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?
The time duration of a satellite orbit (around Jupiter for example) is
T = 2*pi * sqrt(R^3/(G*M))
where R is the radius of a satellite orbit, G is the universal gravitational constant and M is the mass of the central body (we are interested in Jupiter's mass).
The density is given by
d = M / (4*pi * r^3 / 3) where r is the planet's (Jupiter's) radius.
Observable through the binoculars, knowing the binocular's power and field of view, you can observe the angular measure of both R and r. Since they are at the same distance from Earth, the angle they subtend (which is tiny, though magnified by the binoculars) is proportional to the the actual values (in km for example) of those measures. That is, you know R/r based on what you see in the binoculars.
You need to take the observations over several nights and hours per night to catch at least one of the four Galilean satellites at greatest elongation and preferably several of these satellites to crosscheck. Do this around time of opposition so distance varies little from night to night (or base calculations on Jupiter's apparent size on the same night as measurement of distance to the satellite that's at its greatest elongation).
From the first equation above
M = (2*pi/T)^2 * R^3 / G
Substituting into the second equation:
d = (2*pi/T)^2 * R^3 / (G * 4 * pi * r^3 / 3)
= (2/T)^2 * R^3 / (G * 4 * r^3 / 3)
= (3 * (2/T)^2 / G) / (4 * r^3 / R^3)
= (3 / (T^2 * G)) * (R/r)^3
Since T is known (OK, you need a clock in addition to the binoculars), as well as G and R/r, you can figure out the density, d.
These are based on the assumption the satellites of Jupiter travel in nearly circular orbits.
Edited on July 5, 2018, 7:57 pm

Posted by Charlie
on 20180629 13:40:52 