|
Orbital Energy
The energy of an object, such as a planet, moving in a gravitational field, such as the Sun's, can be divided into two parts -- kinetic energy, or energy of motion, and potential energy, or energy of position.
The kinetic energy depends upon the mass and velocity of the object, according to the formula Ekinetic = m v2 / 2
where m is the mass of the object, and v is its velocity, or speed of motion (velocity also includes the direction of motion, but that is not important in this situation). This is always a positive value, and since planets move faster in smaller orbits, their kinetic has a larger positive value, the smaller the orbit.
The gravitational potential energy is defined as a negative value, equal to the kinetic energy that the object would gain by falling from an infinite distance to its current position. At very large distances from the Sun, the object would have zero potential energy (since it wouldn't have picked up any speed, by falling). Objects close to the Sun have large (albeit negative) potential energies, corresponding to the speed they would gain by falling a long way. Because the gravitational force of the Sun changes with distance, deriving the formula for potential energy requires calculus; but the result is a simple alteration of the Law of Gravity:
Epotential = - G m M / r
where m and M are the masses of the planet and the Sun, r is the distance from the planet to the Sun, and G is the constant of gravitation. Remember that unlike kinetic energy, which has a larger positive value in smaller orbits, potential energy has a larger negative value in smaller orbits.
The total energy of an object is the sum of its kinetic and potential energies, and since for a falling object, the kinetic energy increases at exactly the same rate that the potential energy decreases, the total energy of a planetary orbit is a constant:
Etotal = Ekinetic + Epotential = m v2 / 2 - G m M / r = (a constant)and the total energy per unit of planetary mass, which is what we will use below, would be the same, but without the m's:
Etotal / m = v2 / 2 - G M / r
Application to Circular Orbits
Kepler's Third Law of Planetary Motion, which relates the size of an orbit to its orbital period, can be stated as P2 = 4 p2 a3 / G Mwhere P is the orbital period of a planet, a is the semi-major axis of the planet's orbit, or its "average" distance from the Sun, and M is the combined mass of the planet and Sun (which is treated as just the mass of the Sun in this discussion because (1) the planets have negligible masses compared to the Sun and (2) taking planetary masses into account here would require taking them into account in the calculation of the potential energy, and once done, would yield the same result as simply ignoring them).
For a circular orbit, Kepler's Second Law of Planetary Motion, which relates the speed of the planet to its distance from the Sun, requires the speed to be as constant as the distance from the Sun. The speed is, in fact, simply the circumference of the orbit, divided by its orbital period: v = 2 p a / P
Squaring this yields v2 = 4 p2 a2 / P2Squaring this yields
and using Kepler's Third Law to replace the square of the period,
v2 = (4 p2 a2) / (4 p2 a3 / G M)
Division cancels most of the terms, leaving (for circular orbits)
v2 = G M / a
or Ekinetic / m = v2 / 2 = G M / 2 a
Since r and a are the same for a circular orbit, this is exactly half the (negative) value of the potential energy per unit of mass. As a result, adding the kinetic and potential energy gives a total energy which is also half the potential energy, and equal to the kinetic energy, but negative. In other words, for a circular orbit, Etotal = Epotential / 2 = - Ekinetic = (a constant, for an orbit of a given size)
Application to Elliptical Orbits
In elliptical orbits, the velocity is not constant, so the simplifications of the preceding section do not apply, and as a planet moves away from or toward the Sun, the kinetic energy and potential energy continually rise and fall; but since one rises exactly as much as the other falls, their sum (the total energy of the orbit) is always the same, and in fact, all orbits of a given size (or semi-major axis) have the same total energy per unit of mass, regardless of the eccentricity of the orbit.
(no other discussion of elliptical orbits is planned at this time)
Deriving the Effect of the Sun's Loss of Mass
With the above as background, we are now ready to see how the Sun's loss of mass, as it grows old, affects the size of planetary orbits. In doing so, we will assume that the orbits are sufficiently circular that we can use the simplifications applicable to that case, and that any friction due to gases lost by the Sun passing the various planetary orbits produces negligible effects on their motion. In that case, the only thing we need worry about is that as the Sun loses mass, it exerts less gravitational force, and cannot hold the planets in the orbits that their current velocities require. As a result, the planets would gradually move away from the Sun, attaining higher orbits and in the process, reducing their speeds to values which the Sun's reduced gravity could maintain in such orbits.
To see how this works, let us start with the formulae for the Sun's current ("original") mass:
Etotal = Epotential / 2 = - Ekinetic = - G Moriginal / 2 aoriginal
As the Sun loses mass, a planet's orbital potential energy will become smaller (although, since negative, it will actually represent a higher energy), but the kinetic energy will (at first) remain as it was, so the total energy will increase by the same amount as the potential energy. This causes the orbit to become larger, reducing the (negative) value of the potential energy still further, and at the same time (because objects move more slowly in larger orbits) reducing the kinetic energy. This continues until a stable state is reached, in which the (new) kinetic energy is numerically equal to the (new) total energy, and half the size of the (new) potential energy.
In numerical terms, the original kinetic energy
Ekinetic (original) = G Moriginal / 2 aoriginalplus the new potential energy Epotential (new) = - G Mnew / aoriginalyield a new total energy
Etotal (new) = Ekinetic (original) + Epotential (new) = G Moriginal / 2 aoriginal - G Mnew / aoriginaland in a new nearly circular orbit, this total energy equals half the new potential energy Etotal (new) = Epotential (new) / 2 = - G Mnew / 2 anewor, combining the terms,
G Moriginal / 2 aoriginal - G Mnew / aoriginal = - G Mnew / 2 anewMultiplying each term by 2 anew / G Moriginal, we obtain
anew / aoriginal - 2 Mnew anew / Moriginal aoriginal = - Mnew / MoriginalFinally, rearranging terms, we obtain anew / aoriginal = (Mnew / Moriginal) / (2 Mnew / Moriginal - 1)which is the result shown in The Fate of the Earth.
|
