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Gravitation: Orbits

Problems on Orbits

Escape Velocity and Viscous Drag

How to Cite This SparkNote

Problem : What is the escape velocity from the earth? ( M e = 5.98×1024 kilograms, r e = 6.38×106 meters)

The escape velocity is given by:

v = = 1.12×104m/s    

Thus the escape velocity from the earth is roughly 11.2 kilometers per second.

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 1.99×1030 kilograms and the speed of light is 3×108 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 = = 2.95×106 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 6.6×103 kilometers, what is its new orbital radius? ( M e = 5.98×1024 kilograms).

We can calculate the energy of the old orbit by summing the potential and kinetic energies. T o = 1/2mv 2 = 1×1011 Joules. U o = - = - 1.2×1011 Joules. The E = T o + U o = - 0.2×1011 Joules. The new kinetic energy is T n = 1/2mv 2 = 2.25×1011 Joules. The potential energy in this new arrangement is given by U n = ET = (- 0.2–2.25)×1011 = - 2.45×1011 Joules. We can now calculate the corresponding radius as r = = 3.2×106 metres. As we would expect, the satellite is now in a lower orbit.

Problem : At a height of 2×107 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 5.98×1024 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 = - = - (2.0×107)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 = (1.0×107)m . This is half the potential, so this satellite will take on a circular orbit. The other two satellites will take elliptical orbits.

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