GravitationMedium
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
The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 ...Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projectedfrom the midpoint of t...If the earth were to cease rotating about its own axis. The increase in the value of g in C.G.S. system at a place of la...The angular speed of earth in rads/s, so that bodies on equator may apper weightless is : [Use g = 10m/ s2 and the radiu...Height of geostationary satellite is...