A satellite of mass m is launched vertically | vedclass

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Question

(B) $1$. Apply energy conservation from the surface to height $R$ (distance $2R$ from the center):$\frac{1}{2}mu^2 - \frac{GMm}{R} = \frac{1}{2}mv^2 -

Vedclass · Accepted Answer

$5 m\left(u^{2}-\frac{119}{200} \frac{GM}{R}\right)$

Vedclass · Answer

(B) $1$. Apply energy conservation from the surface to height $R$ (distance $2R$ from the center):$\frac{1}{2}mu^2 - \frac{GMm}{R} = \frac{1}{2}mv^2 - \frac{GMm}{2R}$$\frac{1}{2}v^2 = \frac{1}{2}u^2 - \frac{GM}{2R} \Rightarrow v = \sqrt{u^2 - \frac{GM}{R}}$$2$. At height $R$, the satellite (mass $m$) ejects a rocket (mass $m/10$). The remaining satellite (mass $9m/10$) enters a circular orbit at distance $2R$. The orbital velocity is $v_o = \sqrt{\frac{GM}{2R}}$.$3$. By conservation of linear momentum in the radial and tangential directions:Radial: $\frac{m}{10} v_r = m v = m \sqrt{u^2 - \frac{GM}{R}} \Rightarrow v_r = 10 \sqrt{u^2 - \frac{GM}{R}}$Tangential: $\frac{m}{10} v_T = \frac{9m}{10} v_o = \frac{9m}{10} \sqrt{\frac{GM}{2R}} \Rightarrow v_T = 9 \sqrt{\frac{GM}{2R}}$$4$. Kinetic energy of the rocket $(K_r = \frac{1}{2} (m/10) (v_r^2 + v_T^2))$:$K_r = \frac{m}{20} \left( 100(u^2 - \frac{GM}{R}) + 81(\frac{GM}{2R}) \right)$$K_r = \frac{m}{20} \left( 100u^2 - 100\frac{GM}{R} + 40.5\frac{GM}{R} \right) = \frac{m}{20} \left( 100u^2 - 59.5\frac{GM}{R} \right)$$K_r = 5m \left( u^2 - 0.595 \frac{GM}{R} \right) = 5m \left( u^2 - \frac{119}{200} \frac{GM}{R} \right)$.

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