Question

Two bodies of mass 9kg and 4kg are saperated by a distance of 200m at what distance from 9kg
the field is zero​

Answer 1

Let no net gravitational field is experienced at a distance 'x' (in m) from the center of 9 kg mass, so that this position will be at a distance (200 - x) from the 4 kg mass.

Equation for gravitational field at a point provided by the mass M which is at a distance R from here is,

$\longrightarrow\sf{F=\dfrac{GM}{R^2}}$

The gravitational field provided by the 9 kg mass at this moment is,

$\longrightarrow\sf{F_1=\dfrac{9G}{x^2}}$

And the gravitational force of attraction provided by the 4 kg mass is,

$\longrightarrow\sf{F_2=-\dfrac{4G}{(200-x)^2}}$

Negative sign is because both forces are opposite to each other.

As said earlier, the net gravitational field at this point is zero.

$\longrightarrow\sf{F_1+F_2=0}$

$\longrightarrow\sf{\dfrac{9G}{x^2}-\dfrac{4G}{(200-x)^2}=0}$

$\longrightarrow\sf{\dfrac{9G}{x^2}=\dfrac{4G}{(200-x)^2}}$

$\longrightarrow\sf{\dfrac{9}{x^2}=\dfrac{4}{(200-x)^2}}$

By rule of alternendo,

$\longrightarrow\sf{\dfrac{x^2}{(200-x)^2}=\dfrac{9}{4}}$

$\longrightarrow\sf{\dfrac{x}{200-x}=\dfrac{3}{2}}$

$\longrightarrow\sf{\dfrac{x}{200-x}=\dfrac{3}{5-3}}$

Then by rule of dividendo,

$\longrightarrow\sf{\dfrac{x}{200}=\dfrac{3}{5}}$

$\longrightarrow\sf{x=200\times\dfrac{3}{5}}$

$\longrightarrow\sf{\underline{\underline{x=120\ m}}}$

Hence the net gravitational field is zero at 120 m away from the center of 9 kg mass or 80 m away from the center of 4 kg mass.

Answer 2

Answer:

$\large\boxed{\sf{120\;m}}$

Explanation:

Given that, two bodies of masses 9 kg and 4 kg respectively are separated by a distance of 200 m.

Let, the net gravitational field is zero at a distance of 'x' m from 4 kg mass.

Therefore, the field will be zero at a distance of (200-x) m from 9 kg mass.

$\red{Note:-}Refer\:to\:the\: attachment.$

Also, if the gravitational field is zero at that point, then the gravitational force will also be zero.

And, we know that, gravitational force is given by,

$\large\boxed{F=G\dfrac{m}{{r}^{2}}}$

where, G is universal gravitational constant, m is the mass of object and r is the separation between the masses.

Now, clearly both the forces will be opposite and equal in magnitude to each other, so that the net force is zero.

So, we can now write,

$= > G \dfrac{4}{ {x}^{2} } = G\dfrac{9}{ {(200 - x)}^{2} } \\ \\ = > 4 {(200 - x)}^{2} = 9 {x}^{2} \\ \\ = > 2(200 - x) = 3x \\ \\ = > 400 - 2x = 3x \\ \\ = > 5x = 400 \\ \\ = > x = \frac{400}{5} \\ \\ = > x = 80$

Therefore, gravitational field will be zero at a distance of (200-80) i.e., 120 m