Gravitation

JEE Main level, single correct, medium

223 of 223

Question 223 of 223

India's Mangalyan was sent to the Mars by launching it into a transfer orbit EOM around the sun. It leaves the earth at EE and meets Mars at MM. If the semi-major axis of Earth's orbit is ae=1.5×1011m{\text{a}}_{\text{e}}=\text{1.5}\times 1{0}^{11}\text{m} , that of Mar's orbit am=2.28×1011m{\text{a}}_{\text{m}}=\text{2.28}\times 1{0}^{11}\text{m}, taking Kepler's laws, give the estimate of time for Mangalyan to reach Mars from Earth.

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Answer: (C) 260260 days.

Semi major axis of mangalyan,

aMn=ae+am2{\text{a}}_{\text{Mn}}=\dfrac{{\text{a}}_{\text{e}}+{\text{a}}_{\text{m}}}{2}

=1.89×1011m=\text{1.89}\times 1{0}^{11}\text{m}

From Kepler's law,

T2∝R3{\text{T}}^{2}\propto {\text{R}}^{3}

∴(TMnTe)2=(RMRe)3∴{\left(\dfrac{{\text{T}}_{\text{Mn}}}{\text{Te}}\right)}^{2}={\left(\dfrac{{\text{R}}_{\text{M}}}{{\text{R}}_{\text{e}}}\right)}^{3}

∴TMn≈518days∴{\text{T}}_{\text{Mn}}\text{≈}518\text{days}

∴∴ Required time =5182=260days=\dfrac{518}{2}=260\text{days}.

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