Example

Question:

Two uniform solid spheres of equal radii \(R\), but mass \(M\) and \(4M\), have a centre-to-centre separation \(6R\). 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 of the projectile so that it reaches the surface of the second sphere.

Solution:

The projectile is acted upon by two mutually opposing gravitational forces of the two spheres. The neutral point \(N\) is defined as the position where the two forces cancel each other exactly. If \(ON = r\), then

\[ \frac{GMm}{r^2} = \frac{4GMm}{(6R - r)^2} \] Solving, we get \(r^2 = 4r^2\) and \(r = 2R\) (the other root is not relevant).
The mechanical energy at the surface of \(M\) is: \[ E_i = \frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{4GMm}{5R} \] At the neutral point \(N\), the speed is zero, so the total energy is: \[ E_N = -\frac{GMm}{2R} - \frac{4GMm}{4R} \] Using conservation of mechanical energy: \[ \frac{1}{2}v^2 - \frac{GM}{R} - \frac{4GM}{5R} = -\frac{GM}{2R} - \frac{GM}{R} \] \[ v^2 = \frac{2GM}{R}\left(\frac{4}{5}-\frac{1}{2}\right) \] \[ v = \sqrt{\frac{3GM}{5R}} \]
Note: The speed of the projectile is zero at \(N\), but it is nonzero when it strikes the heavier sphere \(4M\).