Two uniform solid spheres of equal radii R, but mass M and 4 M have a center to centre separation 6 R, 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. Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere.

Explanation

`implies` The gravitational force of spheres of mass M and 4M acted on the projectile body of mass m. Suppose, the resultant force on point N will be zero.
Suppose `DN = r " " :. NC = 6R – r`
Because, distance between centres of spheres is 6R.
The resultant force on point N is zero. `:. (GMm)/r^2=(4GMm)/((6R -r)^2)`
`:. (6R -r)^(2) =4r^(2) `
`:. 6R - r = 2r " or " 6R - r = -2r`
`:. 6R = 3r " " :. 6R =-r`
`:. r = 2R " " :. r = - 6R`
The kinetic energy of body of mass m on the surface of mass `MK_(i) =1/2 mv^(2)`
where velocity of body of mass m is v and potential energy
`U_(i) =-(GMm)/R-(4GMm)/((6R-R))`
`= -(GMm)/R-(4GMm)/(5R)=(9GMm)/(5R)`
`:.` Mechanical energy `E_(i) = K_(i) +U_(i)`
`=1/2mv^(2) mv^(2) -(9GMm)/(5) " "...(1)`
Now at neutral point N, the velocity of projectile body is zero. Hence, mechanical energy at this point,
`E_(N) =-(GMm)/r-(4GMm)/((6R-r))" "[:. K_(N)=0]`
`=(GMm)/(2r)-(4GMm)/(4R)" "[ :. r = 2R]`
`=-(6GMm)/(4R)`
`:. E_N=-(3GMm)/(2R) " "...(2)`
From conservation of mechanical of mechanical energy,
`E_(i) = E_N`
`1/2mv^(2)-(9GMm)/(5R)=-(3GMm)/(2R)`
`" "[:.` From equaiton (1) and (2)]
`:. 1/2mv^(2)=(9GMm)/(5R)-(3GMm)/(2R)`
`=(3GMm)/(10R)`
`:. v^(2) =3/5(GM)/R`
`:. v = sqrt((3GM)/(5R))`