An earth satellite $S$ has an orbit radius which is $4$ times that of a communication satellite $C$. The period of revolution of $S$ is ........ $days$.

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?

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 $............. \times 10^8 \, km$.

According to Kepler, the period of revolution of a planet $(T)$ and its mean distance from the sun $(r)$ are related by the equation:

An earth satellite $X$ is revolving around the earth in an orbit whose radius is one-fourth of the radius of the orbit of a communication satellite. The time period of revolution of $X$ is ..........

$A$ satellite is launched into a circular orbit of radius $R$ around the Earth. $A$ second satellite is launched into an orbit of radius $1.02 R$. The period of the second satellite is larger than the first one by approximately ........ $\%$