| - G m M |
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| R |
Imagine that we have an object of mass m in a circular orbit around an object of mass M. An example could be a satellite orbiting the Earth. What is the total energy associated with this object in its circular orbit?
As usual, E = U + K.
| U | = |
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and K = ½ mv2 |
The only force acting on the object is the force of gravity. Applying Newton's Second Law gives:
ΣF = ma
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= |
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| Therefore: mv2 | = |
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and K | = | ½ mv2 | = |
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The kinetic energy is positive, and half the size of the potential energy.
| E | = |
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+ |
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= |
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A negative total energy tells us that this is a bound system. Much like an electron is bound to a proton in a hydrogen atom with a negative binding energy, the satellite is bound to the Earth - energy would have to be added to each system to remove the electron or the satellite.
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Again we have two masses, m and M, with m << M. The smaller mass will be placed at a particular distance from the larger one and given an initial velocity directed perpendicular to the line joining the masses. We'll examine a few different cases, giving the mass initial velocities of various speeds and seeing what kind of orbit we get in each case.
Case 1: A circular orbit. Let's say this happens to require an initial velocity of 1 unit.
Case 2: v < 1.0. With even less kinetic energy, the mass follows an elliptical path. The starting point is the aphelion, the point furthest from the Sun.
Case 3: v = 0.0. The object simply gets sucked in to the large mass.
Case 4: v > 1.0 but the total energy is still negative. We still have a bound system. The orbit is elliptical again, but this time the starting point is the perihelion - the point closest to the Sun.
Case 5: v is larger by a factor of the square root of two than the speed needed to go in a circle. This is actually the escape speed - the orbit is parabolic, and the object never comes back.
Case 6: v is larger than the escape speed, so the total energy is positive. The orbit is hyperbolic - note that it's much straighter than the parabolic curve.