GravitationHard
Question
Two bodies of masses 1 m and 2 m are initially at rest at infinite distance apart. They are then allowed to move towerds each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is
Options
A.
B.
C.
D.
Solution
By applying law of consservation of monteum,
m1v1 - m2v2 = 0 ⇒ m1v1 ....(i)
Where v1 and v2 are the velocities of masses m1 and m2 at a distance r from each other By conservation of energy,
Change in P.E = change in K.E.
m1v12 = 
m2v22 ....(ii)
Soveling eqn.(i)and (ii) we get
and v2 = 
Relative velocity of approeach, VR
= |v1| + |v2| =
m1v1 - m2v2 = 0 ⇒ m1v1 ....(i)
Where v1 and v2 are the velocities of masses m1 and m2 at a distance r from each other By conservation of energy,
Change in P.E = change in K.E.
m1v12 = m2v22 ....(ii) Soveling eqn.(i)and (ii) we get
and v2 = Relative velocity of approeach, VR
= |v1| + |v2| =
Create a free account to view solution
View Solution FreeMore Gravitation Questions
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the...A particle of mass m is transferred from the centre of the base of a uniform solid hemisphere of mass M and radius R to ...If a satellite orbits as close to the earth′s surface as possible...The kinetic energy needed to project a body of mass m from the earth surface to infinity is :-...A spherical planet has a mass Mp and diameter Dp. A particle of mass m falling freely near the surfaceof this planet wil...