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Two uniform solid spheres of equal radii $R$, but mass $M$ and $4M$ have a centre separation of $6R$. The two spheres are held fixed. A projectile of mass $m$ is projected from the surface of the sphere $M$ and towards the second sphere along the line joining the centres of the two spheres. Obtain an expression for the minimum speed $v$ of the projectile so it reaches the surface of the second sphere.

I was looking at the answer of this problem and noticed that the neutral point (i.e. the point where the forces between the two spheres exactly cancel out) had been calculated and conservation of energy had been applied at the neutral point $N$ and at the surface ($E_s$ being the the mechanical energy at the surface).

$$E_s= \frac 12 mv^2-\frac {GMm}R-\frac{4GMm}{5R} $$ $$E_N= -\frac{GMm}{2R}-\frac{4GMm}{4R}$$

Equating $E_s$ and $E_N$ gives $v=\sqrt\frac{3GM}{5R}$.

What I did not understand was that while writing the mechanical energy at the neutral point $N$ they assumed the kinetic energy of the projectile to be zero.

If the kinetic energy of the particle at $N$ is zero, implying that the particle is stationary, then how would it reach the surface of the second sphere since there is no force pulling it towards $4M$?

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  • $\begingroup$ You start with minimum velocity v to overcome the distance 4R , the ending velocity is zero $\endgroup$
    – Eli
    Commented Jan 16, 2020 at 14:38
  • $\begingroup$ You're saying that the projectile moves with a constant velocity v throughout it's journey from M to 4M? @Eli $\endgroup$
    – Quantum-meatball
    Commented Jan 16, 2020 at 14:54
  • $\begingroup$ Not necessary, the requirement is that the ending velocity is zero $\endgroup$
    – Eli
    Commented Jan 16, 2020 at 14:59
  • $\begingroup$ I don't get it. Can you clarify what you are saying? @Eli $\endgroup$
    – Quantum-meatball
    Commented Jan 16, 2020 at 15:17
  • $\begingroup$ You are right, my mistake $\endgroup$
    – Eli
    Commented Jan 16, 2020 at 15:35

2 Answers 2

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The null point...

... it is unstable i.e. the force on that point is indeed zero but what if we displace the object by $\mathbf{\delta x}$?

What I mean to say is that the system is in unstable equilibrium.

The force on null point $N$ may be zero but the potential energy at the point is maximum suggesting unstable equilibrium.


What is unstable equilibrium?

Assume a pendulum in two positions

(i) normal equilibrium where hinge is above free end

(ii) equilibrium where hinge is below free end and it is perfectly inverted

The first case is stable equilibrium and the second state is unstable equilibrium.


In unstable equilibrium if we displace the object by very small displacement it will not return to its original state or position.



Similarly

In your question we just need the object to reach the unstable point after that even a small negligible displacement toward other sphere (4M) will result in the gravitational force of 4M to take over.

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  • $\begingroup$ If any mistake in answer please notify $\endgroup$
    – Dr_Paradox
    Commented Jan 16, 2020 at 17:46
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In this situation, the minimum speed is a limiting value. Anything above that will carry the projectile beyond the null point.

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  • $\begingroup$ So anything more than the calculated value of v written above. Right? I was confused since the book just writes v=(3GM/5R)^1/[email protected]. Bird $\endgroup$
    – Quantum-meatball
    Commented Jan 16, 2020 at 15:58