Answer:
(dr/dt) = √(v₀² + k²/r - k²/R)
Explanation:
v² = v₀² + k²[(1/r) - (1/R)]
v = velocity of Body launched from the surface of the earth
v = initial velocity of body
k = a constant
r = distance from the centre of the earth
R = radius of the earth
v² = v₀² + k²/r - k²/R
v = √(v₀² + k²/r - k²/R)
But v = dr/dt
(dr/dt) = √(v₀² + k²/r - k²/R)
(dr/√(v₀² + k²/r - k²/R)) = dt
To go one step beyond and integrate the differential equation
∫ (dr/√(v₀² + k²/r - k²/R)) = ∫ dt
Integrating the left hand side from 0 to r and the right hand side from 0 to t
Note, v₀, k and R are all constants
(- 4r²/k²)[√(v₀² + k²/r - k²/R)] = t
√(v₀² + k²/r - k²/R) = (- k² t/4r²)
(v₀² + k²/r - k²/R) = (- k² t/4r²)²
(v₀² + k²/r - k²/R) = (k⁴t²/16r⁴)
(k⁴t²/16r⁴) + (k²/r) = [v₀² - (k²/R)]
k² [(k²t²/16r⁴) + (1/r)] = [v₀² - (k²/R)]
[(k²t²/16r⁴) + (1/r)] = [(v₀²/k²) - (1/R)]
