Question

Sove this problem pls..................

Answer 1

Let the point where gravitational potential is zero be "x" distance from the mass "m''

AT this point ! 

$\frac{Gm}{x^2} = \frac{GM}{(d-x)^2}$

$\frac{m}{x^2} = \frac{m}{(d-x)^2}$

$\frac{d-x}{x} = \sqrt{ \frac{M}{m} }$

$\frac{d}{x} = 1 + \sqrt{ \frac{M}{m} }$

$x = \frac{d}{1 + \sqrt{ \frac{m}{n} } }$

$x = \frac{d \sqrt{m} }{ \sqrt{m} + \sqrt{M} }$

Now,

$d -x = \frac{d \sqrt{M} }{ \sqrt{m} + \sqrt{M} }$

And , 

$V = - \frac{Gm}{x} - \frac{GM}{d-x}$

$V = -\frac{Gm}{ \frac{d \sqrt{m} }{ \sqrt{m} + \sqrt{m} } } -\frac{GM}{ \frac{d \sqrt{M} }{ \sqrt{m} + \sqrt{m} } }$

$V = - \frac{Gm . ( \sqrt{m}+ \sqrt{M}) . \sqrt{M} }{d \sqrt{m} \sqrt{M} } - \frac{GM . ( \sqrt{m}+ \sqrt{M}) . \sqrt{m} }{d \sqrt{m} \sqrt{M} }$

$V = -\frac{G}{d \sqrt{m} \sqrt{M} } (m ( \sqrt{m} +\sqrt{M} ) \sqrt{M} + M( \sqrt{m} \sqrt{M} ) \sqrt{m} )$

$V = - \frac{G}{d \sqrt{m} \sqrt{M} } (m \sqrt{mM} + \sqrt{mM} + \sqrt{mM} + M \sqrt{mM} )$

$V = - \frac{G \sqrt{m} \sqrt{M} }{d \sqrt{m} \sqrt{M} } (m + M + 2 \sqrt{m} \sqrt{M} )$

$V = - \frac{G}{d} (m + M + 2 \sqrt{mM} )$

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Note In your case d = r 

m1 = m 
m2 = M 
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But the concept its same