Search any question & find its solution
Question:
Answered & Verified by Expert
Two bodies of masses $m$ and $9 m$ are placed at a distance $r$. The gravitational potential at a point on the line joining them, where gravitational field is zero, is ( $G$ is universal gravitational constant)
Options:
Solution:
1787 Upvotes
Verified Answer
The correct answer is:
$\frac{-16 G m}{r}$
Rom the body of mass $m$, the distance of the point where gravitational field is zero
$\begin{aligned}
& x=\frac{d}{\sqrt{\frac{m_2}{m_1}}+1} \Rightarrow x=\frac{r}{\sqrt{\frac{9 m}{m}}+1} \quad \therefore \quad x=\frac{1}{4} \\
& \text { Again }(r-x)=r-\frac{r}{4}=\frac{3 r}{4} \\
& \therefore \text { Potential, } v=v_1+v_2 \\
& =\frac{G m_2}{X}-\frac{G m_1}{r-x}=-\frac{G m}{r / 4}-\frac{G(9 m)}{3 r / 4} \\
& =-\frac{4 G m}{r}-\frac{12 G m}{r}=-\frac{16 G m}{r}
\end{aligned}$
$\begin{aligned}
& x=\frac{d}{\sqrt{\frac{m_2}{m_1}}+1} \Rightarrow x=\frac{r}{\sqrt{\frac{9 m}{m}}+1} \quad \therefore \quad x=\frac{1}{4} \\
& \text { Again }(r-x)=r-\frac{r}{4}=\frac{3 r}{4} \\
& \therefore \text { Potential, } v=v_1+v_2 \\
& =\frac{G m_2}{X}-\frac{G m_1}{r-x}=-\frac{G m}{r / 4}-\frac{G(9 m)}{3 r / 4} \\
& =-\frac{4 G m}{r}-\frac{12 G m}{r}=-\frac{16 G m}{r}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.