The escape velocity of a particle of mass m varies as

The correct option is m⁰.

To understand why the escape velocity of a particle is independent of its mass, we can derive the formula for escape velocity using the principle of conservation of energy.

Escape velocity is the minimum speed required for a particle to escape the gravitational influence of a celestial body (like Earth) and reach infinity with zero kinetic energy.

Let:
- m be the mass of the escaping particle.
- M be the mass of the celestial body (e.g., Earth).
- R be the radius of the celestial body.
- G be the universal gravitational constant.
- v_e be the escape velocity.

The total mechanical energy of the particle at the surface of the celestial body is the sum of its kinetic energy and gravitational potential energy:

$E_{\mathrm{initial}}=\frac{1}{2}m{v}_e^2-\frac{GMm}{R}$

For the particle to just escape to infinity, its total energy at infinity must be at least zero (where both kinetic and potential energy become zero):

$E_{\mathrm{final}}=0$

By the law of conservation of energy:

$E_{\mathrm{initial}}=E_{\mathrm{final}}$

$\frac{1}{2}m{v}_e^2-\frac{GMm}{R}=0$

We can divide the entire equation by the mass of the particle m, which shows that the mass of the particle cancels out:

$\frac{1}{2}{v}_e^2=\frac{GM}{R}$

Solving for escape velocity, we get:

$v_e=\sqrt{\frac{2GM}{R}}$

Since the mass of the particle m does not appear in this final expression, the escape velocity is independent of the mass of the escaping particle. Therefore, the escape velocity varies as m⁰ (which is equal to 1).