$A$ mass '$m$' on the surface of the Earth is shifted to a height equal to the radius of the Earth. If '$R$' is the radius and '$M$' is the mass of the Earth,then the work done in this process is:

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.

An object is projected vertically upward from the earth's surface with a velocity $u = \sqrt{Rg}$,where $R$ is the radius of the earth and $g$ is the acceleration due to gravity on the earth's surface. Find the maximum height reached by the object.

$A$ 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 planets have masses $M$ and $16\, M$ and their radii are $a$ and $2\, a$ respectively. The separation between the centres of the planets is $10\, a.$ $A$ body of mass $m$ is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach the surface of the smaller planet,the minimum firing speed needed is

$A$ missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energy is