Final answer:
The dimensional relation of escape velocity with the acceleration due to gravity and Earth's radius can be derived by equating kinetic and gravitational potential energies, leading to the formula v = √(2gR). The correct relation is option A: v = √gR.
Step-by-step explanation:
To derive the dimensional relation of escape velocity (v) with the acceleration due to gravity (g) and the radius of the Earth (R), we can start by equating the gravitational force that provides the necessary centripetal force for an object to maintain its orbit with the force needed to overcome Earth's gravitational pull. When an object reaches escape velocity, it has enough kinetic energy to break free from the gravitational attraction of the Earth without further propulsion.
The formula for escape velocity (v) is derived from the concept that the kinetic energy (1/2 mv², where m is the mass of the object and v is the velocity) should be equal to the gravitational potential energy (GMm/R, where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth) required to overcome the Earth's gravity. Therefore, setting 1/2 mv² = GMm/R and solving for v gives us v = √(2GM/R). Now, because the acceleration due to gravity (g) is defined as GM/R², we can substitute GM/R² for g, which yields v = √(2gR). Hence, the correct option in the final part is A. v = √gR, not √g/R, gR, or g/R².