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 Mar's orbit $a_m=2.28\times 10^{11}m$, taken 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 1

Answer: (C)

Consider the trajectory to be an Ellipse, Semi-major axis, $a=\frac{a_e+a_m}{2}=1.89\times 10^{11}m$
Semi-minor axis, $b\simeq a_e=1.5\times 10^{11}m$
Now, We consider a circle whose area is equal to the path covered by ellipse, $\therefore \pi R^2=\frac{\pi ab}{2}$
$\therefore R=\sqrt{\frac{ab}{2}}$
$=\sqrt{\frac{1.89\times 10^{11}\times 1.5\times 10^{12}}{2}}$
$=1.19\times 10^{11}m$
Applying kepler's law, $\therefore T^2\propto r^3$
$\therefore T^2\propto r^3$
$\therefore {\left(\frac{T}{365}\right)}^2={\left(\frac{R}{a_e}\right)}^3$
$\therefore T^2={\left(\frac{1.19}{1.5}\right)}^3\times 3.65^2$
$\therefore T\simeq 260$ days.
Velocity $=\frac{2\pi R}{V}=\frac{2\pi \times 1.19\times 10^{11}}{260\times 24\times 3600}=33.28Km/s$