Problem : What is the escape velocity from the earth? ( M _{e} = 5.98×10^{24} kilograms, r _{e} = 6.38×10^{6} meters)
The escape velocity is given by:
v = = 1.12×10^{4}m/s |
Problem : If a black hole contains a singularity (all the mass is concentrated at a point) with a mass 1000 times the mass of the sun, what is the radius beyond which light cannot escape? This is called the Schwartzchild radius. The mass of the sun is 1.99×10^{30} kilograms and the speed of light is 3×10^{8} m/s.
The escape velocity is dependent on the radius as well as the mass. The escape velocity is given by: v = . Solving for r , we have: r = . Substituting in we find: r = = 2.95×10^{6} meters. This is less than half the radius of the earth. Closer than this radius the escape velocity will be greater than the speed of light.Problem : We would expect a satellite orbiting the earth to be slowed down by friction with the atmosphere. We also know that velocity is inversely proportional to the radius of the orbit, so when the satellite slows down, it should spiral away from the earth. But it is observed that satellites spiral in towards the earth. How can we explain this paradox?
The satellite is actually caused to speed up by the viscous drag. This causes the satellite to gain kinetic energy, but lose potential energy as it spirals towards the earth. Overall, energy is dissipated into heat by the frictional force.Problem : Suppose the viscous drag causes a 2000 kilogram satellite to increase its speed from 10000 m/s to 15000 m/s. If the satellite had an initial orbital radius of 6.6×10^{3} kilometers, what is its new orbital radius? ( M _{e} = 5.98×10^{24} kilograms).
We can calculate the energy of the old orbit by summing the potential and kinetic energies. T _{o} = 1/2mv ^{2} = 1×10^{11} Joules. U _{o} = - = - 1.2×10^{11} Joules. The E = T _{o} + U _{o} = - 0.2×10^{11} Joules. The new kinetic energy is T _{n} = 1/2mv ^{2} = 2.25×10^{11} Joules. The potential energy in this new arrangement is given by U _{n} = E–T = (- 0.2–2.25)×10^{11} = - 2.45×10^{11} Joules. We can now calculate the corresponding radius as r = = 3.2×10^{6} metres. As we would expect, the satellite is now in a lower orbit.Problem : At a height of 2×10^{7} meters above the center of the earth, three satellites are ejected from a space shuttle. They are given tangential speeds of 5.47 kilometers/sec, 4.47 km/sec, and 3.47 km/sec respectively. Which one(s) will assume circular orbits? Which elliptical? The mass of the earth is 5.98×10^{24} kilograms.
We can calculate the total energy of each satellite in terms of its mass m . In fact, the potential energy of each orbit will be the same: U = - = - (2.0×10^{7})m . For a circular orbit, the kinetic energy must be exactly half the absolute value of the potential energy. For the 4.47 km/sec satellite, T = 1/2mv ^{2} = (1.0×10^{7})m . This is half the potential, so this satellite will take on a circular orbit. The other two satellites will take elliptical orbits.Take a Study Break!