Getting back to the old things #4

Since the mass density is uniform over the whole cylindrical planet and the length of this cylinder is very large, so the gravitational field lines will be cylindrically-radially outward w.r.t. the axis of the cylinder and the gravitational field will be same at all points at a distance $r$ from the axis of the cylinder.

Taking an element of length $\ell$ of this cylinder and using the Gauss's law for gravitation, we have

$\begin{aligned} & \oint \mathbf{\vec{E}} \cdot d\mathbf{\vec S} = 4\pi G \mathbf{M} \\ \implies & \mathbf{E} \oint d \mathbf{S} = 4 \pi G ( \rho \pi R^2 \ell ) \\ \implies & \mathbf{E} (2 \pi r \ell ) = 4 \pi G ( \rho \pi R^2 \ell ) \\ \implies & \mathbf{E} = \dfrac{2 \rho \pi G R^2}{r} \end{aligned}$

Let the mass of the particle projected be $m$ . By work energy theorem, we have

$\begin{aligned} & \displaystyle \mathbf{W}_{\text{force of gravity}} = \int_{r_1}^{r_2} F \cdot dr = \Delta K \\ \implies & \displaystyle - \int_{r_1}^{r_2} m \mathbf{E} \cdot dr = 0 - \dfrac{mv^2}{2} \\ \implies & \displaystyle 2 \rho \pi G R^2 m \int_{R}^{3R} \dfrac 1r \,dr = \dfrac{mv^2}{2} \\ \\ \implies & 4 \rho \pi G R^2 \ln (3) = v^2 \\ \implies & v = \boxed{2R \sqrt{\pi \rho G \ln(3)}} \end{aligned}$

Thus $a+b = 2+3 = 5$ .