Two small bodies of masses 10 kg and 20 kg are kept a distnce 1.0 m apart and released. Assuming that only mutual gravitational force are acting, find the speeds of the particles when the separation decreases to 0.5 m.
Answers
Given :
Mass of Body 1 (M)= 20 Kg
Mass of Body 2 (m)= 10 kg
distance between them (R) is 1 m
To Find:
speed of two bodies when sepration is 0.5 m
Solution :
•Gravitational force on two bodies of Mass M & m is
F = (GMm)/R²
•For body 2
•Also, F is mass × acceleration
ma = GMm/R²
a = GM/R²
dV/dt = GM/R²
•Multiplying & dividing by dR
(dV/dR).(dR/dt) = GM/R²
•As dR/dt = V
(V).dV/dR = GM/R²
V.dV = (GM)(dR/R²)
V²/2 = (GM)(-1/R)
from R = 1m to R = 0.5 m
V² = 2GM[-1/1.0-(-1/0.5) ]
V² = 2GM [2-1]
V² = 2GM
V² = 40G
where G is universal constant
V = 2√(10G)
•Now , For body 1
•Also, F is mass × acceleration
Ma = GMm/R²
a = Gm/R²
dV/dt = Gm/R²
•Multiplying & dividing by dR
(dV/dR).(dR/dt) = Gm/R²
•As dR/dt = V
(V).dV/dR = Gm/R²
V.dV = (Gm)(dR/R²)
V²/2 = (Gm)(-1/R)
from R = 1m to R = 0.5 m
V² = 2Gm[-1/1.0-(-1/0.5) ]
V² = 2Gm [2-1]
V² = 2Gm
V² = 20G
where G is universal constant
V = √(20G)