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JEE Main 2014 · 9 Apr · Shift 1, Physics question 7

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=1.5×1011a_e=1.5\times10^{11} m, that of Mar's orbit am=2.28×1011a_m=2.28\times10^{11} m, taken Kepler's laws give the estimate of time for Mangalyan to reach Mars from Earth to be close to :

The Sun at the centre of two circles, Earth's orbit (radius a_e) inside Mars' orbit (radius a_m). The transfer orbit is half an ellipse from E on Earth's orbit (right) round through O at the top to M on Mars' orbit (left, opposite E).
Choose one option
Answer and worked solution

Official answer: C

The transfer ellipse has semi-major axis a=1.5+2.282×1011=1.89×1011a=\frac{1.5+2.28}2\times10^{11}=1.89\times10^{11} m. By Kepler's third law,
T=1 yr×(1.891.5)3/2=1.41 yr.T=1\text{ yr}\times\left(\frac{1.89}{1.5}\right)^{3/2}=1.41\text{ yr}.
The trip E → M is half the orbit: 0.707×365≈2580.707\times365\approx258 days, close to 260 days.

Topics:Kepler's lawsSatellites and orbital motion
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