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(B) The initial orbital speed of the satellite is $V_0 = \sqrt{\frac{GM}{R_e}}$.After firing the rockets,the new speed is $V = \sqrt{\frac{3}{2}} V_0

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$3$

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(B) The initial orbital speed of the satellite is $V_0 = \sqrt{\frac{GM}{R_e}}$.After firing the rockets,the new speed is $V = \sqrt{\frac{3}{2}} V_0 = \sqrt{\frac{3GM}{2R_e}}$.Using the conservation of angular momentum at the perigee $(R_e)$ and apogee $(R_{max} = R)$:$L_{initial} = L_{final} \implies m V R_e = m V' R$$V' = \frac{V R_e}{R} = \frac{R_e}{R} \sqrt{\frac{3GM}{2R_e}}$Using the conservation of mechanical energy:$-\frac{GMm}{R_e} + \frac{1}{2} m V^2 = -\frac{GMm}{R} + \frac{1}{2} m V'^2$$-\frac{GM}{R_e} + \frac{1}{2} \left(\frac{3GM}{2R_e}\right) = -\frac{GM}{R} + \frac{1}{2} \left(\frac{R_e^2}{R^2} \cdot \frac{3GM}{2R_e}\right)$$-\frac{1}{R_e} + \frac{3}{4R_e} = -\frac{1}{R} + \frac{3R_e}{4R^2}$$-\frac{1}{4R_e} = \frac{-4R + 3R_e}{4R^2}$$-R^2 = -4R R_e + 3R_e^2 \implies R^2 - 4R R_e + 3R_e^2 = 0$$(R - 3R_e)(R - R_e) = 0$Since $R > R_e$,we get $R = 3R_e$.

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