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March 6, 2015

March 6, 2015

Posted by **POD** on Monday, April 7, 2008 at 4:28pm.

please help!!!

- physics(please help) -
**drwls**, Monday, April 7, 2008 at 5:02pmThis 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]

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