A spaceship is stationed on Mars. How much en | vedclass

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(D) Mass of the spaceship,$m_{s} = 1000 \; kg$. Mass of the Sun,$M = 2 	imes 10^{30} \; kg$. Mass of Mars,$m_{m} = 6.4 	imes 10^{23} \; kg$. Orbital

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$6 	imes 10^{11} \; J$

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(D) Mass of the spaceship,$m_{s} = 1000 \; kg$. Mass of the Sun,$M = 2 	imes 10^{30} \; kg$. Mass of Mars,$m_{m} = 6.4 	imes 10^{23} \; kg$. Orbital radius of Mars,$R = 2.28 	imes 10^{8} \; km = 2.28 	imes 10^{11} \; m$. Radius of Mars,$r = 3395 \; km = 3.395 	imes 10^{6} \; m$. Gravitational constant,$G = 6.67 	imes 10^{-11} \; N m^{2} kg^{-2}$. Total potential energy of the spaceship is the sum of potential energy due to the Sun and Mars: $U = -\frac{GMm_{s}}{R} - \frac{Gm_{m}m_{s}}{r}$. Since the spaceship is stationary on Mars,its kinetic energy is zero. Total energy $E = -Gm_{s} \left( \frac{M}{R} + \frac{m_{m}}{r} ight)$. Energy required to escape the solar system is $E_{req} = -E = Gm_{s} \left( \frac{M}{R} + \frac{m_{m}}{r} ight)$. $E_{req} = 6.67 	imes 10^{-11} 	imes 1000 	imes \left( \frac{2 	imes 10^{30}}{2.28 	imes 10^{11}} + \frac{6.4 	imes 10^{23}}{3.395 	imes 10^{6}} ight)$. $E_{req} = 6.67 	imes 10^{-8} 	imes (8.772 	imes 10^{18} + 1.885 	imes 10^{17}) \approx 6.67 	imes 10^{-8} 	imes 8.96 	imes 10^{18} \approx 5.97 	imes 10^{11} \; J \approx 6 	imes 10^{11} \; J$.

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