Quick questions on Gravitational potential energy and fields - TCE Physics (Tasmania)

The work needed to move a mass between two points equals the change in its gravitational potential energy:

A particularly important application is the escape velocity, the minimum launch speed for an object to reach infinity with no kinetic energy left. Setting the kinetic energy at the surface equal to the energy needed to lift the object to zero potential energy gives $\tfrac12 m v_\text{esc}^2 = \dfrac{GMm}{r}$, so $v_\text{esc} = \sqrt{\dfrac{2GM}{r}}$. For Earth this is about $11\ \text{km s}^{-1}$. Notice the mass of the projectile cancels, so the escape speed depends only on the planet, exactly as for orbital speed.

If you plot gravitational field strength $g$ against distance $r$, the area under the graph between two radii gives the change in gravitational potential per unit mass. Because the field follows an inverse-square curve, this area is found by integration or by counting squares, and it confirms that most of the energy to escape a planet is spent close to the surface where the field is strongest.