Irodov 1.231: Where Earth and Moon Pulls Cancel – Amit & Deepali Sinha Foundation

Problem Irodov 1.231

The Earth (mass $M_1$ ) and the Moon (mass $M_2$ ) are separated, center to center, by $nR$ , where $n=60$ and $R$ is the Earth’s own radius. The Earth’s mass is $\eta=81$ times the Moon’s, $M_1=\eta M_2$ . Find the distance $r$ from the Moon’s center at which the resultant of the Earth’s and the Moon’s gravitational field strengths is zero.

The Earth (left, mass M_1) and the Moon (right, mass M_2), not to scale, their centers a distance nR apart. — The Earth (left, mass $M_1$ ) and the Moon (right, mass $M_2$ ), not to scale, their centers a distance $nR$ apart.

Given

$\eta=81$
$n=60$
$R$

Find

r

Solution

Step 1Equal and opposite field strengths

The balance point P on the line of centers, a distance r from the Moon's center within the separation nR. The pull F_1 toward the Earth and the pull F_2 toward the Moon are drawn equal in length. — The balance point $P$ on the line of centers, a distance $r$ from the Moon’s center within the separation $nR$ . The pull $F_1$ toward the Earth and the pull $F_2$ toward the Moon are drawn equal in length.

At the sought point $P$ , distance $r$ from the Moon’s center and $nR-r$ from the Earth’s center, the two field strengths are equal in magnitude; writing each as $\gamma M/(\text{distance})^2$ and using $M_1=\eta M_2$ eliminates both masses in favor of the ratio $\eta$ .

\begin{align*}\frac{\gamma M_1}{(nR-r)^2} &= \frac{\gamma M_2}{r^2} \\ \left(\frac{nR-r}{r}\right)^2 &= \frac{M_1}{M_2}\end{align*}

\left(\frac{nR-r}{r}\right)^2=\eta

(1)

Step 2Solve for

r

, then substitute the numbers

Taking the positive square root, since $nR-r$ and $r$ are both lengths, turns the balance condition (1) into a linear equation for $r$ .

\begin{align*}\sqrt\eta &= \frac{nR-r}{r} \\ r\sqrt\eta &= nR-r \\ r\left(1+\sqrt\eta\right) &= nR\end{align*}

With $\eta=81$ , $\sqrt\eta=9$ , so $r$ is one tenth of the full separation $nR$ ; substituting the Earth’s radius $R$ from the standard table gives the numeric distance.

r=6R=3.82\times10^7\ \text{m}

r=\frac{nR}{1+\sqrt\eta}\approx3.8\times10^4\ \text{km}

(2)

Check

As a limiting case, equal masses ( $\eta=1$ ) give $r=nR/2$ , the midpoint of the separation, exactly as symmetry demands. Dimensionally $r$ is $nR$ divided by a pure number, so it is a length, as required.

Remark

Because the divisor is $1+\sqrt\eta$ rather than $1+\eta$ , the balance point sits far farther from the Moon than dividing by $1+\eta$ would place it: that alternative divisor gives $r=nR/82=4.66\times10^6\ \text{m}$ , eight times closer to the Moon than the true $r=nR/10=3.82\times10^7\ \text{m}$ , so taking the square root of the mass ratio, rather than the ratio itself, pushes the null point much farther out toward the heavier body. As $\eta\to\infty$ the divisor grows only as $\sqrt\eta$ , so $r\to0$ and the point closes in on the lighter body’s own center rather than receding without bound, a general feature of this balance for any two-body mass ratio (for a real, finite-sized Moon the point instead reaches the surface once $\eta$ is large enough, and then ceases to exist, rather than migrating into the interior).