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Question: The distance between the centres of the Moon and the earth is \[D\] . The mass of the earth is \(81\...

The distance between the centres of the Moon and the earth is DD . The mass of the earth is 8181 times the mass of the Moon. At what distance from the centre of the earth, the gravitational force will be zero?
A. D2\dfrac{D}{2}
B. 2D3\dfrac{{2D}}{3}
C. 4D3\dfrac{{4D}}{3}
D. 9D10\dfrac{{9D}}{{10}}

Explanation

Solution

Hint If net gravitational field intensity due to some bodies at a point is zero, then the gravitational force will also be zero at that point.
The gravitational field intensity due to a body of MM at a distance rr from its centre is given by GMr2\dfrac{{GM}}{{{r^2}}} where GG is called Gravitational constant.

Complete step-by-step solution :As in the question we are asked to find at what distance from the centre of the earth, the gravitational force will be zero
We know that, if net gravitational field intensity due to some bodies at a point is zero, then the gravitational force will also be zero at that point. So, we first calculate the gravitational field intensity due to the earth and the moon at a particular point.
Let the net field at any point P from the earth is zero which is at distance ‘x’ from the centre of the earth. As the distance between the centres of the Moon and the earth is given DD then point P will be at a distance DxD - x from the centre of the moon.
We know that the gravitational field intensity due to a body of MM at a distance rr from its centre is given by GMr2\dfrac{{GM}}{{{r^2}}} where GG is called Gravitational constant.
As given in the question that the mass of the earth is 8181 times the mass of the Moon i.e. Me=81Mm{M_e} = 81{M_m}
Then the gravitational field intensity at point P due to earth
Ee=GMex2=81GMmx2{E_e} = \dfrac{{G{M_e}}}{{{x^2}}} = \dfrac{{81G{M_m}}}{{{x^2}}} towards centre of earth
Now, the gravitational field intensity at point P due to moon
Em=GMm(Dx)2\Rightarrow {E_m} = \dfrac{{G{M_m}}}{{{{\left( {D - x} \right)}^2}}} towards centre of moon
Now, for zero intensity at point P, Ee=Em{E_e} = {E_m} i.e.
81GMmx2=GMm(Dx)2\Rightarrow \dfrac{{81G{M_m}}}{{{x^2}}} = \dfrac{{G{M_m}}}{{{{\left( {D - x} \right)}^2}}}
On solving the equation we get
81(Dx)2=x2\Rightarrow 81{\left( {D - x} \right)^2} = {x^2}
On further solving this quadratic equation we have
9(Dx)=x\Rightarrow 9\left( {D - x} \right) = x
On simplifying we get the final answer
x=9D10\Rightarrow x = \dfrac{{9D}}{{10}}

Hence, option D is correct.

Note: The direction of Gravitation field intensity due to a body at a point is towards the centre of gravity of that body.
The gravitational force is a force that attracts any two objects with mass. We call the gravitational force attractive because it always tries to pull masses together, it never pushes them apart. In fact, every object, including you, is pulling on every other object in the entire universe! This is called Newton's Universal Law of Gravitation.