{2}}{1.50 \times 10^{11}}\)
03

Repeating for planet Mercury - Calculation of orbital velocity

Using the formula for orbital velocity as before, and substituting the radius \(r = 5.79 \times 10^{10} \mathrm{~m}\) and period \(t = 88.0 \times 24 \times 3600 \mathrm{~s}\), we get \(v = \frac{2 \pi (5.79 \times 10^{10})}{88.0 \times 24 \times 3600}\)
04

Repeating for planet Mercury - Calculation of radial acceleration

Using the formula for radial acceleration as before, and substituting the orbital velocity and radius for Mercury, we get \(a = \frac{v^{2}}{5.79 \times 10^{10}}\)

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Key Concepts

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

Orbital Velocity Calculation
Understanding how celestial bodies like planets move requires calculating their orbital velocity, which is the speed at which they travel around another body in space. To calculate orbital velocity, we use the basic formula
\( v = \frac{2 \pi r}{t} \)
where \( v \) is the velocity, \( r \) is the radius of the orbit, and \( t \) is the time it takes to complete one orbit, known as the orbital period.

For the Earth, this calculation takes the radius of its orbit and divides it by the time it takes to go around the Sun once, which is one year. In our specific example, when we plug in the Earth's radius of \( 1.50 \times 10^{11} \) meters and the orbital period of 365 days, we determine Earth's orbital velocity. This velocity tells us how fast Earth is traveling in its path around the Sun, which is essential not only for understanding Earth's motion but also for space missions that send satellites into orbit or travel to other planets.
Radial Acceleration
Radial acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. It's what keeps planets in their orbits around the Sun, instead of flying off into space. The formula to calculate radial acceleration is
\( a = \frac{v^{2}}{r} \)
where \( a \) is the radial acceleration, \( v \) is the orbital velocity, and \( r \) is the radius of the orbit.

In the context of Earth's motion around the Sun, once we've calculated Earth's orbital velocity, we use it to find the radial acceleration by squaring the velocity and dividing by the radius of the orbit. This gives us a value that basically quantifies the 'pull' that Earth feels towards the Sun, allowing it to maintain a stable orbit and not veer off into the void of space. It's a critical component of orbital mechanics that explains not just the motion of planets, but also of satellites and other orbiting objects in space.
Planetary Motion
The motion of planets around their stars, such as Earth's journey around the Sun, follows certain physical laws that were first described by Johannes Kepler and later explained by Isaac Newton's theory of gravitation. Planetary motion involves both linear movement along the orbit and radial acceleration towards the center of the orbit.

Repeating the calculations of Earth's orbital velocity and radial acceleration for another planet, like Mercury, involves the same principles but requires adapting the formulae to Mercury's specific orbital radius and period. This not only demonstrates the universality of the laws governing planetary motion but also highlights variations due to different orbital characteristics. For instance, the shorter orbital period of Mercury compared to Earth reflects its greater velocity and changes its radial acceleration due to its closer proximity to the Sun. All these different factors interplay to create the complex dance of planetary motion observed in our solar system.

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Most popular questions from this chapter

A Ferris wheel with radius \(14.0 \mathrm{~m}\) is turning about a horizontal axis through its center (Fig. E3.31). The linear speed of a passenger on the rim is constant and equal to \(6.00 \mathrm{~m} / \mathrm{s}\) What are the magnitude and direction of the passenger's acceleration as she passes through (a) the lowest point in her circular motion and (b) the highest point in her circular motion? (c) How much time does it take the Ferris wheel to make one revolution?

In the middle of the night you are standing a horizontal distance of \(14.0 \mathrm{~m}\) from the high fence that surrounds the estate of your rich uncle. The top of the fence is \(5.00 \mathrm{~m}\) above the ground. You have taped an important message to a rock that you want to throw over the fence. The ground is level, and the width of the fence is small enough to be ignored. You throw the rock from a height of \(1.60 \mathrm{~m}\) above the ground and at an angle of \(56.0^{\circ}\) above the horizontal. (a) What minimum initial speed must the rock have as it leaves your hand to clear the top of the fence? (b) For the initial velocity calculated in part (a), what horizontal distance beyond the fence will the rock land on the ground?

Firefighters use a high-pressure hose to shoot a stream of water at a burning building. The water has a speed of \(25.0 \mathrm{~m} / \mathrm{s}\) as it leaves the end of the hose and then exhibits projectile motion. The firefighters adjust the angle of elevation \(\alpha\) of the hose until the water takes \(3.00 \mathrm{~s}\) to reach a building \(45.0 \mathrm{~m}\) away. Ignore air resistance; assume that the end of the hose is at ground level. (a) Find \(\alpha\). (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

A small object is projected from level ground with an initial velocity of magnitude \(16.0 \mathrm{~m} / \mathrm{s}\) and directed at an angle of \(60.0^{\circ}\) above the horizontal. (a) What is the horizontal displacement of the object when it is at its maximum height? How does your result compare to the horizontal range \(R\) of the object? (b) What is the vertical displacement of the object when its horizontal displacement is \(80.0 \%\) of its horizontal range \(R ?\) How does your result compare to the maximum height \(h_{\max }\) reached by the object? (c) For when the object has horizontal displacement \(x-x_{0}=\alpha R,\) where \(\alpha\) is a positive constant, derive an expression (in terms of \(\alpha\) ) for \(\left(y-y_{0}\right) / h_{\max }\). Your result should not depend on the initial velocity or the angle of projection. Show that your expression gives the correct result when \(\alpha=0.80,\) as is the case in part (b). Also show that your expression gives the correct result for \(\alpha=0, \alpha=0.50\), and \(\alpha=1.0\)

In a World Cup soccer match, Juan is running due north toward the goal with a speed of \(8.00 \mathrm{~m} / \mathrm{s}\) relative to the ground. A teammate passes the ball to him. The ball has a speed of \(12.0 \mathrm{~m} / \mathrm{s}\) and is moving in a direction \(37.0^{\circ}\) east of north, relative to the ground. What are the magnitude and direction of the ball's velocity relative to Juan?

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