lew@ihuxr.UUCP (Lew Mammel, Jr.) (12/07/84)
Here are the formulas I derived for my Venusian brightness program. (I derived all of these from first principles.) The size of the illuminated portion of the disk of Venus as viewed from earth is given by: S = (PI/2) * a^2 * ( 1 + cos(alpha) ) where "a" is the radius of the disk and "alpha" is the angle E-V-S. This formula is based on the projected area of the terminator, which is an ellipse. The distance from Earth to Venus is given by the Law of Cosines: X = (Rv^2 + Re^2 - 2*Re*Rv*cos(theta) )^.5 where "theta" is the angle V-S-E. By using the Law of Cosines again, I expressed cos(alpha) as a function of "X". We know that the radius of the disk is inversely proportional to its distance from us, so taking distance in A.U. we get: S = (PI/2) * a0^2 * ( (Rv + X)^2 - 1 )/( 2 * Rv * X^3 ) where a0 is the radius of the Venusian disk at 1 A.U. This is the formula I used in my program, (with (PI/2) * a0^2 == 1 ). By taking derivatives and solving a few quadratic equations I found the formula for theta at maximum brightness: theta(max) = acos( 2 * (Rv^2 + 3)^.5 - 2 * Rv - 1/Rv ) This comes to 22.387 degrees for Rv = .723 . The argument of acos in this formula is less than -1 for Rv less than about .2 (The root of a cubic equation.) This means that an observer farther than 5 times the distance of Venus from the sun will see maximum brightness at opposition. Lew Mammel, Jr. ihnp4!ihuxr!lew P.S. This is my last day on ihuxr. I will be at ihnp4!ihlpa soon, but mail and netnews availability is questionable for a while.