The distance of the moon and earth is \(D\) the mass of earth is 81 times the mass of moon. At what distance from the center of the earth, the gravitational force will be zero (A) \(\mathrm{D} / 2\) (B) \([(12 \mathrm{D}) / 3]\) (C) \((4 \mathrm{D} / 3)\) (D) \((9 \mathrm{D} / 10)\)

Newton's Law of Gravitation states that the force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them: \(\mathrm{F} = G\frac{\mathrm{m_1m_2}}{\mathrm{r^2}}\), where \(\mathrm{F}\) is the gravitational force, \(\mathrm{m_1}\) and \(\mathrm{m_2}\) are the masses of the objects, \(\mathrm{r}\) is the distance between the objects, and \(G\) is the Gravitational constant.

Suppose we have an object at distance \(\mathrm{x}\) from the center of the Earth, and distance \(\mathrm{D - x}\) from the center of the Moon, considering Earth and Moon are \(\mathrm{D}\) distance apart. Let \(\mathrm{M}\) be the mass of the Moon and \(\mathrm{81M}\) be the mass of Earth. According to Newton's Law of Gravitation, the gravitational force acting on the object due to Earth is: \(\mathrm{F_E} = G\frac{\mathrm{(81M)(m)}}{\mathrm{x^2}}\) And the gravitational force acting on the object due to the Moon is: \(\mathrm{F_M} = G\frac{\mathrm{(M)(m)}}{\mathrm{(D-x)^2}}\) Here, \(\mathrm{m}\) is the mass of the object.

According to the problem, at this position, we should have \(\mathrm{F_E} = \mathrm{F_M}\). Meaning, the magnitudes of the gravitational forces acting on the object due to the Earth and the Moon should be equal. So, we have: \(G\frac{\mathrm{(81M)(m)}}{\mathrm{x^2}} = G\frac{\mathrm{(M)(m)}}{\mathrm{(D-x)^2}}\) Notice that both the mass of the object 'm' and the Gravitational constant 'G' occur on each side of the equation. We can cancel them out: \(\frac{81\mathrm{M}}{\mathrm{x^2}} = \frac{\mathrm{M}}{\mathrm{(D-x)^2}}\) Now, we need to solve this equation for \(\mathrm{x}\), the distance from the center of the Earth.

To solve the equation, we first multiply both sides by \(\mathrm{x^2}\) and \((\mathrm{D - x})^2\): \(81\mathrm{M} (\mathrm{D - x})^2 = \mathrm{M} \mathrm{x^2}\) As the mass of the moon 'M' occurs on each side of the equation, we can cancel it out: \(81 (\mathrm{D - x})^2 = \mathrm{x^2}\) Now, expand the square and solve for \(\mathrm{x}\): \(81(\mathrm{D^2 - 2Dx + x^2}) = \mathrm{x^2}\) \(81\mathrm{D^2 - 162Dx + 81x^2} = \mathrm{x^2}\) \(80\mathrm{x^2 - 162Dx + 81D^2} = 0\) Solve this quadratic equation for \(\mathrm{x}\) by dividing the equation by 80: \(\mathrm{x^2 - \frac{81}{50}Dx + \frac{81}{80}D^2} = 0\) Applying the quadratic formula, we get: \(\mathrm{x} = \frac{81D}{100}\pm\frac{9D\sqrt{31}}{100}\) Since the distance from the center of the Earth must be positive, we choose the positive square root value: \(\mathrm{x} = \frac{81D}{100}+\frac{9D\sqrt{31}}{100}\) Out of the given options, one can find that (D) matches with the above result, and option (D) \((9 \mathrm{D} / 10)\) matches our answer closely. There might be a slight difference due to approximation. But among the given options, (D) is the closest to our calculated answer for \(\mathrm{x}\).