A satellite of mass m revolves around a planet of mass M in a radius n times the radius of the moon (R). Due to cosmic dust, it experiences a resistance F = αv². Find how long it will stay in the orbit?

For a satellite in a circular orbit of radius $r$ around a body of mass $M$, the gravitational force provides the centripetal force:

$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$

From this, the orbital velocity $v$ is:

$$v = \sqrt{\frac{GM}{r}}$$

The total mechanical energy $E$ of the satellite is the sum of its kinetic energy ($K$) and potential energy ($U$):

$$E = K + U = \frac{1}{2}mv^2 - \frac{GMm}{r}$$

Substituting the expression for $v^2$:

$$E = \frac{1}{2}m\left(\frac{GM}{r}\right) - \frac{GMm}{r} = -\frac{GMm}{2r}$$

The resistance force due to cosmic dust is given as $F_r = \alpha v^2$. This force dissipates energy. The rate of energy dissipation (power $P$) is:

$$P = F_r \cdot v = (\alpha v^2)v = \alpha v^3$$

The rate of change of the satellite's total energy is equal to the negative of the power dissipated (since energy is being lost):

$$\frac{dE}{dt} = -P = -\alpha v^3$$

Substitute the expressions for $E$ and $v$ into the energy loss equation:

$$\frac{d}{dt}\left(-\frac{GMm}{2r}\right) = -\alpha \left(\sqrt{\frac{GM}{r}}\right)^3$$ $$\frac{GMm}{2r^2} \frac{dr}{dt} = -\alpha \frac{(GM)^{3/2}}{r^{3/2}}$$

Rearrange the equation to solve for $\frac{dr}{dt}$:

$$\frac{dr}{dt} = -\frac{2\alpha}{GMm} r^2 \frac{(GM)^{3/2}}{r^{3/2}}$$ $$\frac{dr}{dt} = -\frac{2\alpha (GM)^{3/2}}{GMm} r^{2 - 3/2}$$ $$\frac{dr}{dt} = -\frac{2\alpha \sqrt{GM}}{m} \sqrt{r}$$

Separate the variables and integrate from the initial radius $r_i = nR$ to the final radius $r_f = R$ (assuming the satellite eventually crashes on the Moon's surface):

$$\frac{dr}{\sqrt{r}} = -\frac{2\alpha \sqrt{GM}}{m} dt$$ $$\int_{nR}^{R} r^{-1/2} dr = -\frac{2\alpha \sqrt{GM}}{m} \int_{0}^{T} dt$$ $$\left[2\sqrt{r}\right]_{nR}^{R} = -\frac{2\alpha \sqrt{GM}}{m} T$$ $$2\sqrt{R} - 2\sqrt{nR} = -\frac{2\alpha \sqrt{GM}}{m} T$$ $$2\sqrt{R}(1 - \sqrt{n}) = -\frac{2\alpha \sqrt{GM}}{m} T$$

Solve for $T$:

$$T = -\frac{m}{2\alpha \sqrt{GM}} \cdot 2\sqrt{R}(1 - \sqrt{n})$$ $$T = \frac{m}{\alpha \sqrt{GM}} \sqrt{R}(\sqrt{n} - 1)$$

This can be rewritten as:

$$T = \frac{m}{\alpha} (\sqrt{n} - 1) \sqrt{\frac{R}{GM}}$$

Let's examine option c: $\frac{m}{\alpha}\frac{(\sqrt{n}-1)}{v}$.
If $v$ in option c represents the final orbital velocity at the Moon's surface, $v_f = \sqrt{\frac{GM}{R}}$, then substituting this into option c gives:

$$T = \frac{m}{\alpha} \frac{(\sqrt{n}-1)}{\sqrt{GM/R}}$$ $$T = \frac{m}{\alpha} (\sqrt{n}-1) \sqrt{\frac{R}{GM}}$$

This exactly matches our derived result. (Note: We assume 'a' in the options refers to $\alpha$, the resistance coefficient).

The final answer is $\text{c)} \frac{m}{\alpha}\frac{(\sqrt{n}-1)}{v}$.