The escape speed from a star




Think first about the properties of our own sun. Its mass M = 1.99 * 1030 kg and radius R = 6.96 * 108 m are much larger than those of any planet, but compared to other stars, our sun is not exceptionally massive. You can find the sun’s average density r in the same way we found the average density of the earth in Section 13.2:

The escape speed from a star

 

 

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The sun’s temperatures range from 5800 K (about 5500°C or 10,000°F) at the surface up to 1.5 * 107 K (about 2.7 * 107 °F) in the interior, so it surely contains no solids or liquids. Yet gravitational attraction pulls the sun’s gas atoms together until the sun is, on average, 41% denser than water and about 1200 times as dense as the air we breathe

Now think about the escape speed for a body at the surface of the sun. In Example 13.5 (Section 13.3) we found that the escape speed from the surface of a spherical mass M with radius R is v = (2)1/2GM/R. Substituting M = rV = r( 4/3 pR3) into the expression for escape speed gives

The escape speed from a star

 

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Using either form of this equation, you can show that the escape speed for a body at the surface of our sun is v = 6.18 * 105 m/s (about 2.2 million km/h, or 1.4 million mi/h). This value, roughly 1 500 the speed of light in vacuum, is independent of the mass of the escaping body; it depends on only the mass and radius (or average density and radius) of the sun.

Now consider various stars with the same average density r and different radii R. Equation (13.29) shows that for a given value of density r, the escape speed v is directly proportional to R. In 1783 the Rev. John Mitchell, an amateur astronomer, noted that if a body with the same average density as the sun had about 500 times the radius of the sun, its escape speed would be greater than the speed of light in vacuum, c. With his statement that “all light emitted from such a body would be made to return toward it,” Mitchell became the first person to suggest the existence of what we now call a black hole.

 

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Frequently Asked Questions

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Ans: In 1916 Albert Einstein presented his general theory of relativity, which included a new concept of the nature of gravitation. In his theory, a massive object actually changes the geometry of the space around it view more..
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Ans: Because the earth rotates on its axis, it is not precisely an inertial frame of reference. For this reason the apparent weight of a body on earth is not precisely equal to the earth’s gravitational attraction, which we will call the true weight w 0 of the body. Figure 13.26 is a cutaway view of the earth, showing three observers. Each one holds a spring scale with a body of mass m hanging from it. view more..
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Ans: We assumed at the beginning that the point mass m was outside the spherical shell, so our proof is valid only when m is outside a spherically symmetric mass distribution. view more..
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Ans: Think first about the properties of our own sun. Its mass M = 1.99 * 1030 kg and radius R = 6.96 * 108 m are much larger than those of any planet, but compared to other stars, our sun is not exceptionally massive view more..
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Ans: The first expression for escape speed in Eq. (13.29) suggests that a body of mass M will act as a black hole if its radius R is less than or equal to a certain critical radius. view more..
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Ans: At points far from a black hole, its gravitational effects are the same as those of any normal body with the same mass. If the sun collapsed to form a black hole, the orbits of the planets would be unaffected. But things get dramatically different close to the black hole. view more..
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Ans: If light cannot escape from a black hole and if black holes are small . how can we know that such things exist? The answer is that any gas or dust near the black hole tends to be pulled into an accretion disk that swirls around and into the black hole, rather like a whirlpool view more..
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Ans: HERE ISA SUMMARY OF GRAVITATION , FOR QUICK REVISION view more..
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Ans: Many kinds of motion repeat themselves over and over: the vibration of a quartz crystal in a watch, the swinging pendulum of a grandfather clock, the sound vibrations produced by a clarinet or an organ pipe, and the back-and-forth motion of the pistons in a car engine. This kind of motion, called periodic motion or oscillation view more..
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Ans: n. A body with mass m rests on a frictionless horizontal guide system, such as a linear air track, so it can move along the x-axis only. The body is attached to a spring of negligible mass that can be either stretched or compressed. The left end of the spring is held fixed, and the right end is attached to the body. The spring force is the only horizontal force acting on the body; the vertical normal and gravitational forces always add to zero view more..
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Ans: Here are some terms that we’ll use in discussing periodic motions of all kinds: view more..
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Ans: The simplest kind of oscillation occurs when the restoring force Fx is directly proportional to the displacement from equilibrium x. This happens if the spring in Figs. 14.1 and 14.2 is an ideal one that obeys Hooke’s law view more..
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Ans: To explore the properties of simple harmonic motion, we must express the displacement x of the oscillating body as a function of time, x1t2. view more..
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Ans: the period and frequency of simple harmonic motion are completely determined by the mass m and the force constant k. In simple harmonic motion the period and frequency do not depend on the amplitude A. view more..
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Ans: We still need to find the displacement x as a function of time for a harmonic oscillator. Equation (14.4) for a body in SHM along the x-axis is identical to Eq. (14.8) for the x-coordinate of the reference point in uniform circular motion with constant angular speed v = 2k/m view more..
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Ans: We can learn even more about simple harmonic motion by using energy considerations. The only horizontal force on the body in SHM in Figs. 14.2 and 14.13 is the conservative force exerted by an ideal spring. The vertical forces do no work, so the total mechanical energy of the system is conserved. We also assume that the mass of the spring itself is negligible. view more..
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Ans: the energy quantities E, K, and U at x = 0, x = ±A/2, and x = ±A. Figure 14.15 is a graphical display of Eq. (14.21); energy (kinetic, potential, and total) is plotted vertically and the coordinate x is plotted horizontally. The parabolic curve in Fig. 14.15a represents the potential energy U = 1/2 kx2 . The horizontal line represents the total mechanical energy E, which is constant and does not vary with x. view more..
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Ans: PROBLEM SOLVING STRATEGY ON ENERGY MOMENTUM OF SHM view more..




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