Question

Check the correctness of the relation

V = (2GM/R)1/2

Where V is velocity, G is gravitational constant, M is the mass of the earth and R is the radius

of earth​

Answer 1

Answer:

this relation is not correct

Answer 2

SOLUTION :

We need to check the correctness of the relation

$\purple{\bigstar}\;\;\;\;\sf{V=\sqrt{\dfrac{2\;G\;M}{R}}}$

[ Where V is velocity, G is gravitational constant, M is mass of the Earth and R is the Radius of Earth ]

so,

Taking LHS and solving dimensionally

$\implies\sf{LHS = V}$

$\implies\sf{LHS =[ M^0 L^1 T^{-1}]}$

Now,

Taking RHS and solving dimensionally

$\implies\sf{RHS=\sqrt{\dfrac{2\;G\;M}{R}}}$

( putting dimensions of G as [ M⁻¹L³T⁻² ], dimensions of M as [ M¹ L⁰ T⁰ ], dimensions of R as [ M⁰ L¹ T⁰ ] since Radius is a length )

$\implies\sf{RHS=\sqrt{\dfrac{[M^{-1}L^3T^{-2}] \;[M^1L^0T^0]}{[M^0L^1T^0]}}}$

$\implies\sf{RHS=\sqrt{\dfrac{[L^3T^{-2}]}{[L^1]}}}$

$\implies\sf{RHS=\sqrt{[L^2T^{-2}]}}$

$\implies\sf{RHS=[L^1T^{-1}]}$

$\implies\sf{RHS=[M^0L^1T^{-1}]}$

Hence,

     $\;\;\;\;\;\;\;\;\;\large{\sf{\purple{LHS=RHS}}}$  dimensionally.

therefore,

Relation given for escape velocity is correct.