If the distance of the earth from the Sun is $1.5 \times 10^8 \, km$. Then the distance of an imaginary planet from the Sun,if its period of revolution is $2.83$ years,is $............. \times 10^8 \, km$.

The mean radius of the earth's orbit around the sun is $1.5 \times 10^{11} \ m$. The mean radius of the orbit of Mercury around the sun is $6 \times 10^{10} \ m$. Mercury will revolve around the sun in:

Let the speed of the planet at the perihelion $P$ in the figure be $v_{P}$ and the Sun-planet distance $SP$ be $r_{P}$. Relate $\{r_{P}, v_{P}\}$ to the corresponding quantities at the aphelion $\{r_{A}, v_{A}\}$. Will the planet take equal times to traverse $BAC$ and $CPB$?

Two planets are at mean distances $d_1$ and $d_2$ from the sun,and their orbital frequencies are $n_1$ and $n_2$ respectively. Then:

$A$ satellite $S_1$ of mass $m$ is moving in an orbit of radius $r$. Another satellite $S_2$ of mass $2m$ is moving in an orbit of radius $2r$. The ratio of the time period of satellite $S_2$ to that of $S_1$ is

$A$ planet takes $200$ days to complete one revolution around the Sun. If the distance of the planet from the Sun is reduced to one-fourth of the original distance,how many days will it take to complete one revolution?