Question

Two uniform solid spheres of equal radii R but mass M and 4M have a centre of separation 6R.The two spheres are held fixed.A projectile of mass m is projected from the surface of sphere of mass M is directly towards center of the second sphere. Find the expression for minimum speed v of the projectile so that it reaches the surface of second sphere.

Answer 1

In this case, we see a neutral point (N) where two gravitational forces of both bodies cancel each other. Now, the position of this neutral point will depend on the distribution of the public. As shown in the mouth, it will always be kept close to the light body. According to the example, due to both the people (which represents a neutral point), the position of the point can be found by the gravitational force equation. It was found that R = 2R body mass is projected from M to 4 M, it will have some mechanical energy, which is the sum of potential and dynamic energy. The dynamic energy of the body is simply (1/2) MV, where V is the motion of the object and M is its mass. Gravitational potential energy potential (GMM / R) will be in the form of energy.

Therefore, due to M and 4 M, the particle particle on the surface of mass M + gravitational potential is the dynamic energy of energy.

It seems, EI = (1/2) MV - GMM / R - G4 mm / R

Similarly, we can exclude the mechanical energy of the particle at point N, which will be here, we will consider two people and V = 0, Ke = 1/2 MV = 0, N here, here

Therefore, N = -GMM ​​/ 2R - G4mm / 4R

Now, in fact we can use the law of conservation of energy, so ei = n

Or (1/2) MV-GMM / R-G4mm / R = -GMM ​​/ 2RG-4mm / 4R

How can we calculate v

V = (3 GM / 5R) 1/2

To avoid gravity group M, the minimum speed required by the particles goes up to 4 meters.

Answer 2

Given:

Two uniform solid spheres

Radius R

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.