 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Astronomy: Binoculars, for fun and profit (Posted on 2018-06-29) True story: In a tough astronomy Oral Exam, my grinning professors posed this problem:
"How would you determine the density of Jupiter using only a pair of binoculars? Assume you don't know the planet's distance."
"Umm..." But, to my surprise, I figured it out.

Once outside the exam room, I boasted of my accomplishment to my friend Rico. "Impossible!" he proclaimed. I bet him 10\$ and won. How did I do it?

 Submitted by Steven Lord No Rating Solution: (Hide) Jupiter has four inner moons whose obits may be followed with binoculars. The key is to use Kepler's 3rd Law recognizing there is a distance-cubed term on either side of the equation which cancels out. Here "J" refers to Jupiter, and "moon" refers to any one of the four Galilean satellites, each viewable in good binoculars. The observables are: J's disk's angular diameter, the J-moon's maximum angular separation and the moon's orbital period. All observed angles may be determined using the diameter of the field of view of the binoculars. OMEGA= J's half diameter angle, R= J's radius, M= J's mass, dist. = J's distance from Earth, d = the J-moon separation, omega = J-moon angular separation, p = moon's orbit period, v = moon's circular velocity, G is the Gravitational constant, m = the moon's massrho = J's density. Note, rho should come out something like 1.3 g/cm^3, i.e., a little denser than water. 1) Derive Kepler's 3rd Law by equating gravitational and centripetal forces: m v^2/d = G M m/d^2, (note: the m's cancel)and use v=2 pi d/p to obtain the 3rd Law: d^3 = [GM/(2 pi)^2] p^2. (Whereas Kepler used this for planets around the Sun, we are using it for moons around Jupiter.) 2) Recall that for an angle "omega" subtending a distant disk or measured between two distant objects, distance x omega [radians] = linear extent. Also, M = rho 4/3 pi R^3 3) Plugging into the 3rd Law, we get (omega dist.)^3 = G rho 4/3 pi (OMEGA dist.)^3 p^2/(2 pi)^2. The "dist." term cancels. This gives rho = (omega/OMEGA)^3 3 pi/p^2, our answer. Plugging in observables, say for Jupiter and Calisto (expressing angles in arc sec rather than radians) omega = 556 arc sec, (6/18)OMEGA = 20.6 arc sec, (6/18)p = 16.69 days, G=6.67E-11, which yields rho = 1328 kg/m^3 = 1.3 g/cm^3, right on the mark. Subject Author Date No Subject Steven Lord 2018-07-05 20:26:00 re(2): Hint Charlie 2018-07-05 19:58:20 re: Hint Charlie 2018-07-05 15:55:18 Hint Steven Lord 2018-07-05 15:10:18 How to... Charlie 2018-06-29 13:40:52 Please log in:

 Search: Search body:
Forums (0)