Specific orbital energy ε = −GM - 2a

WHAT is specific orbital energy?

WHY "specific"? We divide by $m$ so the satellite's own mass cancels — a feather and a boulder in the same orbit have the same $\varepsilon$. This is the same reason all objects fall at the same rate.

WHY is total energy constant? (first principles)

Gravity is a conservative force: work done depends only on start/end positions, not the path. So kinetic + potential energy is conserved.

• Kinetic energy per mass: $\dfrac{1}{2}v^2$.
• Potential energy per mass: $-\dfrac{GM}{r}$ (negative, zero at infinity — see Mistake 1).

Their sum $\varepsilon$ never changes during free orbital motion. WHY this matters: as the satellite falls inward ($r\downarrow$), PE drops, so KE rises ($v\uparrow$) — it speeds up at perigee, slows at apogee, exchanging the two while the total stays put.

DERIVATION: why $\varepsilon = -GM/2a$

We need two ingredients. We derive both.

Step 1 — Energy at perigee and apogee. At perigee ($r_p$) and apogee ($r_a$) the velocity is perpendicular to the radius, so all speed is "useful." Energy is the same at both: $\frac{v_p^2}{2}-\frac{GM}{r_p} = \frac{v_a^2}{2}-\frac{GM}{r_a}.$ Why this step? $\varepsilon$ is constant, so I can equate it at two convenient points.

Step 2 — Conserve angular momentum. At those two points $\mathbf v \perp \mathbf r$, so $L/m = h = r_p v_p = r_a v_a$. Hence $v_a = v_p\,\frac{r_p}{r_a}.$ Why this step? Torque from a central force is zero, so $h$ is conserved — gives a second equation linking the speeds.

Step 3 — Solve for $v_p^2$. Substitute $v_a$: $\frac{v_p^2}{2}-\frac{GM}{r_p} = \frac{v_p^2}{2}\frac{r_p^2}{r_a^2}-\frac{GM}{r_a}.$ Collect $v_p^2$: $\frac{v_p^2}{2}\left(1-\frac{r_p^2}{r_a^2}\right) = GM\left(\frac{1}{r_p}-\frac{1}{r_a}\right).$ The left bracket factors: $1-\frac{r_p^2}{r_a^2}=\frac{(r_a-r_p)(r_a+r_p)}{r_a^2}$. The right bracket $=\frac{r_a-r_p}{r_p r_a}$. Cancel $(r_a-r_p)$: $\frac{v_p^2}{2}\cdot\frac{r_a+r_p}{r_a^2} = \frac{GM}{r_p r_a}\;\Rightarrow\; v_p^2 = \frac{2GM\,r_a}{r_p(r_a+r_p)}.$ Why this step? Algebraic clean-up isolates the perigee speed in terms of geometry only.

Step 4 — Plug into $\varepsilon$. $\varepsilon = \frac{v_p^2}{2}-\frac{GM}{r_p} = \frac{GM\,r_a}{r_p(r_a+r_p)}-\frac{GM}{r_p} = \frac{GM}{r_p}\left(\frac{r_a-(r_a+r_p)}{r_a+r_p}\right) = \frac{GM}{r_p}\cdot\frac{-r_p}{r_a+r_p}.$ $\varepsilon = -\frac{GM}{r_a+r_p}.$ Why this step? Substituting kills $v_p$ and leaves pure geometry.

Step 5 — Use the definition of $a$. By geometry of the ellipse, perigee $+$ apogee distances span the major axis: $r_p+r_a = 2a$. Therefore $\boxed{\varepsilon = -\frac{GM}{2a}}.\qquad\blacksquare$

The vis-viva equation (a free bonus)

Equate the two formulas for $\varepsilon$ at any point $r$ with speed $v$: $\frac{v^2}{2}-\frac{GM}{r} = -\frac{GM}{2a}\;\Rightarrow\; \boxed{v^2 = GM\left(\frac{2}{r}-\frac{1}{a}\right)}.$ WHY useful: it gives speed anywhere from just $r$ and $a$ — no need to track the angle.

Orbit type from the sign of $\varepsilon$

Worked examples

Common mistakes (steel-manned)

Active-recall flashcards

#flashcards/physics

Connections

• Vis-viva equation — derived directly from $\varepsilon$.
• Kepler's Third Law — $T^2\propto a^3$, same $a$ governing energy.
• Conservation of angular momentum — used in the derivation.
• Escape velocity — the $\varepsilon=0$ limiting case.
• Hohmann transfer orbit — uses energy/vis-viva to size burns.
• Conic sections in orbits — sign of $\varepsilon$ ↔ ellipse/parabola/hyperbola.
• Gravitational potential energy −GM/r — the PE term's origin.

Concept Map

Hinglish (regional understanding)