A graph of kinetic energy $K$ of an asteroid versus $\frac{1}{r}$ (where $r$ is the distance from the centre of the planet $A$) is shown in the figure. Asteroid falls directly towards the centre of the planet $A$. What is the value of mass of the asteroid? ($M$ is the mass of planet $A, G$ is universal gravitational constant in SI units)

From a solid sphere of mass $M$ and radius $R\text{,}$ a spherical portion of radius $\left(\frac{R}{2}\right)$ is removed as shown in the figure. Taking gravitational potential $V=0$ at $r=\infty ,$ the potential at the centre of the cavity thus formed is ($G=$gravitational constant)

Consider two solid spheres of radii $R_1=1 \mathrm{m}$,$R_2=2 \mathrm{m}$ and masses $M_1$ and $M_2,$ respectively. The gravitational field due to sphere $\left(1\right)$ and $\left(2\right)$ are shown. The value of $\frac{M_1}{M_2}$ is:

A spherically symmetric gravitational system of particles has mass density $\rho =\left\{\begin{array}{l}{\rho }_0\text{ for }r\le R\\ 0\text{ for }r>R\end{array}\right.$ where, ${\rho }_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $v$ as a function of distance $r(0<r<\infty )$ from the centre of the system is represented by