Deriving Escape Velocity

Deriving Escape Velocity – What Planet Is This?What Planet Is This?

We can derive escape velocity from Newton’s gravity force law:

$\displaystyle F=-G\cdot \frac{{{m}_{1}}\cdot {{m}_{2}}}{{{r}^{2}}}$

If we replace force, F, with the classic physics definition of Newton’s second law, m·a (or in this case m₂·a), then we get:

$\displaystyle {{m}_{2}}\cdot a=-G\cdot \frac{{{m}_{1}}\cdot {{m}_{2}}}{{{r}^{2}}}$

Cancelling terms, we have the general equation for the radial (centripetal) acceleration of a single, point mass (here we replace m₁ with m):

$\displaystyle a=-G\cdot \frac{m}{{{r}^{2}}}$

Escape velocity is the velocity that lets us leave the surface of a mass and never return. This means that we always have a positive radial velocity, and that radial velocity only approaches zero as distance from the mass approaches infinity. We obtain escape velocity by integrating this equation with respect to r from r = r_surface to r = ∞:

$\displaystyle \int_{r={{r}_{surface}}}^{r=\infty }{a\cdot dr=}\int_{r={{r}_{surface}}}^{r=\infty }{-G\cdot \frac{m}{{{r}^{2}}}\cdot dr}$

The expression on the right-hand side is straightforward; however, the expression on the left-hand side requires some adjustment. We start by replacing acceleration, a, with its definition, dv/dt:

$\displaystyle \int_{r={{r}_{surface}}}^{r=\infty }{\frac{dv}{dt}\cdot dr=\int_{r={{r}_{surface}}}^{r=\infty }{-G\cdot \frac{m}{{{r}^{2}}}\cdot dr}}$

Rearranging, we get:

$\displaystyle \int_{r={{r}_{surface}}}^{r=\infty }{\frac{dr}{dt}\cdot dv=\int_{r={{r}_{surface}}}^{r=\infty }{-G\cdot \frac{m}{{{r}^{2}}}\cdot dr}}$

The derivative dr/dt is simply velocity, v. This now gives us:

$\displaystyle \int_{r={{r}_{surface}}}^{r=\infty }{v\cdot dv=\int_{r={{r}_{surface}}}^{r=\infty }{-G\cdot \frac{m}{{{r}^{2}}}\cdot dr}}$

To complete the adjustment, we must alter the limits of integration for the change of variable. At r = r_surface, we have v = v_escape; and for r = ∞, we have v = 0; (the definition of an escape velocity at infinity). This final change gives us:

$\displaystyle \int_{v={{v}_{escape}}}^{v=0}{v\cdot dv=\int_{r={{r}_{surface}}}^{r=\infty }{-G\cdot \frac{m}{{{r}^{2}}}\cdot dr}}$