Two uniform solid spheres of equal radii R, but mass M & 4M have a centre to centre separation 6R. The two spheres are held fixed. A projectile of mass m is projected frm the surface of the sphere of mass M directly towrds 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.

(diagram of que. is given in the textbook eg.8.4 pg 193)

Hello!

In this example, we define a neutral point (N) where the gravitational forces of the two bodies cancel each other out. Now, the position of this neutral point would depend upon the distribution of masses. It will always be placed closer to the lighter body, as shown in the figure.

The position of the point can be found by equation the gravitational forces due to the two masses (which what a neutral point represents), as done in the example. It was found out to be that r= 2R

A body is projected from mass M to 4M it will have certain mechanical energy, which is the sum of potential and kinetic energy.

The kinetic energy of the body is simply (1/2)mv^{2}, where v is the velocity of the object and m is its mass.

the potential energy would of the form of Gravitational Potential energy of the form - (GMm/r).

So, the mechanical energy of the particle at the surface of mass M is the kinetic energy of the particle + gravitational potential energy due to M and 4M as well.

this comes out to be, **Ei = (1/2)mv**^{2}** - GMm/r - G4Mm/r**

similarly we can find out the mechanical energy of the particle at point N, which would be

*here we will consider the distance of N from the two masses and as v = 0, KE = 1/2mv ^{2} = 0, here*

so,

**En = -GMm/2R - G4Mm/4R**

Now, finally we can use the law of conservation of energy, so

Ei = En

or** **(1/2)mv^{2} - GMm/r - G4Mm/r = -GMm/2R - G4Mm/4R

we can thus calculate v, which would be

v = (3GM/5R)^{1/2}

which is the minimum speed required by the particle to escape the gravitational field of M and go to 4M.

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