Question

How far from the centre of the Moon is the Earth- Moon neutral point, where the earth and the moon’s gravitational field strengths are equal in magnitude but opposite in direction?
${M_E} = 6.0 \times {10^{24}}\;{\rm{kg}}$ ${M_M} = 7.4 \times {10^{22}}\;{\rm{kg}}$
The radius of Moon’s orbit (assumed to be circular) is:
A. $3.8 \times {10^2}\;{\rm{m}}$
B. $38 \times {10^6}\;{\rm{m}}$
C. $3.8 \times {10^4}\;{\rm{m}}$
D. $28 \times {10^2}\;{\rm{m}}$

Answer 1

The above problem can be resolved using the earth's gravitational strength and the moon's mathematical relation. These gravitational strengths at equilibrium get equated and provide the value for the distance of the equilibrium point. Then the final answer is calculated by resolving the equation containing the values of gravitational strength.

Complete step by step answer:
Given:
The mass of the earth is, ${M_E} = 6.0 \times {10^{24}}\;{\rm{kg}}$.
The mass of the moon is, ${M_M} = 7.4 \times {10^{22}}\;{\rm{kg}}$.
Let d be the distance between the earth and the moon such that its standard value is $d = 380 \times {10^6}\;{\rm{m}}$.
Let the equilibrium point form the earth be at x distance. Then the strength of gravity of earth is equivalent to the strength of gravity at moon.
The strength of gravity at earth is, ${g_e} = 4 \times {10^{14}}\;{{\rm{m}}^{\rm{3}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}$.
The strength of gravity at moon is, ${g_m} = 5 \times {10^{12}}\;{{\rm{m}}^{\rm{3}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}$.
Then at equilibrium,

\dfrac{{{g_e}}}{{{x^2}}} = \dfrac{{{g_m}}}{{{{\left( {d - x} \right)}^2}}}\\\ {g_e}\left( {{x^2} - 2xd + {d^2}} \right) = {g_m}{x^2}\\\ \left( {{g_e} - {g_m}} \right){x^2} - 2d{g_e}x + {d^2}{g_e} = 0 \end{array}$$ Further solving the above quadratic equation as, $$\begin{array}{l} x = \dfrac{{2d{g_e} \pm \sqrt {4{d^2}g_e^2 - 4{d^2}{g_e}\left( {{g_e} - {g_m}} \right)} }}{{2\left( {{g_e} - {g_m}} \right)}}\\\ x = \dfrac{{d{g_e} \pm d\sqrt {{g_m}{g_e}} }}{{{g_e} - {g_m}}}\\\ x = \dfrac{{\left( {380 \times {{10}^6}\;{\rm{m}}} \right) \times \left( {4 \times {{10}^{14}}\;{{\rm{m}}^{\rm{3}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \right) + 380 \times {{10}^6}\;{\rm{m}}\sqrt {5 \times {{10}^{12}}\;{{\rm{m}}^{\rm{3}}}{\rm{/}}{{\rm{s}}^{\rm{2}}} \times 4 \times {{10}^{14}}\;{{\rm{m}}^{\rm{3}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} }}{{4 \times {{10}^{14}}\;{{\rm{m}}^{\rm{3}}}{\rm{/}}{{\rm{s}}^{\rm{2}}} - 5 \times {{10}^{12}}\;{{\rm{m}}^{\rm{3}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}}}\\\ x = 342 \times {10^6}\;{\rm{m}} \end{array}$$ The distance from the moon is given as, $$\begin{array}{l} {d_1} = d - x\\\ {d_1} = 380 \times {10^6}\;{\rm{m}} - 342 \times {10^6}\;{\rm{m}}\\\ {d_1} = 38 \times {10^6}\;{\rm{m}} \end{array}$$ Therefore, the required distance from the centre of the moon is $$38 \times {10^6}\;{\rm{m}}$$ and option B is correct. **Note:** To resolve the given problem, it is necessary to go through the concept of gravity and the gravitational field strength around the surface of the earth and the surface of the moon. These concepts provide a clear idea regarding the equilibrium points along with the positions of equilibrium points. Moreover, the gravitational force's basic knowledge between the two heavenly masses can also bring about the conclusion. Furthermore, it is to be remembered that the gravitational strength depends on the distance of the equilibrium point from the center of the planet.