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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 \(\mathrm{m} / \mathrm{s} ?\) (b) What is 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
(a) Earth's velocity: 29,865 m/s, (b) Earth's acceleration: 5.93 x 10^-3 m/s², (a) Mercury's velocity: 47,872 m/s, (b) Mercury's acceleration: 3.82 x 10^-2 m/s².

Step by step solution

01

Convert Earth’s Orbital Radius to Meters

The Earth's orbit radius is given in kilometers: \(1.50 \times 10^{8}\) km. To convert it into meters, we multiply by 1000 (since 1 km = 1000 m):\[\text{Earth's radius in meters} = 1.50 \times 10^{8} \times 1000 = 1.50 \times 10^{11} \text{ m}.\]
02

Calculate Earth’s Orbital Velocity

The formula for orbital velocity \(v\) is \(v = \frac{2\pi r}{T}\), where \(r\) is the radius of the orbit and \(T\) is the orbital period. First, convert the period from days to seconds: \(365 \text{ days} = 365 \times 24 \times 3600 \text{ seconds}.\)\[T = 3.1536 \times 10^{7} \text{ s}.\] Now, plug the values into the formula:\[v = \frac{2\pi \times 1.50 \times 10^{11}}{3.1536 \times 10^{7}} \approx 29,865\text{ m/s}.\]
03

Calculate Earth’s Radial Acceleration

Radial (centripetal) acceleration \(a_r\) is given by \(a_r = \frac{v^2}{r}\). Using the velocity from Step 2, \[v = 29,865 \text{ m/s}\], and the radius \(r = 1.50 \times 10^{11} \text{ m}\), we find:\[a_r = \frac{(29,865)^2}{1.50 \times 10^{11}} \approx 5.93 \times 10^{-3} \text{ m/s}^2.\]
04

Convert Mercury’s Orbital Radius to Meters

Mercury's orbit radius is \(5.79 \times 10^{7}\) km. Converting to meters:\[\text{Mercury's radius in meters} = 5.79 \times 10^{7} \times 1000 = 5.79 \times 10^{10} \text{ m}.\]
05

Calculate Mercury’s Orbital Velocity

Using the same formula for orbital velocity \(v = \frac{2\pi r}{T}\). Convert Mercury's period to seconds: \(88 \times 24 \times 3600\).\[T = 7.6032 \times 10^{6} \text{ s}.\] Plug the values into the formula:\[v = \frac{2\pi \times 5.79 \times 10^{10}}{7.6032 \times 10^{6}} \approx 47,872\text{ m/s}.\]
06

Calculate Mercury’s Radial Acceleration

Using the radial acceleration formula \(a_r = \frac{v^2}{r}\), with \(v = 47,872 \text{ m/s}\) and \(r = 5.79 \times 10^{10} \text{ m}\), we find:\[a_r = \frac{(47,872)^2}{5.79 \times 10^{10}} \approx 3.82 \times 10^{-2} \text{ m/s}^2.\]

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Key Concepts

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

Orbital Velocity Calculation
Orbital velocity is a critical concept when studying how planets and other celestial bodies move through space. To find the orbital velocity of a planet like Earth, we use its orbit's radius and the time it takes to complete one orbit around the sun, known as the orbital period.
To begin, we transform the radius of Earth's orbit from kilometers to meters, because standard units in physics are required. This involves multiplying the given kilometers by 1000 (since 1 km = 1000 m). Then, using the formula for orbital velocity:
  • \( v = \frac{2\pi r}{T} \),
where \( r \) is the orbit's radius and \( T \) is the orbital period in seconds, we calculate the velocity. This formula denotes that the speed of an object in a circular orbit is directly affected by the circumference of the orbit \(2\pi r\) and inversely affected by the time it takes to complete that orbit. Understanding these calculations is crucial for taking the first steps into orbital mechanics and planetary motion.
Centripetal Acceleration
Centripetal acceleration, often referred to as radial acceleration in the context of circular motion, is a key factor that keeps an object moving along a curved path. For a planet like Earth, this acceleration is directed toward the center of its orbit, which in this case, is the Sun.
The formula for calculating centripetal acceleration is:
  • \( a_r = \frac{v^2}{r} \),
where \( v \) is the orbital velocity and \( r \) is the radius of the orbit. This acceleration is necessary to constantly change the direction of the velocity vector of the planet, keeping it on its elliptical path around the sun. Without it, the planet would move off in a straight line due to inertia, leaving its orbit.
The centripetal acceleration can be understood as the force required to "pull" the planet back toward the center with each passing moment, allowing it to follow its orbital trajectory.
Planetary Motion
The study of planetary motion provides insight into the dynamics and interactions of celestial bodies within our solar system. It encompasses not just how planets move, but why they maintain certain speeds and paths.
The Earth's orbit around the Sun is a classic example of such motion, primarily dictated by two mechanisms: gravity and inertia.
  • Gravity: The gravitational pull from the sun acts as the central force that governs the elliptical orbits of planets.
  • Inertia: Due to its momentum, the Earth strives to move in a straight line but is continually deflected into its orbit by the Sun's gravity.
Together, these forces create the stable orbits seen in our solar system. This section of physics often uses principles laid by Kepler's Laws, which describe the motions of the planets. For instance, the orbital period is related to the distance from the sun, so planets like Mercury, which are closer, have shorter orbital periods compared to Earth.
Physics Problem Solving
Physics problem solving, especially in orbital mechanics, involves a step-by-step approach that builds understanding of mathematical relationships in systems. Tackling problems should always begin with a clear understanding of the knowns and unknowns in terms of quantities and formulas.
Computations, such as those determining a planet's orbital velocity or radial acceleration, require:
  • Proper conversion of units to maintain consistency.
  • A systematic application of physics formulas to extract desired quantities.
  • Verification of the results against expected behavior or known physical principles.
Approaching physics problems with clarity, such as separating problems into smaller steps and consistently checking your work, can greatly enhance one’s problem-solving skills. This discipline ensures that margins of error are minimized and understanding deepens with practice, crucial for mastering topics like orbital mechanics and beyond.

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Most popular questions from this chapter

A shot putter releases the shot some distance above the level ground with a velocity of \(12.0 \mathrm{m} / \mathrm{s}, 51.0^{\circ}\) above the horizontal. The shot hits the ground 2.08 \(\mathrm{s}\) later. You can ignore air resistance. (a) What are the components of the shot's acceleration while in flight? (b) What are the components of the shot's velocity at the beginning and at the end of its trajectory? (c) How far did she throw the shot horizontally? (d) Why does the expression for \(R\) in Example 3.8 not give the correct answer for part (c)? (e) How high was the shot above the ground when she released it? Draw \(x-t\) , \(y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.

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