Two planets orbit a star in circular paths with radii \(R\) and \(4R,\) respectively. At a specific time, the two planets and the star are aligned in a straight line. If the orbital period of the planet closest to the star is \(T,\) what is the minimum time after which the star and the planets will again be aligned in a straight line?

The kinetic energies of a planet in an elliptical orbit around the Sun, at positions \(A,B~\text{and}~C\) are \(K_A, K_B~\text{and}~K_C\) respectively. \(AC\) is the major axis and \(SB\) is perpendicular to \(AC\) at the position of the Sun \(S\), as shown in the figure. Then:

1. \(K_A <K_B< K_C\)
2. \(K_A >K_B> K_C\)
3. \(K_B <K_A< K_C\)
4. \(K_B >K_A> K_C\)

Kepler's third law states that the square of the period of revolution (\(T\)) of a planet around the sun, is proportional to the third power of average distance \(r\) between the sun and planet i.e. \(T^2 = Kr^3\), here \(K\) is constant. If the masses of the sun and planet are \(M\) and \(m\) respectively, then as per Newton's law of gravitation, the force of attraction between them is \(F = \frac{GMm}{r^2},\) here \(G\) is the gravitational constant. The relation between \(G\) and \(K\) is described as:
1. \(GK = 4\pi^2\)
2. \(GMK = 4\pi^2\)
3. \(K =G\)
4. \(K = \frac{1}{G}\)

Two astronauts are floating in gravitation-free space after having lost contact with their spaceship. The two will:

Two spherical bodies of masses \(M\) and \(5M\) and radii \(R\) and \(2R\) are released in free space with initial separation between their centres equal to \(12R.\) If they attract each other due to gravitational force only, then the distance covered by the smaller body before the collision is:

A spherical planet has a mass \(M_p\) and diameter \(D_p\)${\mathrm{}}_{}$. A particle of mass \(m\) falling freely near the surface of this planet will experience acceleration due to gravity equal to: