A hot air balloon is carrying some passengers, and a few sandbags of mass $1\ \mathrm{kg}$ each so that its total mass is $480\ \mathrm{kg}$. Its effective volume giving the balloon its buoyancy is $V$. The balloon is floating at an equilibrium height of $100\ \mathrm{m}$. When $N$ number of sandbags are thrown out, the balloon rises to a new equilibrium height close to $150\ \mathrm{m}$ with its volume $V$ remaining unchanged. If the variation of the density of air with height $h$ from the ground is $\rho(h)=\rho_0 e^{-h/h_0}$, where $\rho_0=1.25\ \mathrm{kg\,m^{-3}}$ and $h_0=6000\ \mathrm{m}$, the value of $N$ is ________.

Step 1: Identify the given parameters.

The initial total mass, heights, and density parameters are given as follows:

$$M_1 = 480\text{ kg}$$ $$h_1 = 100\text{ m}$$ $$h_2 = 150\text{ m}$$ $$\rho(h) = \rho_0 e^{-h/h_0}$$ $$h_0 = 6000\text{ m}$$

Step 2: Apply the equilibrium condition at the initial height.

At the initial equilibrium height $h_1$, the buoyant force equals the total gravitational force.

$$\rho(h_1) V g = M_1 g$$ $$\rho(h_1) V = M_1$$

Step 3: Apply the equilibrium condition at the final height.

After throwing out $N$ sandbags of mass $1\text{ kg}$ each, the new mass is $M_2 = M_1 - N$.

$$\rho(h_2) V = M_2$$ $$\rho(h_2) V = M_1 - N$$

Step 4: Relate the two equilibrium states.

Divide the second equilibrium equation by the first to eliminate the volume $V$.

$$\frac{M_1 - N}{M_1} = \frac{\rho(h_2)}{\rho(h_1)}$$

Step 5: Calculate the ratio of the densities.

Substitute the density function into the ratio.

$$\frac{\rho(h_2)}{\rho(h_1)} = \frac{\rho_0 e^{-h_2/h_0}}{\rho_0 e^{-h_1/h_0}}$$ $$\frac{\rho(h_2)}{\rho(h_1)} = e^{-(h_2 - h_1)/h_0}$$

Substitute the given values $h_1 = 100\text{ m}$, $h_2 = 150\text{ m}$, and $h_0 = 6000\text{ m}$.

$$\frac{\rho(h_2)}{\rho(h_1)} = e^{-(150 - 100)/6000}$$ $$\frac{\rho(h_2)}{\rho(h_1)} = e^{-1/120}$$

Step 6: Solve for the number of sandbags $N$.

Equate the mass ratio to the density ratio.

$$\frac{480 - N}{480} = e^{-1/120}$$ $$N = 480 \left(1 - e^{-1/120}\right)$$

Using the approximation $e^{-1/120} \approx 0.9917$:

$$N \approx 480 (1 - 0.9917)$$ $$N \approx 480 (0.0083)$$ $$N \approx 3.98$$

Since the number of sandbags $N$ must be an integer, we round to the nearest whole number.

$$N = 4$$

Answer: 4