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Two bodies of masses $\mathrm{\mathrm{m}\: and \: \mathrm{M}}$ are placed a distance $\mathrm{d}$ apart. The gravitational potential at the position where the gravitational field due to them is zero is $\mathrm{V}$ .

Option: 1
$\mathrm{\quad V=-\frac{G}{d}(m+M)}$

Option: 2
$\mathrm{\quad V=-\frac{G m}{d}}$

Option: 3
$\mathrm{\mathrm{V}=-\frac{\mathrm{GM}}{\mathrm{d}}}$

Option: 4
$\mathrm{\quad V=-\frac{G}{d}(\sqrt{m}+\sqrt{M})^2}$

Answers (1)

best_answer

Let gravitational field be zero at a point lying at distance $\mathrm{x}$ from $\mathrm{M}$ . Then,

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

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

$\mathrm{ \Rightarrow \frac{d}{x}-1=\sqrt{\frac{m}{M}} }$

$\mathrm{ \Rightarrow x=\left(\frac{\sqrt{M}}{\sqrt{M}+\sqrt{m}}\right) d }$ ..............(1)

$\mathrm{\Rightarrow(d-x)=\left(\frac{\sqrt{M}}{\sqrt{M}+\sqrt{m}}\right) d }$ ..............(2)

Since, $\mathrm{V_p=-\frac{G m}{d-x}-\frac{G m}{x}}$ .............(3)

Substituting (1) and (2) in (3), we get

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

Hence option 4 is correct.

Posted by

shivangi.bhatnagar

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