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JEE Main 20149 AprilOnline
Exam
JEE Main
Year
2014
Date
9 April
Mode
Online
Subject
Physics
Chapter
Gravitation

JEE Main 2014 · 9 April · Online · Physics · Gravitation

India's Mangalyan was sent to the Mars by launching it into a...

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}^{1 1} \text{m} , that of Mar's orbit am=2.28×1011m{\text{a}}_{\text{m}} = \text{2.28} \times 1 {0}^{1 1} \text{m}, taking Kepler's laws, give the estimate of time for Mangalyan to reach Mars from Earth.
India's Mangalyan was sent to the Mars by launching it into a transfer orbit EOM around the sun. It leaves the earth at

Options

  1. A220220 days.
  2. B500500 days.
  3. C260260 days.
  4. D320320 days.

Correct answer

C. 260260 days.

Step-by-step solution

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}^{1 1} \text{m}

From Kepler's law,

T2∝R3{\text{T}}^{2} ∝ {\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\approx518days∴ {\text{T}}_{\text{Mn}}\text{\approx}518\text{days}

∴∴ Required time =5182=260days= \dfrac{5 1 8}{2} = 2 6 0 \text{days}.
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