India's Mangalyan was sent to the Mars by launching it into a transfer orbit EOM around the sun. It leaves the earth at E and meets Mars at M. If the semi-major axis of earth's orbit is ae = 1.5 × 1011 m, that of Mars' orbit am = 2.28 × 1011 m, then Kepler's laws give the estimate of time for Mangalyan to reach Mars from Earth to be close to: (a) 500 days (b) 320 days (c) 260 days (d) 220 days

Answer: (c). The transfer orbit EOM is half an ellipse with semi-major axis a = (ae + am)/(2) = (1.5 + 2.28)/(2) × 10¹¹ = 1.89 × 10¹¹ m. Kepler's third law: T = Te(a/ae)3/2 = 365 × (1.26)3/2 days. (1.26)3/2 ≈ 1.414, so T ≈ 516 days. Earth to Mars is half the orbit: t = T/2 ≈ 258 days. So the answer is (c) 260 days.

India's Mangalyan was sent to the Mars by launching it into a… | JEE Main 2014 (9 Apr, Shift 1)

Q26JEE Main 2014 · 9 APR · SHIFT 1Difficult

Kepler's Laws & Planetary Motion

India's Mangalyan was sent to the Mars by launching it into a transfer orbit EOM around the sun. It leaves the earth at E and meets Mars at M. If the semi-major axis of earth's orbit is ae = 1.5 $\times$ 1011 m, that of Mars' orbit am = 2.28 $\times$ 1011 m, then Kepler's laws give the estimate of time for Mangalyan to reach Mars from Earth to be close to:

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Answer: (c)

The transfer orbit EOM is half an ellipse with semi-major axis $a = \frac{a_e + a_m}{2} = \frac{1.5 + 2.28}{2} \times 10^{11} = 1.89 \times 10^{11}$ m.
Kepler's third law: $T = T_e\left(\frac{a}{a_e}\right)^{3/2} = 365 \times (1.26)^{3/2}$ days.
$(1.26)^{3/2} \approx 1.414$ , so $T \approx 516$ days.
Earth to Mars is half the orbit: $t = \frac{T}{2} \approx 258$ days.
So the answer is (c) 260 days.

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