To derive the expression for $x$ where the net gravitational force is zero, we need to analyze the forces acting on an object placed at a distance $x$ from the center of the planet along the line connecting the planet and the moon.

Step-by-step derivation:

1. Gravitational force exerted by the planet: The gravitational force between the planet (mass $M$) and the object (mass $m_o$) at a distance $x$ from the planet is given by Newton's law of gravitation: $F_p = \frac{GMm_o}{x^2}$ where $G$ is the gravitational constant.

2. Gravitational force exerted by the moon: The moon is at a distance $D$ from the planet. The distance between the object and the moon is $D - x$. The gravitational force between the moon (mass $m$) and the object is: $F_m = \frac{Gmm_o}{(D - x)^2}$

3. Condition for equilibrium (Net force = 0): At the point where the object feels no net gravitational force, the forces due to the planet and the moon must be equal in magnitude but opposite in direction. Thus, we set the forces equal: $\frac{GMm_o}{x^2} = \frac{Gmm_o}{(D - x)^2}$

4. Simplify the equation: Cancel out $G$ and $m_o$ from both sides: $\frac{M}{x^2} = \frac{m}{(D - x)^2}$ Now, take the square root of both sides to simplify further: $\frac{\sqrt{M}}{x} = \frac{\sqrt{m}}{D - x}$

5. Cross multiply and solve for $x$: $\sqrt{M}(D - x) = \sqrt{m}x$ Expanding the left side: $\sqrt{M}D - \sqrt{M}x = \sqrt{m}x$ Combine like terms: $\sqrt{M}D = \sqrt{m}x + \sqrt{M}x$ Factor out $x$ on the right side: $\sqrt{M}D = x(\sqrt{m} + \sqrt{M})$ Finally, solve for $x$: $x = \frac{\sqrt{M}D}{\sqrt{M} + \sqrt{m}}$

Final Expression:

$x = \frac{\sqrt{M}D}{\sqrt{M} + \sqrt{m}}$

This is the distance from the planet where the net gravitational force on the object is zero.

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Would you like further details or have any questions?

Here are 5 related questions for further practice:

1. What happens to the value of $x$ if the mass of the moon is much smaller than the mass of the planet?
2. How does the expression for $x$ change if the object is placed between the moon and the planet but closer to the moon?
3. Derive the expression for the point where the forces due to both the planet and the moon are equal, not necessarily zero.
4. If $D$ increases, how does it affect the position of the zero-force point?
5. What would happen if the moon and planet were of equal mass?

Tip: When dealing with gravitational forces, always remember that the force depends inversely on the square of the distance between two masses.