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)
Answers
Given:
Two uniform solid spheres
Radius R
Center to center separation = 6R
Mass of Sphere 1 = M
Mass of Sphere 2 = 4M
A projectile of mass m is projected from the surface of sphere of mass M is directly towards center of the second sphere
To find:
The expression for minimum speed v of the projectile so that it reaches the surface of second sphere.
Solution:
By formula,
Kinetic energy = 1/2 m v^2
Where,
v - the velocity of the object and
m - mass.
Hence,
To find the gravitational potential energy potential ( GMM / R ),
Mass + gravitational potential energy
Substituting the values,
We get,
Ei = Kinetic energy - GMM / R - G 4mm / R
Consider,
v = 0,
Therefore,
Kinetic energy = 0
Substituting,
n = -GMM / 2R - G 4mm / 4R
By the law of conservation of energy,
ei = n
Hence,
( 1/2 ) m v - GMM / R - G4mm / R = - GMM / 2 R - G4mm / 4R
Calculate v,
v = ( 3 GM / 5R ) 1/2
Hence, v = ( 3 GM / 5R ) 1/2 is the expression for minimum speed v of the projectile so that it reaches the surface of second sphere.
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