Solution 3

Derivation

1. Angular Velocity of the System

The moon orbits the planet with an angular velocity $\omega$. Assuming $M \gg m$, the center of mass is effectively at the planet’s center.

$$ \omega^2 = \frac{GM}{R^3} $$

2. Forces on the Object

Consider an object of mass $m_0$ on the surface of the moon closest to the planet. The distance of this object from the planet’s center is $(R – r)$. For the object to be in “weightlessness” (i.e., normal reaction $N = 0$), the net gravitational force must provide exactly the required centripetal force for it to move in a circle of radius $(R-r)$ with angular velocity $\omega$.

Forces acting on the object:

• Gravitational pull from Planet ($F_M$) acting towards the left (towards planet).
• Gravitational pull from Moon ($F_m$) acting towards the right (towards moon center).

Equation of motion:

$$ F_M – F_m = m_0 \omega^2 (R – r) $$ $$ \frac{GM m_0}{(R – r)^2} – \frac{Gm m_0}{r^2} = m_0 \left( \frac{GM}{R^3} \right) (R – r) $$

3. Approximation and Solution

Canceling $G$ and $m_0$ and using the approximation $R \gg r$ (using binomial expansion $(1-x)^{-2} \approx 1+2x$):

$$ \frac{M}{R^2(1 – r/R)^2} – \frac{m}{r^2} = \frac{M}{R^3}(R – r) $$ $$ \frac{M}{R^2} \left( 1 + \frac{2r}{R} \right) – \frac{m}{r^2} \approx \frac{M}{R^2} \left( 1 – \frac{r}{R} \right) $$

Expanding terms:

$$ \frac{M}{R^2} + \frac{2Mr}{R^3} – \frac{m}{r^2} = \frac{M}{R^2} – \frac{Mr}{R^3} $$

Canceling $\frac{M}{R^2}$ from both sides:

$$ \frac{2Mr}{R^3} – \frac{m}{r^2} = -\frac{Mr}{R^3} $$

Rearranging to solve for $R$:

$$ \frac{3Mr}{R^3} = \frac{m}{r^2} $$ $$ 3M r^3 = m R^3 $$ $$ R^3 = \frac{3M}{m} r^3 $$