Question

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 $a_e=1.5\times {10}^{11}m$, that of Mars's orbit $a_m=2.28\times {10}^{11}m,$ taken Kepler's laws give the estimate of time of Mangalyan to reach Mars from Earth to be close to :

• A. $500days$
• B. $320days$
• C. $260days$
• D. $220days$

Answer 1

The correct option is C $260days$

Consider the trajectory to be an Ellipse,

Semi-major axis, $a=\frac{a_e+a_m}{2}=1.89\times {10}^{11}m$

Radius of the earth = $1.5\times {10}^{11}m$

Now,

We consider a circle whose area is equal to the path covered by ellipse,

$\therefore \pi R^2=\frac{\pi ab}{2}$

$\therefore R=\sqrt{\frac{ab}{2}}$

$=\sqrt{\frac{1.89\times {10}^{11}\times 1.5\times {10}^{11}}{2}}$

$=1.19\times {10}^{11}m$

Applying kepler's law,

$\therefore T^2\propto r^3$

$\therefore {\left(\frac{T}{365}\right)}^2={\left(\frac{R}{a_e}\right)}^3$

$\therefore T^2={\left(\frac{1.19}{1.5}\right)}^3\times {3.65}^2$

$\therefore T\simeq 260days.$