Question

The bodies of masses $100kg$ and $8100kg$ are held at distance of $1m$. The gravitational field at a point on the line joining them is zero. The gravitational potential at that point in $J/kg$ is $(G=6.67\times {10}^{-11}N.m^2/k{g}^2)$

• A. $-6.67\times {10}^{-7}$
• B. $-6.67\times {10}^{-10}$
• C. $-13.34\times {10}^{-7}$
• D. $-6.67\times {10}^{-9}$

Answer 1

Answer: (A)

Distance of null point from $100kg$
$x=\frac{1}{\sqrt{\frac{8100}{100}}+1}=\frac{1}{10}=0.1m$

Gravitational potential at that point is:
$-G(\frac{m_1}{0.1}+\frac{m_2}{9})=-6.67\times {10}^{-11}(\frac{100}{{10}^{-1}}+\frac{8100}{9\times {10}^{-1}})$

$=-6.67\times {10}^{-11}(1000+9000)=-6.67\times {10}^{-7}m$