Particles of masses m-1- and m-2- are a fixed distance apart- If the gravitational field strength m-1- and m-2- are l-1- and l-2- respectively- Then-m-1-l-1-m-2-l-2-0m-1-l-2-m-2-l-1-0m-1-l-2-m-2-l-1-0m-1-l-1-m-2-l-2-0

M1I1-x2212-m2I2By Newton-apos-s Third Law of MotionSince miIi is force on mi due to other particle

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Particles of masses $m_{1}$ and $m_{2}$ are at a fixed distance apart. If the gravitational field strength at $m_{1}$ and $m_{2}$ are $l_{1}$ and $l_{2}$ respectively. Then:
$m_{1} l_{1} + m_{2} l_{2} = 0$
$m_{1} l_{2} + m_{2} l_{1} = 0$
$m_{1} l_{2} - m_{2} l_{1} = 0$
$m_{1} l_{1} - m_{2} l_{2} = 0$

m_{1} l_{1} + m_{2} l_{2} = 0

m_{1} l_{2} + m_{2} l_{1} = 0

m_{1} l_{1} - m_{2} l_{2} = 0

m_{1} l_{2} - m_{2} l_{1} = 0

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Solution

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$m_{1} I_{1} = - m_{2} I_{2}$
By Newton's Third Law of Motion
Since $m_{i} I_{i}$ is force on $m_{i}$ due to other particle

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m1I1=−m2I2By Newton's Third Law of MotionSince miIi is force on mi due to other particle

$m_{1} I_{1} = - m_{2} I_{2}$
By Newton's Third Law of Motion
Since $m_{i} I_{i}$ is force on $m_{i}$ due to other particle