Distance to the Sun

Date: 04/16/99 at 12:12:32
From: Larissa Buchholz
Subject: Calculus

My teachers and I can't figure this one out. We would appreciate it if 
you could help!  Thank you.

According to Kepler's laws, the planets in our solar system move in 
elliptical orbits around the sun. If a planet's closest approach to 
the sun occurs at time t=0, then the distance r from the center of the 
planet to the center of the sun at some later time t can be determined 
from the equation

   r=a(1-ecos(phi))

where a is the average distance between centers, e is the positive 
constant that measures the "flatness" of the elliptical orbit, and phi 
is the solution of Kepler's equation:

   (2*pi*t)/T=phi-esin(phi)

in which T is the time it takes for one complete orbit of the planet.
 
Find the distance from Earth to the sun when t=90 days. {First find 
phi from Kepler's equation, and use this value of phi to find the 
distance. Use a = 150x10^6 km, e = 0.0167 and T = 365 days.}

The problem we had was that we couldn't find a way to isolate phi to 
solve for phi in Kepler's equation.

Date: 04/19/99 at 13:13:20
From: Doctor Fwg
Subject: Re: Calculus

Dear Larissa,

Thank you for sending such an interesting question. This kind of 
problem is a little uncommon, except in more advanced mathematical and 
scientific situations, so it is not surprising that you and your 
teachers were stumped. However, there are at least a few different 
ways of dealing with it successfully.

I am assuming that your own references show all the correct relations 
and definitions between the variables and constants that are needed 
here, along with appropriate diagrams, so I will not go into a lot of 
excessive detail about all that and will instead concentrate on one 
method of solution. The method I will describe here for you involves 
solution by successive approximation (SA) but you might also be able 
to use what is sometimes called Newton's Method (or the Newton-Raphson 
Method), but, in this case, I think Newton's Method will be more 
difficult to apply. When you have time you should look into Newton's 
Method though; it is very powerful, not too difficult to apply in most 
situations, and you will probably enjoy knowing more about it if you 
really like mathematics.

Now the method of SA:

Important Constants and Variables

Phi = Angle (in radians) between center of elliptical orbit and planet
Pi = 3.14159... 
e = Eccentricity (dimensionless) of elliptical orbit = approx. 0.0167 
    for earth
T = Orbital period = approx. 365 for earth
t = Time (in days) after Perihelion
a = Semi major axis of elliptical orbit = approx. 150 x 10^6 km for 
    earth
R = Distance between earth and Sun (will come out with same dimensions 
    as "a")

Equations

Phi = [(2 Pi t)/T] + [e Sin(Phi)], rearranged slightly
R = a [1.0 - e Cos(Phi)]

Notice that I have written the first equation for Phi with Phi on both 
sides. This technique allows us to guess at a value for Phi, place the 
guessed value in the RHS of the equation only (along with values of 
Pi, t, T, and e) and then calculate a new value for Phi. 

In this case, the new value for Phi is on the lefthand side (LHS) of 
the equation. If we happen to guess exactly right, the new value for 
Phi will equal the guessed value (this seldom occurs). If we are lucky 
or clever in our initial guesses and in the way we have chosen to re-
write the original equation, the new value of Phi will be a little 
closer (at least) to the true value we are after. This process is 
repeated by putting every new value of Phi only into the righthand 
side (RHS) of the equation again and repeating the process until there 
is no difference between the calculated and guessed value.  

In some cases, the process of convergence occurs very rapidly, i.e., 
only a few iterations are necessary before the differences between the 
guessed and calculated values agree.

As a side comment, note: agreement of almost any reasonable number of 
significant figures can usually be obtained using this technique over 
and over again, but usually it is not necessary to go beyond the 
number of significant figures known to exist within the least well 
known quantity employed in the calculation. In this case, three 
significant figures are probably all that are needed. Do you see why?

In some applications of this method, what may be referred to as 
divergence occurs, and each calculated answer moves farther and 
farther away from the "correct" solution. If this happens, it will be 
obvious. Sometimes this situation can be corrected by re-expressing 
the original equation in another way. In general, it is best to try to 
put the term or expression that is most sensitive to change on the LHS 
of the equation before attempting the method of SA.  

A hand calculator is okay for this kind of thing but if you know how 
to use a computer spreadsheet you can more easily follow how each 
iteration affects your answer.

Now, a few more comments: the elliptical orbit of the earth is almost 
circular, so in 90 days, Phi will be about 90 degrees or Pi/2 (i.e., 
1.57080) radians because 90 days is very near the time interval needed 
for the earth to complete 1/4 of its orbit around the Sun. So, a good 
1st approximation (i.e., guess) for Phi is 1.57080 radians. Placing 
that value of Phi into the RHS of the equation (above), you should get 
a second (new) approximation for Phi of about 1.56598.

Note: for the time being I am deliberately ignoring my earlier 
comments about significant figures. 

Using the second approximation for Phi of 1.56598, and repeating the 
calculation, you should get a third (new) approximation of about 
1.56598. You can stop here because the new answer is equal (within six 
significant figures) to the previous estimate. If you convert this 
last value of Phi (1.56598 radians) to degrees, you should get about 
89.724 deg. So, you can see that our first approximation (i.e., guess) 
of about 90 deg in 90 days wasn't too bad. We were a little lucky 
here; if the orbit had been "highly elliptical," our first guess might 
not have been so good. But even then it would have probably only taken 
a few more iterations for our solutions to converge. You might want to 
try this kind of calculation with Pluto - it has a highly elliptical 
orbit but you need to remember to use the values of a, e, t, and T 
that apply to Pluto.

In any case, you should check over my results since I cannot guarantee 
that I haven't made some error myself, and you also need to complete 
the calculation to get the distance between the earth and the Sun 90 
days after Perihelion. You may also want to try this calculation for 
Earth at t = 180 days. What would be a good initial guess for Phi in 
this case?

I hope this helps out.

With Best Wishes,

- Doctor Fwg, The Math Forum
http://mathforum.org/dr.math/