Lab 9 - Mike Jacobs

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Collin County Community College District *

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1403

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Astronomy

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Dec 6, 2023

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pdf

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1 PHYS 1403 Lab 9: BLACK HOLES Worksheet Name: _________________________________________________________ CWID: _________________________________________________________ INTRODUCTION Although the lifetimes of stars are very long, all stars will eventually die. There are three possible end results of a star. If the star is smaller than approximately 1.4 solar masses, the star will shed its outer layers to form a shell of gas and dust called a planetary nebula. This will leave the star's core behind as an extremely dense, extremely hot remnant, called a white dwarf. White dwarfs are about the size of the Earth, and a tablespoon of white dwarf matter would weigh several tens of thousands of tons, about as much as an ocean liner. If the star is larger than 1.4 solar masses, but smaller than about 3 solar masses, it will go through a series of contractions and expansions. The death of the star will occur when the nuclear fires in the core no longer generate enough heat to balance the gravitational attractive forces of the star's mass, leading to a collapse of the mass into a super-dense, compact core. During this final collapse an enormous amount of energy is stored within the star's nuclear structure. When this stored energy is high enough to stop the collapse of the star, it will suddenly be released in a spectacular supernova explosion brighter than an entire galaxy. A super-dense core, called a neutron star, will remain where the star once was. Although called stars, neutron stars are not normal stars because there is no fusion taking place within them anymore. Neutron stars have a radius of a few tens of kilometers and a tablespoon of neutron star matter would weigh billions of tons. Initially, both white dwarfs and neutron stars are very bright, due to the extreme surface temperature. But they will gradually cool down over a period of millions of years until they become cold, dark embers. It is speculated that there are countless numbers of these objects traveling through space, too dark for us to observe them. If the star is larger than approximately three solar masses, it will go through the same contraction and expansion as the star that went supernova. But when it goes through the final collapse, the gravitational forces are so strong that they overcome the energy stored in the nuclear structure. This star will continue to collapse upon itself until the entire mass of the star is concentrated at a single point, called a singularity. It has
2 now become so dense that nothing, not even light, will be able to escape from it. Since anything caught in its gravitational field cannot ever hope to escape, it can be thought of as a 'hole' in space. And since no light is emitted by it, it is a 'black hole'. This term was first used by the American physicist John Archibald Wheeler in the 1960's. Let's see how it is that not even light can escape from a black hole. An object on a planet's surface has a potential energy due to the planet's gravity. This potential energy can be found with the equation R GMm U = where M is the mass of the planet, m is the mass of the object, and R is the radius of the planet. G (= 6.67 x 10 -11 Nm 2 /kg 2 ) is the universal constant for gravity discovered by Sir Isaac Newton. If we were to fire this object upward from a planet's surface and make it a projectile, it would rise up with some starting velocity, but would immediately begin to slow down until it momentarily came to rest in mid-air, and then return to the surface. This is something you've experienced before whenever you have thrown a ball into the air. As you know, if we make the starting velocity larger, that is, if we throw the ball harder, the projectile will go higher in the air. The larger we make the starting velocity, the higher the projectile will go, until eventually, we can make the starting velocity so large the projectile will go up and never fall down. In other words, we have now placed our object in orbit. If we were to give it a little more energy it would then be able to leave the planet's gravitational field and never return. This velocity is called the 'escape velocity'. The energy from the projectile’s velocity, its kinetic energy found with the equation 2 2 1 mv K = which must be large enough to overcome the potential energy. So, we have R GMm mv esc > 2 2 1
3 Solving to find the escape velocity gives us (1) R GM v esc 2 > Note that the escape velocity does not depend on the mass of the projectile, but it does depend on one over the radius of the planet. This means that as the planet gets smaller, with its mass staying the same, the escape velocity will get larger. Well, we saw that neutron stars and black holes are the result of stars collapsing. This means that the mass has remained large, but the radius has decreased. By our equation (1) above, we would expect the escape velocity of these objects to increase as they collapse. A black hole is a collapsed star whose v esc is greater than the speed of light, which is equal to 3 x 10 8 m/s, and which we denote with the symbol ' c '. Using equation (1), and substituting the value of the speed of light for v esc , we can then solve for the radius that would be required to have an escape velocity equal to the speed of light. This result is (2) 2 2 c MG R SCH = What we have found here is the minimum radius that a given mass must compress to make its escape velocity equal to the speed of light. This radius was first solved for in 1916 by the German physicist, Karl Schwartzschild, and is known as the Schwartzschild radius. As a collapsing star passes its Schwartzschild radius, it becomes a black hole. This is the radius a black hole would have if it were not rotating. The core would continue to collapse further, but we would not be able to see anything within the black hole. The Schwartzschild radius marks what is known as the 'event horizon', beyond which we can have no knowledge of events that may occur. In other words, if we were close enough to see a black hole, what we would see is the event horizon. The actual star, and anything else that had passed through the event horizon, would be inside the event horizon and will never be seen again. The star remnant becomes a black hole when it reaches this point because of two principles described by Einstein in his Theory of Relativity, (1) nothing can go faster than the speed of light; and (2) light is attracted by gravity. This explains why we call them 'black holes'. The gravitational field is so strong that nothing, not even light, can reach the escape velocity. The light is still emitted by the singularity within, but its trajectory is bent until it falls back onto the singularity without ever passing through the event horizon.
4 It is important to note that the Schwartzschild radius is valid for a non-rotating star only. Since we believe all stars rotate, their singularities would also rotate and the true radius of the event horizon would be different. However, this calculation was so complex that it defied all attempts to conduct it. Finally, in 1963, the Australian mathematician Roy P. Kerr was conducting some calculations on another aspect of relativity when he realized he had calculated the radius of a rotating black hole. Such a black hole is now called a Kerr black hole. However, we will concern ourselves with Schwartzschild black holes only. Next, imagine an object emitting light photons as it approaches a black hole. As this object gets closer and closer to the black hole, but is still outside the event horizon. As it gets continues to get closer, the trajectories of the light photons are bent more and more and more and more of the photons would enter the event horizon. Some of the photons would be radiated away at angles that would allow them to escape from the black hole, even with their trajectories bent towards it. However, some would be radiated at just the right orbit to neither escape, nor pass into the event horizon. These photons would go into orbit around the black hole and form what is called the 'photon sphere'. We can find the radius of the photon sphere via relativity to obtain: (3) 2 3 c MG R PS = This means that a black hole has a sphere of photons orbiting it at this radius and anything approaching a black hole must pass through this sphere before entering the event horizon.
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