$A$ projectile is thrown straight upward from the Earth's surface with an initial speed $v = \alpha v_E$,where $\alpha$ is a constant and $v_E$ is the escape speed. The projectile travels up to a height $h = 800 \ km$ from the Earth's surface before it comes to rest. The value of the constant $\alpha$ is (Radius of the Earth $R = 6400 \ km$):

Two uniform solid spheres of equal radii $R$,but mass $M$ and $4M$ have a centre-to-centre separation $6R$,as shown in the figure. The two spheres are held fixed. $A$ projectile of mass $m$ is projected from the surface of the sphere of mass $M$ directly towards the centre of the second sphere. Obtain an expression for the minimum speed $v$ of the projectile so that it reaches the surface of the second sphere.

The change in the gravitational potential energy when a body of mass $m$ is raised to a height $h = nR$ above the surface of the Earth is (where $R$ is the radius of the Earth).

$A$ body is thrown from the surface of the earth with velocity $u \ m \ s^{-1}$. The maximum height in meters above the surface of the earth up to which it will reach is ($R=$ radius of the earth,$g=$ acceleration due to gravity).

$A$ body is projected vertically upwards from the earth's surface. If its kinetic energy of projection is equal to half of its minimum value required to escape from the gravitational influence,then the height up to which it rises is ($R =$ radius of the earth).

Energy required to move a body of mass $m$ from an orbit of radius $2R$ to $3R$ is