Loose Leaf For Explorations: Introduction To Astronomy
9th Edition
ISBN: 9781260432145
Author: Thomas T Arny, Stephen E Schneider Professor
Publisher: McGraw-Hill Education
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Chapter 17, Problem 6P
To determine
Whether it is possible to directly image the black hole as a dark spot in the accretion disk and compare the results with
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A light of wavelength 620 nm is emitted from the following four places. What wavelength is observed for this light by an observer a long distance away? (The objects are not moving with respect to the observer)
The surface of a 0.84 solar mass white dwarf that has a radius of 708000 km:
?
The surface of a 2.52 solar mass neutron star that has a radius of 14.2 km:
2 Schwarzschild radii from a 20 solar mass black hole:
?
1.048 Schwarzschild radii from a 20 solar mass black hole:
?
Problem 2: Black hole – the ultimate blackbody
A black hole emits blackbody radiation called Hawking radiation. A black hole with mass
M has a total energy of Mc², a surface area of 167G²M² /c*, and a temperature of
hc³/167²KGM.
a) Estimate the typical wavelength of the Hawking radiation emitted by a 1 solar
mass black hole (2 × 103ºkg). Compare your answer to the size of the black hole.
b) Calculate the total power radiated by a one-solar mass black hole.
c) Imagine a black hole in empty space, where it emits radiation but absorbs nothing.
As it loses energy, its mass must decrease; one could say "evaporates". Derive a
differential equation for the mass as a function of time, and solve to obtain an
expression for the lifetime of a black hole in terms of its mass.
What is the orbital period (in s) of a bit of matter in an accretion disk that is located 6 ✕ 105 km from a 99 M black hole?
Hint:
Use the circular orbit velocity formula,
Vc =
GM
r
.
s
Chapter 17 Solutions
Loose Leaf For Explorations: Introduction To Astronomy
Ch. 17 - Prob. 1QFRCh. 17 - Prob. 2QFRCh. 17 - Prob. 3QFRCh. 17 - Prob. 4QFRCh. 17 - (17.3) Why are galaxy collisions of interest?Ch. 17 - Prob. 6QFRCh. 17 - Prob. 7QFRCh. 17 - Prob. 8QFRCh. 17 - Prob. 9QFRCh. 17 - Prob. 10QFR
Ch. 17 - Prob. 11QFRCh. 17 - Prob. 12QFRCh. 17 - Prob. 13QFRCh. 17 - Prob. 14QFRCh. 17 - Prob. 15QFRCh. 17 - Prob. 16QFRCh. 17 - Prob. 17QFRCh. 17 - Prob. 18QFRCh. 17 - Prob. 19QFRCh. 17 - Prob. 20QFRCh. 17 - Prob. 21QFRCh. 17 - Prob. 22QFRCh. 17 - Prob. 1TQCh. 17 - Prob. 2TQCh. 17 - Prob. 3TQCh. 17 - Prob. 4TQCh. 17 - Prob. 5TQCh. 17 - Prob. 6TQCh. 17 - Prob. 1PCh. 17 - Prob. 2PCh. 17 - Prob. 3PCh. 17 - Prob. 4PCh. 17 - Prob. 5PCh. 17 - Prob. 6PCh. 17 - Prob. 7PCh. 17 - Prob. 8PCh. 17 - Prob. 9PCh. 17 - Prob. 10PCh. 17 - Prob. 1TYCh. 17 - Prob. 2TYCh. 17 - Prob. 3TYCh. 17 - Prob. 4TYCh. 17 - Prob. 5TYCh. 17 - Prob. 6TYCh. 17 - Prob. 7TYCh. 17 - Prob. 8TYCh. 17 - Prob. 9TYCh. 17 - Prob. 10TY
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- The Plummer sphere is a model for some star clusters, and it has the gravitational potential as given in the provided image. M and rp are both constant. What is the density profile ρ(r)? When the Plummer sphere is observed from a large distance along the z axis, what is its surface density, Σ(R), where R is the distance from the center?arrow_forward(a) Calculate the photon capture radius and the Schwarzschild radius of M87* (in AU). (b) Why was it not possible for previous telescopes to take such a picture of the black hole? (c) Describe the components and functionality of the event horizon telescope. (d) Explain the two algorithms used to reconstruct the image from the telescope data. (e) What parameters were required for the GRMHD simulations to generate an image? (f) Explain the physical origins of the features (g) How can the image resolution be increased in future observations? (central dark region, ring, shadow).arrow_forwardWhat is the Schwarzschild radius (in km) of a 6Msun black hole? What fraction of the Earth's radius is this? What percent of the speed of light (2.998 x 108 m/s) is the escape velocity at the Schwarzschild radius? Part 1 of 3 The Schwarzschild radius of a black hole is given by: 2GM Rs = c2 so for the given mass, 2G(6)(Msun) Rs c2 where M. Sun = 1.99 x 1030 kg. Then convert this into kilometers using 1 km = 1,000 m. Rs kmarrow_forward
- Calculate the Schwarzschild radius for a uranium-238 nucleus. (r = 7x10-¹5 m).arrow_forwardHow can we calculate Schwarzschild radius?arrow_forwarda double line The velocity curve for spectroscopic binary is shown in the sketch. The system is viewed edge-on, i.e., with an inclination angle of i = 90°, so that the maximum possible Doppler shifts for this system are observed. 400 U, Aini 300 200 oo - v Ain i 100 -100 -200 -300 400 o 1 2 3 +s 7 8 10 Time (daye) Find the orbital period of this binary in days. Doppler Velocity (2esun)arrow_forward
- Calculate the Schwarzschild radius of the Sun.arrow_forwardFor the Galactic Center black hole (Sag. A*), which has mass ~4.1 million solar masses, find the maximum luminosity that could be emitted via spherical accretion of ionized gas (i.e., the Eddington limit LEdd). Express your answer as a multiple of the Qmlaxy's total luminous output, = 2.5 x 1010 L. LMW (i.e., enter the ratio LEdd/LMw).arrow_forwardQ1 (oints): A0620-00 is an X-ray binary system, there is a normal star and a compact object. For the normal star the radial orbital velocity is 457 km s and for the compact object it is 43 km s1.They have an orbital period of 0.3226 day. Calculate the mass function.arrow_forward
- What is the orbital period of a bit of matter in an accretion disk that is located 4 ✕ 105 km from a 87 M black hole? Use the circular orbit velocity formula.arrow_forwardMany galaxies appear to have supermassive black holes in their centers powering active galactic nuclei (also called AGN). The Schwarzschild radius of these supermassive black holes can be estimated in part by watching for changes in the brightness of the surrounding AGN and measuring the timescale of those changes. Assume we observe an AGN and determine it varies with a timescale of 9.85 minutes, which implies a Schwarzschild radius on the order of 1.77x1011 meters. Estimate the mass of this supermassive black hole. kgarrow_forwardIn the mass function f(Ma, M) = MB/(1+ Ma/MB)² sin°i = M/(Ma + Mb) sin°i it gives us a lower limit on the mass of the unseen component of a single-lined spectroscopic binary. Why is the limit is on the unseen component rather than the component that we see?arrow_forward
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