Question

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.

Answer 1

Dimension of velocity v = [LT^-1]

Dimension of G = [M^-1L³T^-2 ]

Dimension of mass M = [M]

Dimension of Radius R = [L]

Now assuming a relation of the form

v α G^xR^yM^z

=> v = k G^xR^yM^z

=> [LT^-1] = k[M^-1L³T^-2 ]^x[L]^y [M^z]

=> [LT^-1] = k[M^-x+zL^3x+y T^-2x ]

Comparing the exponents of the dimensions

L¹ = 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 = M⁰

=> -x+z = 0

=> z = x

=> z = ½

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

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

=> v = k √(GM/R)

By actual theoretical calculation k = √2

v = √2 √(GM/R)

=> v = √(2GM/R)