Question

India's Mangalyan was sent to the Mars 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 $a_m=2.28\times {10}^{11}$, taken Kepler's laws given the estimate of time for 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 $260days$

Solution:

$r=\frac{1.5+2.28}{2}=1.89$

$\frac{T_m}{T_e}=(\frac{1.89}{1.5}{)}^{\frac{3}{2}}$

$t_m=\frac{T_m}{2}={\left(\frac{1.89}{1.5}\right)}^{\frac{3}{2}}$

$\Rightarrow \frac{365}{2}$ x$1.41=257.4days$

257.4 is approximately equal to 260 days and close to option C

Hence the correct option: C