Universe: Stars And Galaxies
6th Edition
ISBN: 9781319115098
Author: Roger Freedman, Robert Geller, William J. Kaufmann
Publisher: W. H. Freeman
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Question
Chapter 17, Problem 69Q
To determine
(a)
The sum of two stars in a binary Steller system.
To determine
(b)
The mass of each individual star.
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Students have asked these similar questions
A certain binary system consists of two stars that have equal masses and revolve in circular orbits around a fixed point half-way between them.
If the orbital velocity of each star is v=186 km/s and the orbital period of each is 11.3 days, calculate the mass M of each star. Give your answer in units of the solar mass, 1.99×1030 kg (e.g. if each planet's mass is 3.98×1030 kg, you would answer "2.00").
A planet (in another galaxy) takes 5 000 Earth days to complete one full revolution around its own star (not the Sun). It is exactly as far away from its star as Earth is to its own Sun. Draw a FBD, then determine how many times more or less massive this star is than our sun (in other words, give a factor of mass, e.g “5x larger” or “5x smaller”)
Problem 4. The mass of the sun is denoted by Mo
1. The star UY Scuti has a mass of 10M. Express the mass of this star in kg.
2 AU
Figure 2.1: Parsec into Km. In the figure B and C are two diametrically opposite positions of the Earth while
it revolves around the sun. The star is located at point A and S is the Sun.
Chapter 17 Solutions
Universe: Stars And Galaxies
Ch. 17 - Prob. 1QCh. 17 - Prob. 2QCh. 17 - Prob. 3QCh. 17 - Prob. 4QCh. 17 - Prob. 5QCh. 17 - Prob. 6QCh. 17 - Prob. 7QCh. 17 - Prob. 8QCh. 17 - Prob. 9QCh. 17 - Prob. 10Q
Ch. 17 - Prob. 11QCh. 17 - Prob. 12QCh. 17 - Prob. 13QCh. 17 - Prob. 14QCh. 17 - Prob. 15QCh. 17 - Prob. 16QCh. 17 - Prob. 17QCh. 17 - Prob. 18QCh. 17 - Prob. 19QCh. 17 - Prob. 20QCh. 17 - Prob. 21QCh. 17 - Prob. 22QCh. 17 - Prob. 23QCh. 17 - Prob. 24QCh. 17 - Prob. 25QCh. 17 - Prob. 26QCh. 17 - Prob. 27QCh. 17 - Prob. 28QCh. 17 - Prob. 29QCh. 17 - Prob. 30QCh. 17 - Prob. 31QCh. 17 - Prob. 32QCh. 17 - Prob. 33QCh. 17 - Prob. 34QCh. 17 - Prob. 35QCh. 17 - Prob. 36QCh. 17 - Prob. 37QCh. 17 - Prob. 38QCh. 17 - Prob. 39QCh. 17 - Prob. 40QCh. 17 - Prob. 41QCh. 17 - Prob. 42QCh. 17 - Prob. 43QCh. 17 - Prob. 44QCh. 17 - Prob. 45QCh. 17 - Prob. 46QCh. 17 - Prob. 47QCh. 17 - Prob. 48QCh. 17 - Prob. 49QCh. 17 - Prob. 50QCh. 17 - Prob. 51QCh. 17 - Prob. 52QCh. 17 - Prob. 53QCh. 17 - Prob. 54QCh. 17 - Prob. 55QCh. 17 - Prob. 56QCh. 17 - Prob. 57QCh. 17 - Prob. 58QCh. 17 - Prob. 59QCh. 17 - Prob. 60QCh. 17 - Prob. 61QCh. 17 - Prob. 62QCh. 17 - Prob. 63QCh. 17 - Prob. 64QCh. 17 - Prob. 65QCh. 17 - Prob. 66QCh. 17 - Prob. 67QCh. 17 - Prob. 68QCh. 17 - Prob. 69QCh. 17 - Prob. 70QCh. 17 - Prob. 71QCh. 17 - Prob. 72QCh. 17 - Prob. 73QCh. 17 - Prob. 74QCh. 17 - Prob. 75QCh. 17 - Prob. 76QCh. 17 - Prob. 77QCh. 17 - Prob. 78QCh. 17 - Prob. 79QCh. 17 - Prob. 80Q
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- Kepler’s third law says that the orbital period (in years) is proportional to the square root of the cube of the mean distance (in AU) from the Sun (Pa1.5) . For mean distances from 0.1 to 32 AU, calculate and plot a curve showing the expected Keplerian period. For each planet in our solar system, look up the mean distance from the Sun in AU and the orbital period in years and overplot these data on the theoretical Keplerian curve.arrow_forwardIn a binary star system, two stars orbit about their common centre of mass, as shown in figure. If 12 = 2r1, what is the ratio of the masses m2/m, of the two stars? CM m2 2arrow_forwardSolve the following problem: Two stars, named A and B, each with a mass equal to the Sun's mass are in orbit around each other. If the distance between the two stars is 1.0 AU. What is the period of their orbit? Describe each step in solving the problem:arrow_forward
- The Algol binary system consists of a 3.7 Msun star and a 0.8 Msun star with an orbital period of 2.87 days. Using Newton’s version of Kepler’s Third Law, calculate the distance, a, between the two stars. Compare that to the size of Betelgeuse (you’ll need to look that up). Newton’s Version of Kepler’s Law: (M1 + M2) P2 = (4p2 /G) a3 Rearrange the equation to solve for a. Pi, p, is equal to 3.14. IMPORTANT NOTE: Google the value of G (the Universal Gravitational Constant) or look it up in your text. NOTICE THE UNITS. You must convert every distance and time in your equation to the same units, otherwise, you’ll get an incorrect answer. That means you must convert distances to meters, solar masses to kilograms, and time to seconds. When you compare your value to the size of Betelgeuse, it will also help that they are in the same units.arrow_forwardConvert 1.39 x 10^9 kilograms to Jupiter Masses, MJ. The mass of Jupiter is known as MJ = 1.898×1027 kg. Mplanet = _________________________ MJ *The accepted mass of this planet HD 209458b is Mplanet = 0.69 MJ. Check your answer for correctness.arrow_forwardUse a distance of R = 1.48x10^11 meters for the distance between the earth and the sun. Use a mass of 1.99x10^30 kg to be 1 solar mass. For each of the different sun masses (as values of solar mass, aka 0.5 solar masses = 1x10^30 kg), as outlined in the lecture, calculate the period of the earth's orbit in days using Kepler's law for circular orbits (I double-checked it with these values and it works) and also calculate the corresponding orbital velocity of the earth. Questions: 1.) Using these values, and 6x10^24 kg for the mass of the earth, what is the strength of the gravitational force between the earth and the sun? 2.) If the earth were twice as far from the sun, what would be its period of orbit? 3.) Mars orbits the sun at a distance of 2.18x10^11 meters. How long is a Martian year, using Kepler's law for circular orbits?arrow_forward
- A planet of mass ?=5.45×10^24 kg is orbiting in a circular path a star of mass ?=4.45×10^29 k . The radius of the orbit is ?=8.35×107^7km. What is the orbital period (in Earth days) of the planet ?planet?arrow_forwardI need help with this hw problem.arrow_forwardConsider a star of mass M₁ that hosts a planet of mass M₂ on a circular orbit with semi- major axis a. The mass of the planet is related to the orbital velocity of the star by the expression G M₂1 Μια Using Kepler's 3rd law show that this expression can be written as V₁ = 2πG (27²) ¹/3 9 M²P where P is the orbital period of the planet around the star. V₁ = M₂arrow_forward
- M6arrow_forwardIn 2004, astronomers reported the discovery of a large Jupiter-sized planet orbiting very close to the star HD179949. The orbit was just 6.4x106 km (about 9 less than the orbit of Mercury) and the planet takes 3.1 days to make one circular orbit. The mass of the star is Answerx_______1030 kg. (Give the number before the exponent.)arrow_forwardTime From this light curve, we can deduce that... O the star has a high mass exoplanet orbiting it O the star has an exoplanet orbiting it that has an eccentric orbit O the star has an exoplanet orbiting it that has an eccentric orbit O the star has an exoplanet that is not on the same orbital plane as the star L Brightnessarrow_forward
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