Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 39, Problem 63PQ
To determine
The Schwarzschild radius of the black hole at the centre of the Milky Way galaxy.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Please mention all theory parts.
Suppose we find an Earth-like planet around one of our nearest stellar neighbors, Alpha Centauri (located only 4.4 light-years away). If we launched a "generation ship" at a constant speed of 1500.00 km/s from Earth with a group of people whose descendants will explore and colonize this planet, how many years before the generation ship reached Alpha Centauri? (Note there are 9.46 ××1012 km in a light-year and 31.6 million seconds in a year.
Convert 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.
Chapter 39 Solutions
Physics for Scientists and Engineers: Foundations and Connections
Ch. 39.1 - Which of the following are (approximately)...Ch. 39.2 - Suppose the primed and laboratory observers want...Ch. 39.7 - Prob. 39.3CECh. 39.10 - Prob. 39.4CECh. 39.12 - Prob. 39.5CECh. 39 - Prob. 1PQCh. 39 - Prob. 2PQCh. 39 - Prob. 3PQCh. 39 - In an airport terminal, there are two fast-moving...Ch. 39 - Prob. 5PQ
Ch. 39 - Prob. 6PQCh. 39 - Prob. 7PQCh. 39 - Prob. 8PQCh. 39 - Prob. 9PQCh. 39 - Prob. 10PQCh. 39 - Prob. 11PQCh. 39 - Prob. 12PQCh. 39 - Prob. 13PQCh. 39 - Prob. 14PQCh. 39 - Prob. 15PQCh. 39 - Prob. 16PQCh. 39 - Prob. 17PQCh. 39 - Prob. 18PQCh. 39 - Prob. 19PQCh. 39 - Prob. 20PQCh. 39 - Prob. 21PQCh. 39 - Prob. 22PQCh. 39 - Prob. 23PQCh. 39 - A starship is 1025 ly from the Earth when measured...Ch. 39 - A starship is 1025 ly from the Earth when measured...Ch. 39 - Prob. 26PQCh. 39 - Prob. 27PQCh. 39 - Prob. 28PQCh. 39 - Prob. 29PQCh. 39 - Prob. 30PQCh. 39 - Prob. 31PQCh. 39 - Prob. 32PQCh. 39 - Prob. 33PQCh. 39 - Prob. 34PQCh. 39 - Prob. 35PQCh. 39 - Prob. 36PQCh. 39 - Prob. 37PQCh. 39 - Prob. 38PQCh. 39 - As measured in a laboratory reference frame, a...Ch. 39 - Prob. 40PQCh. 39 - Prob. 41PQCh. 39 - Prob. 42PQCh. 39 - Prob. 43PQCh. 39 - Prob. 44PQCh. 39 - Prob. 45PQCh. 39 - Prob. 46PQCh. 39 - Prob. 47PQCh. 39 - Prob. 48PQCh. 39 - Prob. 49PQCh. 39 - Prob. 50PQCh. 39 - Prob. 51PQCh. 39 - Prob. 52PQCh. 39 - Prob. 53PQCh. 39 - Prob. 54PQCh. 39 - Prob. 55PQCh. 39 - Prob. 56PQCh. 39 - Consider an electron moving with speed 0.980c. a....Ch. 39 - Prob. 58PQCh. 39 - Prob. 59PQCh. 39 - Prob. 60PQCh. 39 - Prob. 61PQCh. 39 - Prob. 62PQCh. 39 - Prob. 63PQCh. 39 - Prob. 64PQCh. 39 - Prob. 65PQCh. 39 - Prob. 66PQCh. 39 - Prob. 67PQCh. 39 - Prob. 68PQCh. 39 - Prob. 69PQCh. 39 - Prob. 70PQCh. 39 - Joe and Moe are twins. In the laboratory frame at...Ch. 39 - Prob. 72PQCh. 39 - Prob. 73PQCh. 39 - Prob. 74PQCh. 39 - Prob. 75PQCh. 39 - Prob. 76PQCh. 39 - Prob. 77PQCh. 39 - In December 2012, researchers announced the...Ch. 39 - Prob. 79PQCh. 39 - Prob. 80PQCh. 39 - How much work is required to increase the speed of...Ch. 39 - Prob. 82PQCh. 39 - Prob. 83PQCh. 39 - Prob. 84PQCh. 39 - Prob. 85PQ
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- What would be the Schwarzschild radius, in light years, if our Milky Way galaxy of 100 billion stars collapsed into a black hole? Compare this to our distance from the center, about 13,000 light years.arrow_forwardconsider plutoz diameter and mass. (2374)km & (1.303E22kg) and day which js 6.4 dayz long. FIND: 1. please elaborate how would you get the answer to the escappe vel0city from plut0. 2. we would need to find the minimum energy required for an aircraft or ship of some sort with mass (525kg) to escape this planet.. 3. we would also need to find the t0tal energy for a complete orbit around the planet with an airship with a same mass (525) and an altitude of 224 kmarrow_forward1. Consider our Sun - it is in orbit around the center of our Milky Way Galaxy. The velocity of the Sun in its orbit is about 250 km/s. The distance to the center of the galaxy is about 9.1 kpc (kiloparsecs). We can use Kepler's third law to calculate the mass of the galaxy interior to the Sun's orbit. We assume that the orbit is circular so that the semimajor axis is just the radius of the circular orbit = 9.1 kpc. First we need to calculate the number of AU's in 9.1 kpc. (Note that 1 Крс - 1000 рс - 3260 1t yrs and 1 pc - 206,265 AU.) %3D a =r =9.1kpc = (9.1kpc) 1000 pc 206,265AU] 1kpc AU Sun 1pcarrow_forward
- consider plutos diameter and mass. (2374)km & (1.303E22kg) and day which is 6.4 dayz long. FIND: 1. please elaborate how would you get the answer to the escappe vel0city from plut0. 2. we would need to find the minimum energy required for an aircraft or ship of some sort with mass (525kg) to escape this planet.. 3. we would also need to find the t0tal energy for a complete orbit around the planet with an airship with a same mass (525) and an altitude of 224 kmarrow_forwardThe supermassive black hole at the center or our galaxy (Sagittarius A*) has a mass equal to 4.3 million Suns (the mass of the Sun is 1.99 × 1030 kg). The distance from Sgr A* to Earth is 7,940 parsecs, where one parsec is equal to 3.09 × 1016 m. (a) What is the gravitational force that Sgr A* exerts on a 75 kg person on Earth, in units of Newtons? Hint: use Newton's law of universal gravitation. (b) Suppose the same person is sitting 1.0 meter away from a paperclip with a mass of 1.0 grams. What is the gravitational force that the paperclip exerts on the person? (c) Compare the forces from parts (a) and (b). Which is greater?arrow_forwardAstronomical observations of our milky way galaxy indicate that it has a mass of about 8x1011 solar masses. A star orbiting near the galaxy's periphery is 5.6x104 light years from its center. a.) What should be the orbital period (in years) of that star be? b.) If its period is 6.4x107 years instead, what is the mass (in solar masses) of the galaxy? Such calculations are used to imply the existence of "dark matter" in the universe and have indicated, for example, the existence of very massive black holes at the center of some galaxies.arrow_forward
- help asaparrow_forwardA 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”)arrow_forwardThe tidal force is a differential force: dF/dr, which is the difference between the force at one distance and the force at another. If dr happens to correspond to, say, your height, then the tidal force is the difference in gravity felt by your head and by your feet. This amounts to a stretching force, since one end is pulled harder than the other end. Calculate the tidal force experienced by your body at the surface of a neutron star. Assume that the neutron star has a mass of 1.5 solar masses and a radius of 10 km. Assume that your mass is 100 kg and that your dr (height) is 2 m when standing and 0.5 m when prone. What is the tidal force when you are standing? What is the tidal force when you are prone? Based on the above, what do you recommend for minimizing the tidal force?arrow_forward
- Consider the attached light curve for a transiting planet observed by the Kepler mission. If the host star is identical to the sun, what is the radius of this planet? Give your answer in terms of the radius of Jupiter. Brightness of Star Residual Flux 0.99 0.98 0.97 0.006 0.002 0.000 -8-881 -0.06 -0.04 -0.02 0.00 Time (days) → 0.02 0.04 0.06arrow_forwardThe Schwarzschild radius is the distance from an object at which the escape velocity is equal to the speed of light. A black hole is an object that is smaller than its Schwarzschild radius, so not even light itself can escape a black hole. The Schwarzschild radius r depends on the mass m of the black hole according to the equation (See image.) where G = 6.673 × 10-11 (Nm2)/(kg2) is the gravitational constant and c = 2.998 × 108 m/s is the speed of light. 1. Consider a black hole with a mass of 3.70 × 107M.. Use the given equation to find the Schwarzschild radius for this black hole. Remember that 1 M = 1.989 × 1030 kg and 1 N = 1 kg * m/s2 2. What is this radius in units of the solar radius? Remember that 1 R = 6.955 × 108 m.arrow_forwardA non-rotating black hole has an ‘edge’ at what’s called it’s Schwartzschildradius. For a black hole of mass M, the Schwartzschild radius isRSch = 2GM/(c^2) where G is the gravitational constant and c is the speed of light. Close to black holes, we really should use Einstein’s theory of gravity (General Relativity) instead of Newton’s, but Newton’s is still a good approximation.Using Newtonian Gravity, find the gravitational force on a mass m at theSchwartzschild radius of a black hole. (Your answer for this should look likeFgrav = an expression in terms of G, M, m, c and numbers.) Is the forcesmaller or larger for a more massive black hole?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
AstronomyPhysicsISBN:9781938168284Author:Andrew Fraknoi; David Morrison; Sidney C. WolffPublisher:OpenStax
University Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice University
Astronomy
Physics
ISBN:9781938168284
Author:Andrew Fraknoi; David Morrison; Sidney C. Wolff
Publisher:OpenStax
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Kepler's Three Laws Explained; Author: PhysicsHigh;https://www.youtube.com/watch?v=kyR6EO_RMKE;License: Standard YouTube License, CC-BY