The figure shows the orbit of a planet $P$ around the sun $S.$ $AB$ and $CD$ are the minor and major axes of the ellipse,respectively. If $U$ is the potential energy and $K$ is the kinetic energy of the planet,then $|U| > |K|$ at:

$A$ satellite of mass $m$ is at a distance $a$ from a star of mass $M$. The speed of the satellite is $u$. Suppose the law of universal gravity is $F = -G \frac{Mm}{r^{2.1}}$ instead of $F = -G \frac{Mm}{r^2}$. Find the speed of the satellite when it is at a distance $b$ from the star.

The relative uncertainty in the period of a satellite orbiting around the Earth is $10^{-2}$. If the relative uncertainty in the radius of the orbit is negligible,the relative uncertainty in the mass of the Earth is:

In the following four periods:
$(i)$ Time of revolution of a satellite just above the earth's surface $({T_{st}})$
$(ii)$ Period of oscillation of mass inside the tunnel bored along the diameter of the earth $({T_{ma}})$
$(iii)$ Period of simple pendulum having a length equal to the earth's radius in a uniform field of $9.8 \; N/kg \; ({T_{sp}})$
$(iv)$ Period of an infinite length simple pendulum in the earth's real gravitational field $({T_{is}})$

$A$ gas molecule of mass $M$ at the surface of the Earth has kinetic energy equivalent to $0\,^{\circ}C$. If it were to go up straight without colliding with any other molecules,how high would it rise? Assume that the height attained is much less than the radius of the Earth. ($k_B$ is Boltzmann constant)

Three identical spheres of mass $m$ are placed at the vertices of an equilateral triangle of side length $a$. When released,they interact only through gravitational force and collide after a time $T = 4 \text{ s}$. If the sides of the triangle are increased to length $2a$ and the masses of the spheres are made $2m$,then they will collide after . . . . . . seconds.