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(B) Using the principle of conservation of mechanical energy, the total energy at the surface equals the total energy at the maximum height $h$ reache

Vedclass · Accepted Answer

$914.28$

Vedclass · Answer

(B) Using the principle of conservation of mechanical energy, the total energy at the surface equals the total energy at the maximum height $h$ reached by the rocket.Total energy at surface = Total energy at height $h$$-\frac{GMm}{R} + \frac{1}{2}mu^2 = -\frac{GMm}{R+h} + 0$Dividing by $m$ and using $GM = gR^2$:$-\frac{gR^2}{R} + \frac{1}{2}u^2 = -\frac{gR^2}{R+h}$$-gR + \frac{1}{2}u^2 = -\frac{gR^2}{R+h}$Given $u = 4000 \ m/s$, $R = 6.4 	imes 10^6 \ m$, and $g = 10 \ m/s^2$:$-(10)(6.4 	imes 10^6) + \frac{1}{2}(4000)^2 = -\frac{10 	imes (6.4 	imes 10^6)^2}{6.4 	imes 10^6 + h}$$-6.4 	imes 10^7 + 8 	imes 10^6 = -\frac{4.096 	imes 10^{14}}{6.4 	imes 10^6 + h}$$-5.6 	imes 10^7 = -\frac{4.096 	imes 10^{14}}{6.4 	imes 10^6 + h}$$6.4 	imes 10^6 + h = \frac{4.096 	imes 10^{14}}{5.6 	imes 10^7} = 0.7314 	imes 10^7 = 7.314 	imes 10^6 \ m$$h = 7.314 	imes 10^6 - 6.4 	imes 10^6 = 0.914 	imes 10^6 \ m = 914 \ km$.Thus, the correct option is $914.28 \ km$.

$A$ rocket is fired vertically with a speed of $4 \ km/s$ from the Earth's surface. How far from the Earth does the rocket go before returning to the Earth (in $km$)? (Take radius of Earth $R = 6.4 \times 10^6 \ m$ and $g = 10 \ m/s^2$)

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