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Two stationary particles of masses M(1) ...

Two stationary particles of masses `M_(1)` and `M_(2)` are at a distance d apart. A third particle, lying on the line joining the particles experiences no resultant gravitational forces. What is the distance of this particle from `M_(1)`.

Text Solution

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The force on m towards `M_(1)` is `F_(1)=(GM_(1)m)/(r^(2))`
the force on m towards `M_(2)` is `F_(2)=(GM_(2)m)/((d-r)^(2))`
According to question net force on m is zero i.e., `F_(1)=F_(2)`
`implies(GM_(1)m)/(r^(2))=(GM_(2)m)/((d-r)^(2))implies((d-r)/(r))^(2)=(M_(2))/(M_(1))implies(d)/(r)-1=(sqrt(M_(2)))/(sqrt(M_(1)))impliesr=d[(sqrt(M_(1)))/(sqrt(M_(1))+sqrt(M_(2)))]`
this point is called neutral point.
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