Question
Two stationary particles of masses $M_1$ and $M_2$ are a distance d' apart. A third particle lying on the line joining the particles, experiences no resultant gravitational force. What is the distance of this particle from $M_1$?
✓
Answer
The force on the third parcticle, say of mass m towards $M_1$ is
$\text{F}=\text{G}\frac{\text{GMm}}{\text{r}^2}$
The force on m towards $M_2$ is
$\text{F}=\text{G}\frac{\text{M}_2\text{m}}{(\text{d}-\text{r})^2}$
Equating the two forces, we have
$\text{G}\frac{\text{M}_1\text{m}}{\text{r}^2}=\text{G}\frac{\text{M}^2\text{m}}{(\text{d}-\text{r})^2}$
$\Big[\frac{\text{d}-\text{r}}{\text{r}}\Big]^2=\frac{\sqrt{\text{M}_2}}{\sqrt{\text{M}_1}}$
$\Rightarrow\frac{\text{d}}{\text{r}}-1=\frac{\sqrt{\text{M}_2}}{\sqrt{\text{M}}_1}$
$\Rightarrow\frac{\text{d}}{\text{r}}=\frac{\sqrt{\text{M}_2}+\sqrt{\text{M}_1}}{\sqrt{\text{M}_1}}$
$\Rightarrow\text{r}=\text{d}\Bigg[\frac{\sqrt{\text{M}_1}}{\sqrt{\text{M}_1+\sqrt{\text{M}_2}}}\Bigg]$
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