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The sun has a mass of \(1.99 \times 10^{30} \mathrm{kg}\) and is moving in a circular orbit about the center of our galaxy, the Milky Way. The radius of the orbit is \(2.3 \times 10^{4}\) light-years (1 light-year \(=9.5 \times 10^{15} \mathrm{m} )\) and the angular speed of the sun is \(1.1 \times 10^{-15} \mathrm{rad} / \mathrm{s}\) . (a) Determine the tangential speed of the sun. (b) What is the magnitude of the net force that acts on the sun to keep it moving around the center of the Milky Way?

Short Answer

Expert verified
Tangential speed: \(2.4035 \times 10^5\) m/s; Net force: \(2.3393 \times 10^{20}\) N.

Step by step solution

01

Convert the Radius to Meters

First, convert the radius of the Sun's orbit from light-years to meters. Given that 1 light-year is equal to \(9.5 \times 10^{15}\) m, compute the orbital radius \(r\):\[r = 2.3 \times 10^4 \, \text{light-years} \times 9.5 \times 10^{15} \, \text{m/light-year} = 2.185 \times 10^{20} \, \text{m}\]
02

Calculate the Tangential Speed

Using the formula for tangential speed \(v\) which relates the radius \(r\) and angular speed \(\omega\), given by \(v = r\omega\), calculate the tangential speed:\[v = (2.185 \times 10^{20} \, \text{m})(1.1 \times 10^{-15} \, \text{rad/s}) = 2.4035 \times 10^{5} \, \text{m/s}\]
03

Calculate the Net Force on the Sun

To find the net force on the Sun, use the centripetal force formula \(F = m v^2 / r\), where \(m\) is the mass of the Sun, \(v\) is the tangential speed, and \(r\) is the radius:\[F = \frac{(1.99 \times 10^{30} \, \text{kg}) (2.4035 \times 10^{5} \, \text{m/s})^2}{2.185 \times 10^{20} \, \text{m}}\]Calculating this gives:\[F = 2.3393 \times 10^{20} \, \text{N}\]

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Key Concepts

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

Tangential Speed
Tangential speed is an important concept when we look at objects moving in circular paths, like the Sun orbiting the center of the Milky Way. Picture a point on the rim of a spinning wheel — tangential speed measures how fast that point is moving along its circular path.
To calculate tangential speed (\(v\)), we use the formula \(v = r \cdot \omega\), where \(r\) is the radius of the circular path, and \(\omega\) is the angular speed, both measurable quantities. In the case of the Sun's orbit:
  • \(r = 2.185 \times 10^{20} \, \text{m}\)
  • \(\omega = 1.1 \times 10^{-15} \, \text{rad/s}\)
By plugging these values into the formula, we arrive at a tangential speed of \(2.4035 \times 10^{5} \, \text{m/s}\). This tells us how rapid the Sun moves along its incredible circular path within our galaxy!
Angular Speed
Angular speed is a measure of how quickly an object travels through an angle. Consider a merry-go-round: as it spins, its angular speed tells us how fast the ride is turning. For objects like the Sun in a galactic orbit, the angular speed is crucial in understanding how fast they move in their circular paths.
Angular speed \(\omega\) is typically measured in radians per second (rad/s), which implies how many radians an object moves through per unit time. The angular speed of the Sun around the Milky Way is \(1.1 \times 10^{-15} \, \text{rad/s}\). It's a reflection of its massive orbit — where even small angular speeds result in substantial linear or tangential speeds, given the vast size of the orbital radius involved.
Galactic Orbit
A galactic orbit refers to the path an object takes while circling around the center of a galaxy, such as our Milky Way. The Sun, for instance, travels along a colossal orbit around the galaxy's center. Understanding this orbit involves knowing the forces at play to maintain such motion, like centripetal force, which keeps the Sun moving in its path.
The radius of this orbit for the Sun is \(2.3 \times 10^{4}\) light-years, demonstrating the incredible scale of such orbits. Within this orbit, the gravitational forces between the various components of the galaxy play a key role in dictating movement. These orbits demonstrate many physics principles in action, from tangential and angular speeds to the gravitational pull and centripetal force.
Conversion of Units
Conversion of units is crucial for solving physics problems, especially when dealing with large-scale measurements like astronomical distances. In this context, converting the Sun's orbital radius from light-years to meters was necessary to utilize the standard unit system in physics calculations.
We know that 1 light-year equals \(9.5 \times 10^{15} \, \text{m}\). By multiplying, we convert the Sun's orbit radius from \(2.3 \times 10^{4}\) light-years to \(2.185 \times 10^{20} \, \text{m}\). Such conversions allow us to work with measurable and relatable units, enabling further calculations, like tangential speed and net force with more straightforward application of formulas and laws of physics.
  • Always keep track of units to avoid errors!
  • Conversion ensures compatibility in equations for consistent outcomes.

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