For a projectile launched from the vertically from the surface the earth with initial velocity v0, the velocity of the projectile when at distance r from the center of the earth satisifes the equation:

v^2 = v0^2 + k^2 (1/r - 1/R)

Here, k is a positive constant.

Write the differential equation governing the distance r (t)

(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) ]