The escape velocity (v ) of a body depends upon the mass (m) of body, gravitational

acceleration (g) and radius(R) of the planet .Derive the relation for escape velocity

dimensionally.

Dimension of velocity v = [LT^{-1}]

Dimension of G = [M^{-1}L^{3}T^{-2} ]

Dimension of mass M = [M]

Dimension of Radius R = [L]

Now assuming a relation of the form

v α G^{x}R^{y}M^{z}

=> v = k G^{x}R^{y}M^{z}

=> [LT^{-1}] = k[M^{-1}L^{3}T^{-2} ]^{x}[L]^{y }[M^{z}]

=> [LT^{-1}] = k[M^{-x+z}L^{3x+y }T^{-2x} ]

Comparing the exponents of the dimensions

L^{1 } = L^{3x+y}

=> 3x+y = 1

T^{-2x} = T^{-1}

=> 2x = 1

=> x = ½

y = 1 – 3x

=> y = 1 – 3( ½ )

=> y = -½

M^{-x+z} = M^{0}

=> -x+z = 0

=> z = x

=> z = ½

Putting the values of x,y and z in the assumed relation

v = kG^{1/2 }R^{-1/2}M^{1/2}

=> v = k √(GM/R)

By actual theoretical calculation k = √2

v = √2 √(GM/R)

=> v = √(2GM/R)

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