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Chapter 23: Problem 47

Consider a star that orbits around Sagittarius \(A^{*}\) in a circular orbit of radius \(530 \mathrm{AU}\). (a) If the star's orbital speed is \(2500 \mathrm{~km} / \mathrm{s}\), what is its orbital period? Give your answer in years. (b) Determine the sum of the masses of Sagittarius A* and the star. Give your answer in solar masses. (Your answer is an estimate of the mass of Sagittarius A*, because the mass of a single star is negligibly small by comparison.)

Short Answer

Expert verified
a) The orbital period of the star is 5.01 years. b) The sum of the masses of Sagittarius \(A^{*}\) and the star is 3.54 million solar masses.

Step by step solution

01

Convert given quantities to SI units

Convert the given radius from astronomical units (AU) to meters and the given velocity from km/s to m/s. Use the following conversions: 1 AU = \(1.496 \times 10^{11}\) m and 1 km = 1000 m. Thus, the radius \(r = 530AU \times 1.496 \times 10^{11}\) m/AU = \(7.94 \times 10^{13}\) m and the velocity \(v = 2500\) km/s \( \times 1000\) m/km = \(2.5 \times 10^{6}\) m/s.
02

Calculate the orbital period

The orbital period \(T\) (in seconds) of an object moving in a circular path is given by the formula \(T = \frac{2 \pi r}{v}\). Substituting the given values, we find \(T = \frac{2 \pi \times 7.94 \times 10^{13}}{2.5 \times 10^{6}} = 1.58 \times 10^{8}\) s. Convert this to years by using the conversion factor 1 year = \(3.1536 \times 10^{7}\) s, giving \(T = \frac{1.58 \times 10^{8}}{3.1536 \times 10^{7}}\) years = 5.01 years.
03

Determine the sum of the masses

Applying the formula for the centripetal force \(F = \frac{mv^{2}}{r}\), where \(m\) is the mass of the star and \(v\) is the star's velocity, and equating it to the gravitational force \(F = G\frac{Mm}{r^{2}}\), where \(M\) is the mass of Sagittarius \(A^{*}\), and \(G\) is the gravitational constant, we can cancel out \(m\) and \(r\) (as the star's mass is negligible and it appears on both equations) and with some algebric work we get: \(M = \frac{v^{2}r}{G}\). Subsituting the given values: \(M = \frac{(2.5 \times 10^{6})^{2} \times 7.94 \times 10^{13}}{6.674 \times 10^{-11}}\) = \(7.05 \times 10^{36}\) kg. Convert this to solar masses using 1 solar mass = \(1.989 \times 10^{30}\) kg, giving \(M = \frac{7.05 \times 10^{36}}{1.989 \times 10^{30}}\) solar masses = 3.54 million solar masses.

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Key Concepts

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

Astrophysics and Orbital Dynamics
Astrophysics, at its core, is the branch of astronomy that deals with the physical nature of stars and other celestial bodies, and the application of the laws and theories of physics to understand astronomical observations. When we look at stars orbiting around massive objects like Sagittarius A*, a supermassive black hole at the center of our galaxy, we examine the astrophysical processes governing their motion.

In the context of the exercise, understanding the motion of a star orbiting Sagittarius A* requires the analysis of its orbital speed and radius. This information helps us calculate the star's orbital period, which is the time it takes to complete one full orbit around the black hole. Applying equations derived from physics, astrophysicists can determine vital statistics about celestial objects, such as their mass and the nature of their orbit, revealing the incredible forces and dynamics at play in the cosmos.
Gravitational Forces in Space
Gravitational forces are the invisible 'hands' that shape the cosmos. According to Newton's law of universal gravitation, every mass attracts every other mass with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This fundamental principle is critical for calculating the orbital period and the masses of celestial objects like stars and black holes.

The force of gravity keeps the star in a stable orbit around Sagittarius A*, preventing it from flying off into space. By equating the centripetal force needed to keep the star in a circular orbit with the gravitational force exerted by Sagittarius A*, we can unravel the mass of this colossal black hole. Understanding these forces not only allows for calculations concerning individual systems but also helps explain larger scale structures like galaxies and clusters, deepening our knowledge of the universe.
Celestial Mechanics and Orbital Period
Celestial mechanics is the study of the motions of celestial bodies under the influence of gravitational forces. It is a subfield of astrophysics that provides the mathematical underpinning to predict celestial events and phenomena. A key aspect of celestial mechanics is the calculation of an object's orbital period.

The orbital period of a celestial body like a star is determined by the properties of its orbit, which includes its velocity and the distance from the object it is orbiting. In this exercise, the circular orbit assumption simplifies the calculations, as it means the speed of the star remains constant, and the radius of its orbit is fixed. Utilizing these parameters and the formula for circular motion, we can determine the time it takes for the star to orbit Sagittarius A*, providing essential insights into the dynamics of our galaxy's central region.

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Most popular questions from this chapter

Use the Starry Night Enthusiast \(\mathrm{TM}^{\mathrm{M}}\) program to observe the Milky Way. (a) Display the entire celestial sphere by selecting Favourites \(>\) Guides \(>\) Atlas. Select View \(>\) Stars \(>\) Milky Way to display this galaxy. Select Options \(>\) Stars \(>\) Milky Way, move the Brightness slide-bar to the far right to brighten the Milky Way and click OK. In the View menu, ensure that the Scrollbars are activated and use them to look at different parts of the Milky Way. Can you identify the direction toward the galactic nucleus? In this direction the Milky Way appears broadest. Open the Find pane, enter Sagittarius in the Query box and press Enter to center on this constellation to check your identification. (b) Use this full-sky view to determine the orientation of the plane of the Galaxy with respect to the celestial sphere. Move the vertical scrollbar to its central position to display the Celestial Equator as a horizontal line across the lower part of the view. Move the horizontal scrollbar until the Milky Way is centered upon the view. Estimate the angle between the Milky Way and the celestial equator on the screen. How well aligned is the plane of the Milky Way with the plane of the Earth's equator? (c) A third plane of interest is that of the ecliptic, which is shown as a green line. Use the scrollbars to adjust the view so that the ecliptic appears as a straight line rather than as a curve, thereby ensuring that you are viewing in a direction that lies in the ecliptic plane. Use the horizontal scrollbar to move the view to where you can see where the ecliptic crosses the Milky Way. Estimate the angle between the Milky Way and the ecliptic on the screen. How well aligned is the plane of the Milky Way to the ecliptic, the plane of the Earth's orbit around the Sun? (d) Click on Home in the toolbar to return to your home view, stop Time Flow and set the local time to midnight (12:00:00 A.M.). Select Options > Stars > Milky Way, move the Brightness slide-bar to the far right to brighten the Milky Way and click OK. Adjust the date to January 1, then February 1, and so on. In which month is the galactic nucleus highest in the sky at midnight, so that it is most easily seen from your location?

In our Galaxy, why are stars of spectral classes \(\mathrm{O}\) and \(\mathrm{B}\) only found in or near the spiral arms? Is the same true for stars of other spectral classes? Explain why or why not.

Do density waves form a stationary pattern in a galaxy? If not, do they move more rapidly, less rapidly, or at the same speed as stars in the disk?

Show that the form of Kepler's third law stated in Box 23-2, \(P^{2}=4 \pi^{2} a^{3} / G\left(M+\mathrm{M}_{\odot}\right)\), is equivalent to \(M=r v^{2} / G\), provided the orbit is a circle. (Hint: The mass of the Sun (M \(\mathrm{M}_{\odot}\) ) is much less than the mass of the Galaxy inside the Sun's orbit (M).)

(a) Calculate the Schwarzschild radius of a supermassive black hole of mass \(3.7 \times 10^{6} \mathrm{M}_{\odot}\), the estimated mass of the black hole at the galactic center. Give your answer in both kilometers and astronomical units. (b) What is the angular diameter of such a black hole as seen at a distance of \(8 \mathrm{kpc}\), the distance from the Earth to the galactic center? Give your answer in arcseconds. Observing an object with such a small angular size will be a challenge indeed! (c) What is the angular diameter of such a black hole as seen from a distance of \(45 \mathrm{AU}\), the closest that the star SO-16 comes to Sagittarius \(A * ?\) Again, give your answer in arcseconds. Would it be discernible to the naked eye at that distance? (A normal human eye can see details as small as about 60 arcseconds.)
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