Let $M_E$ be the mass of the Earth. At the surface, $g = GM_E/R^2$, so $GM_E = gR^2$. The orbital radius is $r=R+h=2R$.

[Q1] Pre-collision Energy For a single satellite in a circular orbit at $r=2R$: The gravitational force provides the centripetal force: $\frac{GM_E m}{r^2} = \frac{mv^2}{r}$. The orbital speed squared is $v^2 = \frac{GM_E}{r} = \frac{gR^2}{2R} = \frac{gR}{2}$. The kinetic energy is $K = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{gR}{2}\right) = \frac{mgR}{4}$. The potential energy is $U = -\frac{GM_E m}{r} = -\frac{gR^2 m}{2R} = -\frac{mgR}{2}$. The total mechanical energy of one satellite is $E_1 = K + U = \frac{mgR}{4} - \frac{mgR}{2} = -\frac{mgR}{4}$. The total energy for the two-satellite system is $E_{total} = 2E_1 = 2\left(-\frac{mgR}{4}\right) = -\frac{mgR}{2}$.

$E_{total} = -\frac{(200 \text{ kg})(10 \text{ m/s}^2)(6.4 \times 10^6 \text{ m})}{2} = -6.4 \times 10^9 \text{ J}$.

[Q2] Post-collision Speed The satellites have equal mass and opposite velocities, so the total momentum before collision is zero. Since the collision is perfectly inelastic, they stick together and their velocity after the collision is zero. The combined mass $2m$ is momentarily at rest at orbital radius $r=2R$. We use conservation of energy for the fall of the combined mass from $r=2R$ to $r=R$. Initial energy at $r=2R$: $E_i = K_i + U_i = 0 - \frac{GM_E(2m)}{2R} = -\frac{gR^2(2m)}{2R} = -mgR$. Final energy at $r=R$: $E_f = K_f + U_f = \frac{1}{2}(2m)v_f^2 - \frac{GM_E(2m)}{R} = mv_f^2 - \frac{gR^2(2m)}{R} = mv_f^2 - 2mgR$. By energy conservation, $E_i = E_f$:

$$-mgR = mv_f^2 - 2mgR$$ $$mv_f^2 = mgR \implies v_f = \sqrt{gR}$$ $v_f = \sqrt{(10 \text{ m/s}^2)(6.4 \times 10^6 \text{ m})} = \sqrt{64 \times 10^6} = 8 \times 10^3 \text{ m/s}$.