A bullet is fired vertically upwards with velocity $v$ from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is $\frac{1}{4}^{\text{th}}$ of its value of the surface of the planet. If the escape velocity from the planet is $v_{\text{esc}} = v\sqrt{N}$, then the value of $N$ is (ignore energy loss due to atmosphere)

Step 1: Relate the acceleration due to gravity at maximum height to the surface gravity.

Let $R$ be the radius of the planet and $h$ be the maximum height reached by the bullet.

$$g_s = \frac{GM}{R^2}$$ $$g(h) = \frac{GM}{(R+h)^2}$$

Step 2: Determine the maximum height $h$.

The acceleration at maximum height is one-fourth of the surface gravity.

$$\frac{GM}{(R+h)^2} = \frac{1}{4}\frac{GM}{R^2}$$ $$\left(\frac{R}{R+h}\right)^2 = \frac{1}{4}$$ $$R+h = 2R$$ $$h = R$$

Step 3: Apply conservation of mechanical energy to find the projection velocity $v$.

The total mechanical energy at the surface equals the total mechanical energy at the maximum height $h = R$.

$$\frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{R+h}$$ $$\frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{2R}$$ $$\frac{1}{2}mv^2 = \frac{GMm}{2R}$$ $$v^2 = \frac{GM}{R}$$

Step 4: Relate the projection velocity to the escape velocity.

The escape velocity $v_{\text{esc}}$ from the surface of the planet is defined as follows.

$$v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$$ $$v_{\text{esc}} = \sqrt{2}v$$ $$N = 2$$

Answer: $N = 2$