Question

Two point object of masses ${10}^{4}kg,{10}^{6}kg$ are $1.2\times {10}^3$ apart. Find the distance of the point from smaller mass at which the net gravitational force due to them will be zero?

Answer 1

Let null point be x m from ${10}^4$ kg mass. Distance of null point from ${10}^6$ kg mass is $(1.21\times {10}^3-x)$ m
$\therefore \frac{{10}^4}{x^2}=\frac{{10}^6}{(1.21\times {10}^3-x{)}^2}$
Taking square root
$\frac{1}{x}=\frac{10}{(1.21\times {10}^3-x)}$

$10x=1.21\times {10}^3-x$

$11x=1.21\times {10}^3$

$x=\frac{1.21\times {10}^3}{11}=110m$