1. Variable Mass formulation

Consider the spaceship when it has penetrated a distance $x$ into the cloud. The volume of dust swept out is $S \cdot x$. The mass of this dust is $m_{dust} = \rho S x$. Since the dust sticks to the spaceship, the total mass of the system at position $x$ is: $$ M(x) = m + \rho S x $$

2. Conservation of Linear Momentum

Since there are no external forces (engine is off, gravity is neglected), the linear momentum of the system remains conserved. The initial momentum (before entering the cloud) is $P_{initial} = m u$. The momentum at distance $x$ is $P(x) = M(x) v(x)$.

$$ m u = (m + \rho S x) v $$

Solving for velocity $v$:

$$ v = \frac{mu}{m + \rho S x} $$

3. Calculating Time

We write velocity as $v = \frac{dx}{dt}$: $$ \frac{dx}{dt} = \frac{mu}{m + \rho S x} $$ $$ dt = \frac{m + \rho S x}{mu} dx $$

To find the total time $T$ to cross the cloud of length $l$, we integrate from $x = 0$ to $x = l$: $$ \int_0^T dt = \int_0^l \frac{m + \rho S x}{mu} dx $$ $$ T = \frac{1}{mu} \int_0^l (m + \rho S x) dx $$ $$ T = \frac{1}{mu} \left[ mx + \frac{\rho S x^2}{2} \right]_0^l $$ $$ T = \frac{1}{mu} \left( ml + \frac{\rho S l^2}{2} \right) $$

Simplifying the expression:

$$ T = \frac{ml}{mu} + \frac{\rho S l^2}{2mu} $$