The Milky Way galaxy rotates about its center with a period...

The Milky Way galaxy rotates about its center with a period of about 200 million yr. The Sun is $2 \times 10^{20} \mathrm{m}$ from the center of the galaxy. (a) What is the Sun's radial acceleration? (b) What is the net gravitational force on the Sun due to the other stars in the Milky Way?

Key Concepts

Centripetal Acceleration

Centripetal acceleration is the inward acceleration required to keep an object moving in a circular path. It is crucial in orbital mechanics and can be calculated using the formula a = v²/r or, alternatively, a = 4?²r/T² when the orbital period is known. This concept links the object's speed, the radius of its circular path, and the acceleration that keeps it in orbit.

Gravitational Force in Orbital Motion

In a stable orbit, the net gravitational force acting on an object provides exactly the centripetal force required to maintain its circular motion. By applying Newton's law of gravitation and Newton's second law, the gravitational force can be determined using F = ma, where m is the mass of the object and a is its centripetal acceleration. This principle is essential for understanding how gravitational forces govern the motion of bodies in space.

Galactic Dynamics

Galactic dynamics studies the motion of stars and other components within a galaxy under the influence of its gravitational potential. It involves analyzing how the rotation period of the galaxy and the radial distances of stars from the galactic center relate to the overall mass distribution. This understanding is important for interpreting observations of galaxy rotation curves and inferring the presence of dark matter.

Transcript

00:02 In this example, we're going to determine the radial acceleration of the sun as it rotates with the milky way galaxy.

00:14 And knowing that, we're also going to make an estimate for the sum of all the forces of gravity acting on the sun from all the other stars in the milky way galaxy.

00:24 So if we recall the radial acceleration can be expressed a few ways, the way that's going to be most useful to a here is omega squared r, where omega is the angular velocity of the rotation.

00:45 And remember, omega can be expressed in terms of period of rotation as 2 pi over the period.

00:58 So if we plug that into our equation here, we'll end up with 4 pi squared over t squared times r, and in this case, the radius of rotation for the sun is just d sub s here, the distance out to the sun.

01:22 So times d sub s, right, because it's going to rotate in this circle.

01:32 Okay, so now that we have that, all we should have to do is plug in our values, right? but we'll want to convert the period first from years to seconds.

01:47 So the period is 200 million years or 200 times 10 to the 6 years.

01:59 Now we know that there are 365 .25 days in a year.

02:16 We also know that there are 24 hours in a day and we also know there are 3 ,600 seconds in one hour.

02:44 So if we crunch those numbers, the period comes out to be 6 .31 times 10 to 15 seconds.

02:58 So now if we plug in everything back into our equation here, we will get that the magnitude of the radial acceleration is about 3 .5.

03:19 My bad...

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