$A$ satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth's radius $R_e$. By firing rockets attached to it,its speed is instantaneously increased in the direction of its motion so that it becomes $\sqrt{\frac{3}{2}}$ times larger. Due to this,the farthest distance from the centre of the earth that the satellite reaches is $R$. The value of $R$ is $....R_e$.

$A$ planet is moving around the sun in an elliptical orbit at different positions $A, B, C, D$. The maximum kinetic energy of the planet is at position:

$A$ geostationary satellite above the equator is orbiting around the earth at a fixed distance $r_1$ from the center of the earth. $A$ second satellite is orbiting in the equatorial plane in the opposite direction to the earth's rotation,at a distance $r_2$ from the center of the earth,such that $r_1 = 1.21 r_2$. The time period of the second satellite as measured from the geostationary satellite is $\frac{24}{p}$ hours. The value of $p$ is $....$

The distances of two planets $A$ and $B$ from the sun are $r_A$ and $r_B$ respectively. Given that $r_B = 100 r_A$. If the orbital speed of planet $A$ is $v$,then the orbital speed of planet $B$ is:

$A$ satellite is to be placed in an equatorial geostationary orbit around the Earth for communication.
$(a)$ Calculate the height of such a satellite.
$(b)$ Find out the minimum number of satellites that are needed to cover the entire Earth,so that at least one satellite is visible from any point on the equator.
Given: $M = 6 \times 10^{24} \ kg$,$R = 6400 \ km$,$T = 24 \ h$,$G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$.

The correct formula for the height $h$ of a satellite from the Earth's surface is: