Two stationary particles of masses `M_(1)` and `M_(2)` are 'd' distance apart. A third particle lying on the line joining the particles, experiences no resultant gravitational force. What is the distance of this particle from `M_(1)` ?

Two stationary particles of masses M_(1) and M_(2) are at a distance d apart. A third particle, lying on the line joining the particles experiences no resultant gravitational forces. What is the distance of this particle from M_(1) .

Two masses 'M' and '4M' are at a distance 'r' apart on the line joining them. 'P' is point where the resultant gravitational force is zero (such a point is called as null point). The distance of 'P' from the mass '4M' is

Two particles of masses m_(1) and m_(2) (m_(1) gt m_(2)) are separated by a distance d. The shift in the centre of mass when the two particles are interchanged.

.Two particles of masses "m_(1)" and m_(2)(m_(1)>m_(2))" are separated by a distance ' "d" '.The shift in the centre of mass when the two particles are interchanged.

Two particle of masses m and 2m are at a distance 3r apart at the ends of a straight line AB . C is the centre of mass of the system. The magnitude of the gravitational force on a unit mass placed at C due to the masses is

Mass M is distributed uniformly along a line of length 2L . A particle of mass m is at a point that is at a distance a above the centre of the line on the its perpendicular bisector (Point P in figure). The gravitational force that the line exert on the particle is

Two particles of masses m_1 and m_2 are kept at a separation of d. when a third particle is kept at a distance of d/3 from m_1 , then it does not experience any net force. Determine the ratio (m_2)/(m_1) .

Two particles A and B having masses M and 4M respectively are kept at a distance 2.73 m apart. Another small particle of mass m is to be placed so that the net gravitational force on it is zero. What will be its distance from body A ?