3. The bound orbits under an inverse-square central force are ellipses, with the
attracting center in one focus. This fact alone would make ellipses important and
worthy of study. This is, in fact, Kepler's First Law, which he deduced for the
planet Mars, using Tycho Brahe's naked-eye observations. The orbital
parameters of the major planets are given in the table at the right. In this table, a
is the semi-major axis in units of the earth's distance from the sun (the
astronomical unit, A.U.), e its eccentricity, and i the inclination of the plane of the
orbit to the average plane of the earth's orbit. It is remarkable that the planets
revolve in the same direction, and almost in the same plane. Only Pluto is
errant, with its large inclination. All the eccentricities are small, except for
Mercury and Pluto. To the eye, all the orbits would appear to be circles, though
the sun would not be accurately at the centers of the circles, quite obviously
eccentric in the cases of Mercury, Mars and Pluto. In the case of the solar
system, the mass of the sun is so predominant, and the planets so far apart, that
to a first approximation they all revolve independently about a fixed center. The
sun is about 1000 times as massive as Jupiter, which is 4 times as massive as
Saturn. The sun is 328,900 times as as massive as the earth plus moon.
4. Kepler's Second Law was that the radius vector of a planet swept out equal
areas in equal times, or that the areal velocity was constant. Newton showed
that this was a conseqence of the constancy of the angular momentum of the
planet, since the force of attraction had no sideways component, or was
central. The angular momentum is the product of the mass of the planet, the
distance from the sun and the velocity normal to the radius. A quantity
proportional to the angular momentum is h = r(dθ/dt)2, which is twice the areal
velocity. When the equation of the orbit is determined from Newton's Laws, the
parameter p = (h/k)2, where k is the Gaussian gravitational constant, equal to
0.01720209895 when the length unit is the A.U. and the time unit is the mean
solar day. From the polar equation of the ellipse, we also find that p = a (1 - e2),
so we can express h in terms of the orbital constants.
5. Since the total area of the orbit is A = πab = πa2(1 - e2)1/2, the time for one
revolution, the period P = A/(h/2) = 2πa3/2/k. Check this result by applying it
to the earth, where a = 1 and P = 365.25 days. You should get the value of k
given above. In fact, this is how k can be determined accurately, since its
value depends only on the length of the year. This result expresses Kepler's
Third Law, that the square of the period is proportional to the cube of the
semimajor axis. Check this for Jupiter, whose period is 11.86 years. The
mean daily motion of a planet is n = 2π/P radians per day, or 360/P degrees
per day. The problem of finding the true anomaly (the angle of the radius
vector) as a function of time is called Kepler's Problem. Its solution is not only
practical, but quite interesting.
To start, the areal velocity is constant, so the area that has been described by
the radius from the moment of perihelion is proportional to time. The ratio of
this area to the total area of the orbit is equal to the ratio of the mean motion
M in angle in the same time to 2π, and M = n(t - T), where n is the mean daily
motion, T the time of perihelion passage, and t the time. This is the
fundamental relation between uniform time t and the nonuniform dθ/dt.
6. In the diagram, the shaded area AFQ is the area swept out at some time t. As we
have explained above, M/2π = AFQ / πa2
(1 - e2
)1/2
= AFQ' / πa2
, since the ellipse is
the squeezed auxiliary circle. Now area AFQ' is equal to the sector ACQ' less the
triangle CQ'F, with base ae and altitude a sin E, or Ea2
/2 - (1/2)ea2
sin E. When we
divide by πa2
and multiply by 2π, we find that M = E - e sin E, Kepler's Equation.
This is a relation between the mean anomaly M, that increases uniformly with time,
and the eccentric anomaly E, which does not. Unfortunately, it gives M directly in
terms of E, not vice-versa, which would have been more convenient.
From the figure, we can work out how r and w depend on E. r2
= FN2
+ QN2
. FN = a
cos E -ae and QN = a(1 - e2
)1/2
sin E. Working through the algebra, we find r = a (1 -
e cos E) and tan(w/2) = [(1 + e)/(1 - e)]1/2
tan (E/2). This is simply another way to
parameterize the equation of an ellipse. Our procedure is to find M = n(t - T), then
to find E from Kepler's Equation, and finally to compute r and w. With computers,
solving Kepler's Equation is easy. For small e, there are approximations that work
well for the solar system.
7. In the diagram, the shaded area AFQ is the area swept out at some time t. As we
have explained above, M/2π = AFQ / πa2
(1 - e2
)1/2
= AFQ' / πa2
, since the ellipse is
the squeezed auxiliary circle. Now area AFQ' is equal to the sector ACQ' less the
triangle CQ'F, with base ae and altitude a sin E, or Ea2
/2 - (1/2)ea2
sin E. When we
divide by πa2
and multiply by 2π, we find that M = E - e sin E, Kepler's Equation.
This is a relation between the mean anomaly M, that increases uniformly with time,
and the eccentric anomaly E, which does not. Unfortunately, it gives M directly in
terms of E, not vice-versa, which would have been more convenient.
From the figure, we can work out how r and w depend on E. r2
= FN2
+ QN2
. FN = a
cos E -ae and QN = a(1 - e2
)1/2
sin E. Working through the algebra, we find r = a (1 -
e cos E) and tan(w/2) = [(1 + e)/(1 - e)]1/2
tan (E/2). This is simply another way to
parameterize the equation of an ellipse. Our procedure is to find M = n(t - T), then
to find E from Kepler's Equation, and finally to compute r and w. With computers,
solving Kepler's Equation is easy. For small e, there are approximations that work
well for the solar system.
