Question

a planet orbits a star with period T. As seen from that planet the star's diameter subtends an angle of theta radians. Show the density of the star is given by rou=24pi/G(T^2)theta^3

please help!!!

Answer 1

This calls for a combination of Kepler's third law (or Newton's gravitational and motion laws) and trigonometry.

Equating the gravitational attraction to the centripetal force tells you that

GM/R^2 = V^2/R = (2 pi R/T)^2/R
= 4 pi^2 R/T^2
Therefore
(4 pi^2/G M) = T^2/R^3

where R is the distance from the planet to the star. I will let r be the diameter of the star.

The angle subtended by the star in radians is theta = 2 r/R

Now, the star's density = M/[(4/3) pi r^3]
= M/[(4/3) pi R^3]*(r/R)^3
= 4 pi^2/(G T^2)(r/R)^3 / [(4/3) pi]
= 3 pi/(G T^2)/(theta/2)^3
= 24 pi/[G T^2 (theta)^3]