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SOLUTIONS TO EXERCISE 20
SOLUTIONS TO EXERCISE 20 (a) If light is moving in a stable orbit, then the orbital velocity at that distance is the speed of light (c = 3 * 10^8 m/s). So let's set the mathematical expression for orbital velocity equal to the speed of light. sqrt(GM/r) = c Now square both sides to get GM/r = c^2 and rearrange to get r = GM/c^2 So the radius at which light moves in a stable orbit around an object with mass M is a function of that mass, the gravitational constant, and the speed of light squared. (This is called the Schwarzschild radius.) (b) If M = 6 * 10^24 kg, then r = (6.67 * 10^(-11)) * (6 * 10^24) / (3 * 10^8)^2 = 4.4 * 10^(-3) m So the diameter of this black hole would be about one centimeter - about the size of a small marble or gumball. (c) The number of Earth-mass black holes needed to equal the mass of the solar system would be (2 * 10^30 kg) / (6 * 10^24 kg/black hole) = 333,000 or so. Each black hole is a sphere with diameter 0.9 cm. We can thus make a rough estimate and say that, in a container, each black hole would occupy one cubic centimeter. So a container with volume 333,000 cm^3 would hold our solar system's worth of matter. A typical bathtub would do - its dimensions would be about 5 feet long, 2.5 feet wide and 1 foot deep, or (150 cm long) * (75 cm wide) * (30 cm deep) = 337,500 cm^3. |