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:

Identify the correct option.

Match the column $-I$ with column $-II$ for a satellite in circular orbit:

Column $-I$	Column $-II$
$(A)$ Kinetic energy	$(p)$ $-\frac{GM_Em}{2r}$
$(B)$ Potential energy	$(q)$ $\sqrt{\frac{GM_E}{r}}$
$(C)$ Total energy	$(r)$ $-\frac{GM_Em}{r}$
$(D)$ Orbital velocity	$(s)$ $\frac{GM_Em}{2r}$

(where $M_E$ is the mass of the earth,$m$ is the mass of the satellite,and $r$ is the radius of the orbit)

Three particles,each of mass $M$,are situated at the vertices of an equilateral triangle of side length $L$. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original separation $L$. The initial speed that should be given to each particle is

$A$ spherical body of radius $R$ consists of a fluid of constant density $\rho$ and is in equilibrium under its own gravity. If $P(r)$ is the pressure at a distance $r$ from the center $(r < R)$,then the correct option$(s)$ is(are):
$(A) P(r=0) = P_c$ (maximum pressure at center)
$(B) \frac{P(r=3R/4)}{P(r=2R/3)} = \frac{63}{80}$
$(C) \frac{P(r=3R/5)}{P(r=2R/5)} = \frac{16}{21}$
$(D) \frac{P(r=R/2)}{P(r=R/3)} = \frac{20}{27}$

$A$ solid sphere of radius $R$ gravitationally attracts a particle of mass $m$ placed at a distance of $3R$ from its centre with a force $F_{1}$. Now,a spherical cavity of radius $\frac{R}{2}$ is made in the sphere (as shown in the figure) such that the cavity is tangent to the centre of the sphere and the outer surface. The force on the particle becomes $F_{2}$. The value of $F_{1} : F_{2}$ is: