The figure shows the motion of a planet around the sun in an elliptical orbit with sun at the focus. The shaded areas $A$ and $B$ are also shown in the figure which can be assumed to be equal. If ${t_1}$ and ${t_2}$ represent the time for the planet to move from $a$ to $b$ and $d$ to $c$ respectively, then
${t_1} < {t_2}$
${t_1} > {t_2}$
${t_1} = {t_2}$
${t_1} \le {t_2}$
Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ?
A planet takes $200$ days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution
The angular momentum of a planet of mass $M$ moving around the sun in an elliptical orbit is $\overrightarrow{ L }$. The magnitude of the areal velocity of the planet is:
The fig. shows the motion of a planet around the un in an elliptical orbit with sun at the focus. The shaded areas can be assumed to be equal. If $t_1$ and $t_2$ represent the time taken for the planet to move from $A$ to $B$ and $C$ to $D$ respectively, then
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 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 ........ $days$.