Question

Derive by the method of dimension an expression for the escape velocity of the body assuming that this velocity depends
(i) on the radius of planet R
(ii) accleration due to gravity​

Answer 1

Answer:

MARK ME BRAINLIEST

MARK ME BRAINLIEST

Answer 2

Explanation:

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)