On an elliptical orbit (Ellipses), the speed changes all the way around. Orbital Energy and Kepler's Second Law from Angular Momentum give one formula for the speed anywhere, and it is the tool for planning trips between planets. Try transfers in the Orbit Lab.
Why it works
At the nearest point $r_p$ and farthest point $r_a$ the velocity is perpendicular to $\vec{r}$, so angular momentum gives $r_pv_p = r_av_a$. Energy gives $\tfrac{1}{2}v_p^2 - \dfrac{GM}{r_p} = \tfrac{1}{2}v_a^2 - \dfrac{GM}{r_a}$ (per kilogram). Substituting $v_a = v_pr_p/r_a$ and solving, the total energy per kilogram comes out as $-\dfrac{GM}{r_p + r_a} = -\dfrac{GM}{2a}$, where $a = \tfrac{1}{2}(r_p + r_a)$ is the semi-major axis. Energy is the same everywhere on the orbit, so at any distance $r$: $\tfrac{1}{2}v^2 - \dfrac{GM}{r} = -\dfrac{GM}{2a}$.
Key idea
The vis-viva equation: $$v^2 = GM\left(\dfrac{2}{r} - \dfrac{1}{a}\right).$$ For a circle ($r = a$) it gives $v^2 = GM/r$. A Hohmann transfer between circular orbits $r_1$ and $r_2$ uses an ellipse with $a = \tfrac{1}{2}(r_1 + r_2)$: one burn at $r_1$ to get onto it, a second at $r_2$ to circularize.
Worked example: Leaving Earth's orbit for Mars
A Hohmann transfer to Mars has $a = \tfrac{1}{2}(1 + 1.524) = 1.262\ \text{AU}$. How fast must a craft move at 1 AU to be on it, compared with Earth's orbital speed? ($GM_\odot = 1.327 \times 10^{20}\ \text{m}^3/\text{s}^2$, $1\ \text{AU} = 1.496 \times 10^{11}\ \text{m}$.)
- On the transfer orbit at $r = 1\ \text{AU}$: $v^2 = GM\left(\dfrac{2}{1\ \text{AU}} - \dfrac{1}{1.262\ \text{AU}}\right)$, so $v = 32.7\ \text{km/s}$.
- Earth's circular speed: $\sqrt{GM/(1\ \text{AU})} = 29.8\ \text{km/s}$.
- The craft needs about $2.9\ \text{km/s}$ beyond Earth's orbital speed (on top of escaping Earth itself).
Answer: $32.7\ \text{km/s}$, about $2.9\ \text{km/s}$ more than Earth's $29.8\ \text{km/s}$.
Worked example: Up to geostationary orbit
A satellite at $r_1 = 6.671 \times 10^{6}\ \text{m}$ (300 km up) transfers to $r_2 = 4.216 \times 10^{7}\ \text{m}$. Find the first burn. (Earth: $GM = 3.986 \times 10^{14}\ \text{m}^3/\text{s}^2$.)
- Circular speed at $r_1$: $\sqrt{GM/r_1} = 7.73\ \text{km/s}$.
- Transfer ellipse: $a = \tfrac{1}{2}(r_1 + r_2) = 2.441 \times 10^{7}\ \text{m}$; at $r_1$, $v = \sqrt{GM\left(\dfrac{2}{r_1} - \dfrac{1}{a}\right)} = 10.16\ \text{km/s}$.
- First burn: $10.16 - 7.73 = 2.43\ \text{km/s}$. (A second burn of about $1.47\ \text{km/s}$ at the top circularizes the orbit.)
Answer: About $2.43\ \text{km/s}$.
Common mistake
Expecting a burn to change the orbit's height right where it happens. Right after a short burn the craft is still at the same place, so that point stays on the new orbit. What changes is the energy, and with it $a$: the far side of the orbit, half a revolution later, rises or falls. Check with vis-viva: at the burn point $r$ is fixed, so a bigger $v$ means a bigger $a$, and the opposite end, at $2a - r$, moves out.