APOD for 1 June 2005
APOD for 1 June 2005
APOD for 1 June 2005 gives data for a binary star system that leave me very confused. When I apply Kepler's third law using this data I get a mass of ten to the eighth solar masses for the pair! (And it is said to be a white dwarf pair!) Would someone explain this or correct me?
Larry Weaver
Physics
Kansas State University
Manhattan, KS 66506
Physics
Kansas State University
Manhattan, KS 66506
I screwed up: the 400,000km is the earth-MOON distance, and I stupidly used it in Kepler's laws for the earth-SUN distance, a number I know well is NOT 400,000km. The correct numbers give a combined mass of roughly one solar mass. Sorry!
Here is the math: GM/r^2=v^2/r, so GM/4pi^2 = r^3/T^2, where M is the combined mass, r is the separation, and T is the period. Write this twice: once for the white dwarf (wd) system and once for the earth-sun. Divide the two expressions and you get
M(wd)/M(sun)=[r(wd)/r(earth-sun)]^3[T(earth-sun)/T(wd)]^2.
Putting in the numbers gives (and I'm using km for the distance and seconds for the time)
M(wd)/M(sun)=[8x10^4/150x10^6]^3[3x10^7/321]^2
=[150x10^(-12)]x[0.87x10^10] = 1.3
So the data given say the combined mass is around 1.3 solar masses, a quite comfortable figure for a pair of white dwarf stars.
Here is the math: GM/r^2=v^2/r, so GM/4pi^2 = r^3/T^2, where M is the combined mass, r is the separation, and T is the period. Write this twice: once for the white dwarf (wd) system and once for the earth-sun. Divide the two expressions and you get
M(wd)/M(sun)=[r(wd)/r(earth-sun)]^3[T(earth-sun)/T(wd)]^2.
Putting in the numbers gives (and I'm using km for the distance and seconds for the time)
M(wd)/M(sun)=[8x10^4/150x10^6]^3[3x10^7/321]^2
=[150x10^(-12)]x[0.87x10^10] = 1.3
So the data given say the combined mass is around 1.3 solar masses, a quite comfortable figure for a pair of white dwarf stars.
Larry Weaver
Physics
Kansas State University
Manhattan, KS 66506
Physics
Kansas State University
Manhattan, KS 66506