Why does escape velocity increases as radius decreases

Why does escape velocity increase as you compress say earth, while keeping the mass same?

Is it due to energy created by pressure?

It actually doesn't matter what Earth's radius is in the question, only how far away from it's center the object is. If you had an infinitesimally small point with the same mass of Earth, and a rocket was orbiting at a normal Earth radius away, it would have the same escape velocity as a rocket sitting on a normal-sized Earth.

Basically, the total energy (KE + PE) is constant, so 1/2mv^2+ -GMm/r = 0. This gives 1/2*v^2 = GM/r. v = Sqrt(2GM/r)

The r in this equation isn't the radius of the Earth, necessarily, but the radius of the object's location/orbit when you calculate the escape velocity. So, at Earth's surface, the escape velocity is 11.2 km/s. At 4 times that radius, the escape velocity is 5.6 m/s. It's unrelated to the Earth's compression.

the escape velocity is the velocity associated with transforming all the initial potential energy of the rocket (or wtv is escaping the earth's gravity) into kinetic energy

as the earth is shrunk at constant mass, the potential energy of an object resting on its surface increases, and thus a greater kinetic energy (escape velocity) is needed to leave the earth's gravitational field

in more technical terms

the gravitational potential energy at the surface of the earth is -GmM/R

the gravitational potential energy infinitely far from the earth is 0

the escape velocity is the kinetic energy needed to move from the initial potential energy state (at the earth's surface) to the final potential energy state (at infinity)

1/2 m v^2 = GmM/R

v=sqrt(2GM/R)

so as the earth's radius is decreased at constant mass, a greater escape velocity is needed to counter the greater potential