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, $b \simeq a _{ e }= 1 . 5 \times 1 0 ^{11} m$
Now, We consider a circle whose area is equal to the path covered by ellipse, $\therefore \pi R ^{2}=\frac{\pi ab }{2}$
$\therefore R =\sqrt{\frac{ a b }{ 2 }}$
$=\sqrt{\frac{1.89 \times 10^{11} \times 1.5 \times 10^{12}}{2}}$
$=1.19 \times 10^{11} m$
Applying kepler's law, $\therefore T ^{2} \propto r ^{3}$
$\therefore T ^{2} \propto r ^{3}$
$\therefore \left(\frac{ T }{ 3 6 5 }\right)^{2}=\left(\frac{ R }{ a _{e}}\right)^{3}$
$\therefore T ^{2}=\left(\frac{ 1 . 1 9 }{ 1 . 5 }\right)^{3} \times 3.65^{2}$
$\therefore T \simeq 2 6 0$ days.
Velocity $=\frac{2 \pi R }{ V }=\frac{2 \pi \times 1.19 \times 10^{11}}{ 2 6 0 \times 24 \times 3600}= 3 3 . 28\, K m / s$