The gravitational attraction between two bodies is calculated as follows:
F=(G*m_1*m_2)/d^2F=G⋅m1⋅m2d2
with GG the universal gravity constant, m_1m1 and m_2m2 the masses of the two bodies and dd the distance between the two bodies.
If we place a body on a straight line between the earth and the sun, the resulting gravitational attractions will be:
F_(sb)=(G*m_s*m_b)/((d_(sb))^2)Fsb=G⋅ms⋅mb(dsb)2
F_(eb)=(G*m_e*m_b)/((d_(eb))^2)Feb=G⋅me⋅mb(deb)2
We are considering the situation when F_(sb)=F_(eb)Fsb=Feb:
(cancel(G)*m_s*cancel(m_b))/((d_(sb))^2)=(cancel(G)*m_e*cancel(m_b))/((d_(eb))^2)
rarr (m_s)/((d_(sb))^2)=(m_e)/((d_(eb))^2)
rarr (m_s)/(m_e)=((d_(sb))^2)/((d_(eb))^2)=((d_(sb))/(d_(eb)))^2
rarr sqrt((m_s)/(m_e))=(d_(sb))/(d_(eb))
Knowing that m_e=6*10^24kg and m_s=2*10^30kg, and that d_(sb)+d_(eb)=1.5*10^11m we have to solve the following equation:
sqrt((2*10^30)/(6*10^24))=(1.5*10^11-d_(eb))/(d_(eb))
rarr d_(eb)=(1.5*10^11-d_(eb))/(sqrt((2*10^30)/(6*10^24)))=(1.5*10^11-d_(eb))/(sqrt(1/3*10^6))=((1.5*10^11-d_(eb))*sqrt3)/(10^3)
rarr d_(eb)*10^3+sqrt(3)*d_(eb)=1.5sqrt3*10^11
rarr d_(eb)(10^3+sqrt3)=sqrt6.75*10^11
rarr d_(eb)=(sqrt6.75*10^11)/(10^3+sqrt3)~~259358400m
