Question

Two uniform solid spheres of equal radii $R$ but mass $M$ and $4M$ have a centre-to-centre separation $6R$ as shown in 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. $N$ is the point where net gravitational is zero.

With reference to the above situation, match the Column I (quantities) with Column II (mathematical expressions) and select the correct answer from the codes given below.

Column I		Column II
A	Distance $r$ of neutral point $N$	1	$\sqrt{\frac{3GM}{5R}}$
B	Minimum speed of the projectile,$v_{\min }$, so that it reaches the surface of the second sphere.	2	$2R$
C	The speed with which the projectile hits the seconds sphere, if projected with $v_{\min }$.	3	$\sqrt{\frac{27GM}{5R}}$

• A. $A-(1),B-(2),C-(3)$
• B. $A-(3),B-(2),C-(1)$
• C. $A-(2),B-(1),C-(3)$
• D. $A-(3),B-(1),C-(2)$

Answer 1

Answer: (C)

A. The projectile is acted upon by two mutually opposing gravitational forces of the two spheres. Hence, there must be a point on the line $OC$, where $F_{\text{ext }}=0$, i.e. net external force due to gravitational attraction force vanishes. The point is called neutral point $N$. Let it be at a distance $r$ from $O$.
$\Rightarrow ON=r$

At $N,F_1-F_2$
$\Rightarrow \frac{GmM}{r^2}=\frac{Gm(4M)}{(6R-r{)}^2}$
$\Rightarrow r=2R$ or $-6R$
The neutral point $r=-6R$ is not relevant.
Thus, $ON=r=2R$.
B. To project the projectile with minimum speed from $M$. It is sufficient to project the particle with a speed which would enable it to reach $N$. After $N$,
projectile will be attracted by $4M$.
At the neutral point, speed approaches zero,
i.e. $v_N=0$
Total energy of the particle at $N=E_N$
$\Rightarrow E_N=\text{ GPE due to }M+\text{ GPE due to }4M$
$E_N=\frac{GmM}{2R}-\frac{4GMm}{4R}$ ....(i)
$[\because$ GPE due to $M=-\frac{GmM}{2R};r=2R$
GPE due to $4M=-\frac{4GMm}{4R};r=4R$ ]
From the principle of conservation of mechanical energy,
$\frac{1}{2}m{v}^2-\frac{GmM}{R}-\frac{4GMm}{5R}=-\frac{GMm}{2R}-\frac{4GMm}{4R}$
$\Rightarrow v=\sqrt{\frac{3GM}{5R}}$
$\Rightarrow v_{\min }=\sqrt{\frac{3GM}{5R}}$
Therefore, minimum speed of the projectile with which it reaches the surface of second sphere is $\sqrt{\frac{3GM}{5R}}$.
C. Let $v_f$ be the speed with which projectile hits the second sphere. Applying principle of conservation of energy, we get
$\Rightarrow v_f=\sqrt{\frac{27GM}{5R}}$
Hence, $A\to 2,B\to 1$ and $C\to 3$.