Question

two heavenly bodies having mass 'm1' and 'm2' are separated by a distance 'd' what happens if the distance between them isreduced by one third keeping their masses constant?​

Answer 1

Explanation:

The forces of attraction between the particles are the internal forces. Therefore, the center of mass of the system will have no acceleration. The particles move, but the center of mass will continue to be at the same place. At the time of collision, the two particles are at one place and the center of mass will also be at that place. As the center of mass does not move, the collision will take place at the center of mass.

Distance of center of mass from M

1

=

M

1

+M

2

M

1

(0)+M

2

(R)

=

M

1

+M

2

M

2

R

Distance of center of mass form M

2

=R−

M

1

+M

2

M

2

R

=

M

1

+M

2

M

1

R

Ratio of the distances=

M

1

M

2