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6 December, 03:58

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)

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  1. 6 December, 04:31
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    (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) ]
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