# The escape velocity (v ) of a body depends upon the mass (m) of body, gravitationalacceleration (g) and radius(R) of the planet .Derive the relation for escape velocitydimensionally.

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 GxRyMz

=> v = k GxRyMz

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

=> [LT-1] = k[M-x+zL3x+y T-2x ]

Comparing the exponents of the dimensions

L1  = L3x+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 = kG1/2 R-1/2M1/2

=> v = k √(GM/R)

By actual theoretical calculation k = √2

v = √2 √(GM/R)

=> v = √(2GM/R)

• 36

The minimum velocity with which a projectile has to be projected to escape the earth’s gravitational field is called escape velocity.

Let a body of mass ‘m’ be projected with velocity ‘v’.

At the ground, PE = GMm/R

And the KE = ½ mv2

To overcome earth’s gravitational field,

KE>PE

=> ½ mv2 > GMm/R

=> v > (2GM/R)1/2

We know,

g = GM/R2

Therefore,

v > [2(GM/R2)R]1/2

=> v > [2gR]1/2

Thus, to escape the earth’s gravitational field, the minimum velocity must be,

ve = [2gR]1/2

• 3

It is 1/2 in place of

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