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Chapter 13: Problem 40

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting around once every 27 hours and moving at 30,000 km/s. (a) How far are these clumps from the center of the black hole? (b) What is the mass of this black hole, assuming circular orbits? Express your answer in kilograms and as a multiple of our sun's mass. (c) What is the radius of its event horizon?

Short Answer

Expert verified
(a) \( r \approx 5.5 \times 10^{10} \) meters; (b) \( M \approx 1.2 \times 10^{34} \) kg, \( 6000 M_{\odot} \); (c) \( R_s \approx 1.8 \times 10^{7} \) meters.

Step by step solution

01

Define Given Parameters

We have the orbital period \( T = 27 \text{ hours} \ = 27 \times 3600 \text{ seconds} \), and the orbital speed \( v = 30,000 \text{ km/s} \ = 3 \times 10^7 \text{ m/s} \). We are tasked to find the orbit radius \( r \) from the black hole's center, the black hole's mass \( M \), and the radius of its event horizon.
02

Calculate the Orbital Radius

For an object in orbit, the centripetal force \( F_c = \frac{mv^2}{r} \) is provided by the gravitational force \( F_g = \frac{GmM}{r^2} \), where \( G \) is the gravitational constant, \( 6.674 \times 10^{-11} \text{ m}^3\text{kg}^{-1}\text{s}^{-2} \). Since \( F_c = F_g \), we have:\[\frac{mv^2}{r} = \frac{GmM}{r^2}\]Rearranging, \( r = \frac{GM}{v^2} \).Using the relation between period and radius for circular orbits \( v = \frac{2\pi r}{T} \), we rearrange to solve for \( r \):\[ r = \left(\frac{G M T^2}{4\pi^2} \right)^{1/3} \]Plug values in \( r = \left(\frac{6.674 \times 10^{-11} \times M \times (27 \times 3600)^2}{4\pi^2} \right)^{1/3} \), using trial and error or solving both equations to find \( r \approx 5.5 \times 10^{10} \) meters.
03

Calculate the Mass of the Black Hole

Calculate \( M \) using the formula derived from Step 2:\[ M = \frac{r v^2}{G} \]With \( r = 5.5 \times 10^{10} \) meters and \( v = 3 \times 10^7 \text{ m/s} \),\[ M = \frac{5.5 \times 10^{10} \times (3 \times 10^7)^2}{6.674 \times 10^{-11}} \]Calculate \( M \approx 1.2 \times 10^{34} \) kilograms. To express as a multiple of the Sun's mass \( M_{\odot} = 1.989 \times 10^{30} \text{ kg} \):\[ \frac{1.2 \times 10^{34}}{1.989 \times 10^{30}} \approx 6000 M_{\odot} \].
04

Calculate the Radius of the Event Horizon

The radius of the event horizon, or Schwarzschild radius \( R_s \), is given by\[ R_s = \frac{2GM}{c^2} \]Where \( c = 3 \times 10^8 \text{ m/s} \) is the speed of light.Plug in \( M = 1.2 \times 10^{34} \text{ kg} \):\[ R_s = \frac{2 \times 6.674 \times 10^{-11} \times 1.2 \times 10^{34}}{(3 \times 10^8)^2} \]Calculate \( R_s \approx 1.8 \times 10^{7} \) meters.

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Key Concepts

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

Orbital Mechanics
Orbital mechanics is the study of how objects move in space due to various forces, primarily gravity. In the context of this exercise, it helps us understand how chunks of matter orbit around a black hole.
These clumps are following paths determined by the interplay of their speed and gravitational pull from the black hole.
Some important points on orbital mechanics include:
  • Orbital Period: The time taken for one complete orbit around a black hole, which in this scenario is 27 hours.
  • Orbital Speed: This indicates how fast the matter is moving, given as 30,000 km/s.
  • Radius of Orbit: Again, it is the distance from the centre of the black hole to the path of the orbiting matter.
The speed and period of the orbit relate to one another, providing the basis for calculating the radius of the orbit around the black hole. Using these principles allows astronomers to assess and predict the behavior of celestial objects in space.
Gravitational Force
Gravitational force is the attractive force between two masses. It plays a crucial role in black hole physics and orbital mechanics.
This force keeps the matter in orbit around the black hole, creating the gravitational pull needed for circular motion.
The formula for gravitational force (F_g) is given by:
\[ F_g = \frac{GMm}{r^2} \]
Where:
  • \( G \) is the gravitational constant, \(6.674 \times 10^{-11} \text{ m}^3\text{kg}^{-1}\text{s}^{-2} \).
  • \( M \) is the mass of the black hole.
  • \( m \) is the mass of the orbiting matter.
  • \( r \) is the distance between the center of the black hole and the orbiting object.
Gravitational forces are responsible for keeping celestial bodies in their respective orbits. They are also essential in calculating the mass of cosmic entities like black holes using observations and measurements from their effects on nearby objects.
Centripetal Force
The centripetal force is what keeps an object moving in a circular path. In the case of objects orbiting a black hole, this force is provided by gravity.
Without this force, the objects would move in straight lines off into space.
The formula for centripetal force (F_c) is:\[ F_c = \frac{mv^2}{r} \]
Where:
  • \( m \) is the mass of the orbiting object.
  • \( v \) is the orbital speed.
  • \( r \) is the radius of the orbit.
In a stable orbit around a black hole, the centripetal force is balanced by the gravitational force:\[ \frac{mv^2}{r} = \frac{GmM}{r^2} \]
This equation allows us to solve for various unknowns such as the mass of the black hole or the radius of the orbit by setting them equal to each other. Understanding centripetal force helps us make sense of why objects remain in orbit and do not drift off course.
Schwarzschild Radius
The Schwarzschild radius represents the size of a black hole's event horizon. It is the boundary beyond which nothing, not even light, can escape the black hole's gravitational pull.
Essentially, it denotes how "large" a black hole appears to be from the outside.
The formula for the Schwarzschild radius (R_s) is:
\[ R_s = \frac{2GM}{c^2} \]
Where:
  • \( G \) is the gravitational constant.
  • \( M \) is the mass of the black hole.
  • \( c \) is the speed of light, \(3 \times 10^8 \text{ m/s} \).
For the black hole in galaxy Markarian 766, with a mass approximately 6000 times that of our sun, the Schwarzschild radius is a crucial component in understanding its powerful gravitational field. This radius indicates the "point of no return" for matter and light approaching the black hole.

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Most popular questions from this chapter

An astronaut, whose mission is to go where no one has gone before, lands on a spherical planet in a distant galaxy. As she stands on the surface of the planet, she releases a small rock from rest and finds that it takes the rock 0.480 s to fall 1.90 m. If the radius of the planet is 8.60 \(\times\) 10\(^7\) m, what is the mass of the planet?

Two stars, with masses \({M_1}\) and \({M_2}\), are in circular orbits around their center of mass. The star with mass \({M_1}\) has an orbit of radius \({R_1}\); the star with mass \({M_2}\) has an orbit of radius \({R_2}\). (a) Show that the ratio of the orbital radii of the two stars equals the reciprocal of the ratio of their masses\(-\)that is, \({R_1}\)/\({R_2}\) \(=\) \({M_2}\)/\({M_1}\). (b) Explain why the two stars have the same orbital period, and show that the period \(T\) is given by \(T = 2\pi\)(R1 + R2)\(^{3/2}\)/\(\sqrt{G(M1 + M2)}\). (c) The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36.0 km/s. The second star, Beta, has an orbital speed of 12.0 km/s. The orbital period is 137 d. What are the masses of each of the two stars? (d) One of the best candidates for a black hole is found in the binary system called A0620-0090. The two objects in the binary system are an orange star, V616 Monocerotis, and a compact object believed to be a black hole (see Fig. 13.28). The orbital period of A0620-0090 is 7.75 hours, the mass of V616 Monocerotis is estimated to be 0.67 times the mass of the sun, and the mass of the black hole is estimated to be 3.8 times the mass of the sun. Assuming that the orbits are circular, find the radius of each object's orbit and the orbital speed of each object. Compare these answers to the orbital radius and orbital speed of the earth in its orbit around the sun.

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject material as high as 500 km (or even higher) above the surface. Io has a mass of 8.93 \(\times\) 10\(^{22}\) kg and a radius of 1821 km. For this calculation, ignore any variation in gravity over the 500-km range of the debris. How high would this material go on earth if it were ejected with the same speed as on Io?

The dwarf planet Pluto has an elliptical orbit with a semimajor axis of 5.91 \(\times\) 10\(^{12}\) m and eccentricity 0.249. (a) Calculate Pluto's orbital period. Express your answer in seconds and in earth years. (b) During Pluto's orbit around the sun, what are its closest and farthest distances from the sun?

Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is 400 km above the earth's surface; at the high point, or apogee, it is 4000 km above the earth's surface. (a) What is the period of the spacecraft's orbit? (b) Using conservation of angular momentum, find the ratio of the spacecraft's speed at perigee to its speed at apogee. (c) Using conservation of energy, find the speed at perigee and the speed at apogee. (d) It is necessary to have the spacecraft escape from the earth completely. If the spacecraft's rockets are fired at perigee, by how much would the speed have to be increased to achieve this? What if the rockets were fired at apogee? Which point in the orbit is more efficient to use?
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