Question

The escape velocity v of a body depends on 

1. The acceleration due to gravity ‘g’ of the planet 

2. The radius R of the planet. Establish dimensionally the relation for the escape velocity?

Answer 1

\(v ∝ g^aR^b ⇒ v = kg^a R^b,\) K → dimensionally proportionality constant

[v] = [g]^a [R]^b

[M⁰ L¹ T^-1 ] = [M⁰ L¹ T^-2 ] [M⁰ L¹ T¹⁰ ]^b

equating powers

1 = a + b 

-1 = -2a ⇒ a = \(\frac{1}{2}\) 

b = 1 – a = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)

∴ v = k \(\sqrt {gR}\)