The period of a satellite, in a circular orbit near an equatorial plane, will not depend on

A satellite of mass $m$ is in a circular orbit of radius $2R_E$ about the earth. The energy required to transfer it to a circular orbit of radius $4R_E$ is (where $M_E$ and $R_E$ is the mass and radius of the earth respectively)

A spherical planet far out in space has a mass ${M_0}$ and diameter ${D_0}$. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to

$Assertion$ : The escape speed does not depend on the direction in which the projectile is fired.
$Reason$ : Attaining the escape speed is easier if a projectile is fired in the direction the launch site is moving as the earth rotates about its axis.

A satellite moving with velocity $v$ in a force free space collects stationary interplanetary dust at a rate of $\frac{{dM}}{{dt}} = \alpha v$ where $M$ is the mass (of satellite + dust) at that instant . The instantaneous acceleration of the satellite is

A satellite is launched into a circular orbit of radius $R$ around earth, while a second satellite is launched into a circular orbit of radius $1.02\, {R}$. The percentage difference in the time periods of the two satellites is -