Escape Velocity Multiplier for Vertical Launch on Spherical Planet

Let $M$ and $R$ be the mass and radius of the spherical planet, respectively.

Step 1: Determine the maximum height reached by the projectile

The acceleration due to gravity at a distance $r$ from the center of the planet is given by: $g(r) = \frac{GM}{r^2}$

At the surface of the planet ($r = R$), the gravitational acceleration is: $g_0 = \frac{GM}{R^2}$

It is given that at the maximum height (distance $r$ from the center), the gravitational acceleration drops to $\frac{1}{4}^{\text{th}}$ of its surface value: $g(r) = \frac{1}{4} g_0$

Substituting the expressions for gravitational acceleration: $\frac{GM}{r^2} = \frac{1}{4} \frac{GM}{R^2}$

Solving for $r$: $r^2 = 4R^2 \implies r = 2R$

Step 2: Apply the law of conservation of mechanical energy

At the surface of the planet, the mechanical energy $E_i$ of the projectile is: $E_i = \frac{1}{2} m v^2 - \frac{GMm}{R}$

At the maximum height $r = 2R$, the projectile momentarily comes to rest ($v_f = 0$), so its mechanical energy $E_f$ is: $E_f = 0 - \frac{GMm}{2R}$

By conservation of mechanical energy ($E_i = E_f$): $\frac{1}{2} m v^2 - \frac{GMm}{R} = -\frac{GMm}{2R}$

$\frac{1}{2} m v^2 = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$

$v^2 = \frac{GM}{R}$

Step 3: Calculate the escape velocity

The escape speed $v_{\text{esc}}$ from the surface of the planet is defined as: $v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$

Substituting $\frac{GM}{R} = v^2$ into the escape velocity formula: $v_{\text{esc}} = \sqrt{2 v^2} = v \sqrt{2}$

Comparing this with the given relation $v_{\text{esc}} = v\sqrt{N}$: $N = 2$