India's Mangalyaan was sent to Mars by launch | vedclass

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(C) The transfer orbit is an elliptical path with the Sun at one of the foci. The semi-major axis $a_{tr}$ of this transfer orbit is the average of th

Vedclass · Accepted Answer

$260$

Vedclass · Answer

(C) The transfer orbit is an elliptical path with the Sun at one of the foci. The semi-major axis $a_{tr}$ of this transfer orbit is the average of the semi-major axes of Earth's and Mars' orbits:$a_{tr} = \frac{a_e + a_m}{2} = \frac{1.5 	imes 10^{11} + 2.28 	imes 10^{11}}{2} = 1.89 	imes 10^{11} \, m$According to Kepler's third law, $T^2 \propto a^3$. Let $T_e$ be the orbital period of Earth $(1 \, 	ext{year} = 365 \, 	ext{days})$ and $T_{tr}$ be the period of the transfer orbit:$\left( \frac{T_{tr}}{T_e} ight)^2 = \left( \frac{a_{tr}}{a_e} ight)^3$$T_{tr} = T_e 	imes \left( \frac{1.89 	imes 10^{11}}{1.5 	imes 10^{11}} ight)^{3/2} = 365 	imes (1.26)^{1.5} \approx 365 	imes 1.415 \approx 516.5 \, 	ext{days}$The time taken to travel from Earth to Mars is half of the full orbital period of the transfer orbit:$t = \frac{T_{tr}}{2} = \frac{516.5}{2} \approx 258.25 \, 	ext{days}$Rounding to the nearest given option, the time is approximately $260 \, 	ext{days}$.

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