Question

There are 2 bodies of masses 100kg and 10000kg respectively and separated by a distance of 1m. At what distance from the smaller body the intensity of the gravitational field will be 0?

Answer 1

The 2 bodies of masses 100kg and 10000kg respectively and separated by the distance of 1m,we will start by assuming that there is a point ‘O’ at a distance ‘x’ from the smaller mass (100kg). Then it is certain that the distance between the larger mass and the point ‘O’ is given by (1-x)meter, then we will set the direction of forces and solve by using forces at ‘o’ are equal.

Complete answer:
Look at the diagram given in the question, let’s try to understand about the forces F1 and F2.

The forces F1 and F2 are the forces on point ‘O’ because of bodies of mass 100kg and 10000kg respectively. The forces are attractive in nature hence the force due to smaller mass will be towards it (F1) and force because of smaller mass will be towards itself(F2).
Now, for the intensity of gravitational field to be 0 at ‘O’;
F1=F2
Let smaller mass be denoted by m and larger mass be denoted by M.
The field is given by :
$\Rightarrow \dfrac{{Gm}}{{r_1^2}} = \dfrac{{GM}}{{r_2^2}}$
$\Rightarrow \dfrac{m}{{{x^2}}} = \dfrac{M}{{{{\left( {1 - x} \right)}^2}}}$
Substituting m and M we get:
$\Rightarrow \dfrac{{100}}{{{x^2}}} = \dfrac{{10000}}{{{{\left( {1 - x} \right)}^2}}}$
$\Rightarrow \dfrac{1}{{100}} = {\left( {\dfrac{x}{{1 - x}}} \right)^2}$
$\Rightarrow \dfrac{1}{{10}} = \left( {\dfrac{x}{{1 - x}}} \right)$
$\Rightarrow 1 - x = 10x$
$\Rightarrow 11x = 1$
$\Rightarrow x = \dfrac{1}{{11}}$meter.
Hence the point at which the intensity of the gravitational field is 0 will be $\dfrac{1}{{11}}$meter from the smaller mass.

Note: Gravitational field is attractive in nature. Hence the force F1 is towards the smaller mass and F2 id towards the larger mass. If the distance from larger mass where the intensity of gravitational field will be 0 is asked it will be given by:
$1 - x$meters,
Or, $\left( {1 - \dfrac{1}{{11}}} \right)$meters.
$= \dfrac{{10}}{{11}}$meters.