Q4.6: Finding angle of Inclination of a circular orbit approach an ecclipse Assu

Question

Q4.6: Finding angle of Inclination of a circular orbit approach an ecclipse

Assume that two stars are in circular orbits about a mutual center of mass and are separated by a distance a. Assume also that the angle of inclination is i and their stellar radii are r_1 and r_2 (assume r_1 > r_2). Find an expression for the smallest angle of inclination that will just barely produce an eclipse. If a = 2 AU, r_1 = 10 R_, and r|_2 = 1 R_, what minimum value of i will result in an eclipse? Simplify your answer to (a) for the case of a planet orbiting a star. Explain any assumptions you make. What is i_min for the case of the Earth orbiting the Sun? Given this, are the odds of finding a planet via the transit method high or low? Why?

Explanation / Answer

a )

the minimum distance possible between two stars is d = r1 + r2  

the projection of the distance between 2 stars is

d = a cos i

equating the above equations

r1 + r2 = a cos i

i = cos-1( r1 + r2 / a )

b )

i = cos-1( r1 + r2 / a )

given r1 = 10 Ro , r2 = 1 Ro , a = 2 AU

i = cos-1( r1 + r2 / a )

i = cos-1( 11 X 7 X 108 / 2 AU )

i = cos-1( 77 X 108 / 3 X 1011 )

i = 1.3 rad

or

i = 74.50o  

c )

i = cos-1( r1 + r2 / a )

so the angle of incidence is based on the r1 and r2 values

d )

the imin is 74.5o only

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