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?

$M_E=6.0\times {10}^{24}kg{M}_M=7.4\times {10}^{22}kg$
The radius of Moon's orbit (assumed to be circular) is: $3.8\times {10}^{8}m$.

• A. $3.8\times {10}^{2}m$
• B. $38\times {10}^{6}m$
• C. $38\times {10}^{4}m$
• D. $28\times {10}^{6}m$

Answer 1

The correct option is B $38\times {10}^{6}m$

Strength of gravity $\frac{Gm}{r^2}$

Let $g_e$ be the value of $Gm$ of earth $=4\times {10}^{14}\frac{m^3}{s^2}$

$g_m$ be the value of $Gm$ of moon $=5\times {10}^{12}\frac{m^3}{s^2}$

Let $d\to$distance between earth and moon $=380\times {10}^{6}m$

Let equilibirum point from earth be at a distanexe :- $x$

$\frac{g_e}{x^2}=\frac{g_m}{{(d-x)}^2}$

$g_e(x^2-2xd+d^2)=g_m{x}^2(g_e-g_m)x^2-{2dg}_{e}x+d^2{g}_e=0$

Solving the quadratic formula

$x=\frac{{2dg}_e\pm \sqrt{4d^2{g}_e^2-{4d}^2{g}_e(g_e-g_m)}}{2(g_e-g_m)}=\frac{{dg}_e\pm d\sqrt{g_e{g}_m}}{g_e-g_m}\approx 342\times {10}^{6}m$

$\therefore$ Distance from moon :-$d-x=38\times {10}^{6}m$