A small block of superdense material has a mass of 3x10^(24) kg. It is situated at height h (much smaller than the earth's radius) from where it falls on the earth's surface. Find its speed when its height from the earth's surface has reduced to h/2. The mass of the earth is 6x10^(24) kg.
Answers
Thanks for asking the question!
ANSWER::
Given => h < < < R
Earth mass , m₁ = 6 x 10²⁴ kg
Mass of block , m₂ = 3 x 10²⁴ kg
Let velocity of earth be v₁
Let velocity of block be v₂
Two blocks are attracted by gravitational force of attraction.
Gravitational potential energy stored will be kinetic energy of two blocks.
m₁ x m₂ [ 1 / { R + (h/2)} - 1 / (R + h) ] = (1/2) m₁ x v₁² + (1/2) m₂ x v₂² [ Equation 1 ]
As internal force acts ,
m₁v₁ = m₂v₂
v₁ = m₂v₂ / m₁ [ Equation 2 ]
Putting value in Equation 1
m₁ x m₂ [ 2 / (2R + h) - 1 / (R + h) ] = (1/2) x (m₁ x m₂²v₂² / m₁² ) x v₁² + (1/2)m₂ x v₂²
m₁ x m₂ [ 2 / (2R + h) - 1 / (R + h) ] = (1/2) x m₂² x v₂² [ (m₂/m₁) + 1 ]
m₁ x m₂ [ (2R + 2h - 2R - h) / (2R + h)(R + h) ] =(1/2) x v₂² x [ (3 x 10²⁴ / 6 x 10²⁴) +1]
[(m₁ x m₂ x h) / (2R² + 3Rh + h²)] = (1/2) x v₂² x 3/2
As said above h < < < R , it can be neglected
(m₁ x m₂ x h) / (2R²) = (1/2) x v₂² x 3/2
v₂ = √(2gh/3)
Hope it helps!
