Suppose a planet goes around the Sun with a linear speed twice as fast as that of the Earth. What will be its orbit size compared to that of the Earth? (Radius of Earth $= R$)

The time period of a geostationary satellite is $24 \; h$,at a height $6 R_{E}$ ($R_{E}$ is the radius of the Earth) from the surface of the Earth. The time period of another satellite whose height is $2.5 R_{E}$ from the surface will be:

If the earth is at one-fourth of its present distance from the sun,the duration of the year will be

$A$ 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 ........ $\%$

The kinetic energies of a planet in an elliptical orbit about the Sun,at positions $A, B$ and $C$ are $K_A, K_B$ and $K_C$,respectively. $AC$ is the major axis and $SB$ is perpendicular to $AC$ at the position of the Sun $S$ as shown in the figure. Then

The figure shows the elliptical path $abcd$ of a planet around the sun $S$ such that the area of triangle $csa$ is $\frac{1}{4}$ of the area of the ellipse. With $db$ as the major axis and $ca$ as a chord perpendicular to the major axis,if $t_1$ is the time taken for the planet to travel along the path $abc$ and $t_2$ is the time taken for the path $cda$,then: