Q.Derive an expression for escape speed.

Step 1. An object of mass $M$ launched from Earth's surface with speed $v_i$ has initial total energy $E_i=\dfrac12Mv_i^2-\dfrac{GMM_e}{R_e}$.

Step 2. For the object to just barely escape (the minimum-energy condition), it should reach an infinite distance with zero leftover kinetic energy; since $U\to0$ as $r\to\infty$ too, the final total energy is $E_f=0$.

Step 3. By conservation of energy, $E_i=E_f=0$. Setting $v_i=v_e$ (escape speed):

Step 4. Solving for $v_e$: $\dfrac12v_e^2=\dfrac{GM_e}{R_e}\Rightarrow v_e^2=\dfrac{2GM_e}{R_e}\Rightarrow v_e=\sqrt{\dfrac{2GM_e}{R_e}}$.

Step 5. Using $g=GM_e/R_e^2$, this can be rewritten as $v_e=\sqrt{2gR_e}$. …

Set total energy at launch equal to zero total energy at inf …

• Setting only the kinetic energy (not the total energy) to zero at infinity, forgetting potential energy also vanishes there. …