The satellite\'s period in this orbit is exactly one year, so it remains fixed r

Question

The satellite's period in this orbit is exactly one year, so it remains fixed relative to earth. At this point the light form the sun is never blocked by the earth yet the satellite remains 'nearby' so the data are easily transmitted to the earth. What is SOHO's distance from earth?
Use binomial approximation, SOHO's distance from the earth is much less than the earth's distance form the sun.
The distance of the Earth from the sun is 1.5*10^11 m
The mass of the sun is 1.99*10^30 kg
The mass of the earth is 5.98*10^24 kg
The radius of the earth is 6.37*10^6 m
The radius of the sun is 6.96*10^8 m

Explanation / Answer

let M=mass of sun, m1=mass of earth, m2=mass of SOHO r=radius of earth orbit, v=speed of earth r'=radius of SOHO orbit, v'=speed of SOHO equation of motion for earth is m1 v^2/r = Gm1 M /r^2 v^2/r = GM /r^2 >>>> equation A v^2/r = GM /r^2 >>>> equation A equation of motion for SOHO is m2 v'^2/r' = Gm2 M /r'^2 - G m1 m2 /(r-r')^2 v'^2/r' = GM /r'^2 - Gm1/(r-r')^2 >>>>equationB now the period T is same for earth and SOHO, so T = 2r/v = 2r'/v' v=r, v'=r', where = 2/T equations A and B gives ^2 r = GM/r^2 ^2 r' = GM/r'^2 - G m1 /r^2 >>> r =r'+r, r << r dividing second equation by first, r'/r = r^2/r'^2 - (m1/M)(r/r)^2 we can expand r'/r and r^2/r'^2 as r'/r = 1-r/r, r^2/r'^2 ˜ 1+2r/r so we get 1-r/r˜1+2r/r - (m1/M)(r/r)^2 3r/r = (m1/M)(r/r)^2 r/r ˜ (m1/3M)^(1/3) ˜ .01 r ˜.01 AU equation of motion for SOHO is m2 v'^2/r' = Gm2 M /r'^2 - G m1 m2 /(r-r')^2 v'^2/r' = GM /r'^2 - Gm1/(r-r')^2 >>>>equationB now the period T is same for earth and SOHO, so T = 2r/v = 2r'/v' v=r, v'=r', where = 2/T equations A and B gives ^2 r = GM/r^2 ^2 r' = GM/r'^2 - G m1 /r^2 >>> r =r'+r, r << r dividing second equation by first, r'/r = r^2/r'^2 - (m1/M)(r/r)^2 we can expand r'/r and r^2/r'^2 as r'/r = 1-r/r, r^2/r'^2 ˜ 1+2r/r so we get 1-r/r˜1+2r/r - (m1/M)(r/r)^2 3r/r = (m1/M)(r/r)^2 r/r ˜ (m1/3M)^(1/3) ˜ .01 r ˜.01 AU v'^2/r' = GM /r'^2 - Gm1/(r-r')^2 >>>>equationB now the period T is same for earth and SOHO, so T = 2r/v = 2r'/v' v=r, v'=r', where = 2/T equations A and B gives ^2 r = GM/r^2 ^2 r' = GM/r'^2 - G m1 /r^2 >>> r =r'+r, r << r dividing second equation by first, r'/r = r^2/r'^2 - (m1/M)(r/r)^2 we can expand r'/r and r^2/r'^2 as r'/r = 1-r/r, r^2/r'^2 ˜ 1+2r/r so we get 1-r/r˜1+2r/r - (m1/M)(r/r)^2 3r/r = (m1/M)(r/r)^2 r/r ˜ (m1/3M)^(1/3) ˜ .01 r ˜.01 AU

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