The Vis-Viva Equation and Orbital Energy
The Vis-Viva Equation and Orbital Energy
The vis-viva equation — from the Latin for "living force" — is the master equation of orbital mechanics. It relates the speed of an orbiting body at any point along its trajectory to its distance from the central body and the shape of its orbit. Once you know the semi-major axis and your current distance, vis-viva immediately tells you your speed, and vice versa.
The equation comes directly from conservation of energy. The total mechanical energy of an orbit — kinetic plus gravitational potential — is constant and depends only on the semi-major axis: E equals negative mu m over 2a. This means two orbits with the same semi-major axis have the same total energy, regardless of eccentricity. A circular orbit and a highly elliptical orbit with the same value of a have identical total energy.
The vis-viva equation is a powerful diagnostic tool. Given any two of the three quantities — speed, distance, and semi-major axis — you can find the third. Want to know the semi-major axis of an orbit after a burn? Measure your speed and altitude, then solve vis-viva for a. Want to know if you're on an escape trajectory? If v squared exceeds 2 mu over r, then the right side of vis-viva gives a negative 1 over a, meaning a is negative — a hallmark of a hyperbolic (escape) trajectory.
The specific orbital energy, epsilon equals negative mu over 2a, is a conserved quantity and one of the most important parameters for classifying orbits. Circular and elliptical orbits have epsilon less than zero (bound). Parabolic trajectories have epsilon equal to zero (marginal escape). Hyperbolic trajectories have epsilon greater than zero (unbound, with residual velocity at infinity).
$$v^2 = \mu\left(\frac{2}{r} - \frac{1}{a}\right)$$
Solving for $a$: $\frac{1}{a} = \frac{2}{r} - \frac{v^2}{\mu}$
$$\varepsilon = \frac{v^2}{2} - \frac{\mu}{r} = -\frac{\mu}{2a}$$
$\varepsilon < 0$: ellipse/circle (bound). $\varepsilon = 0$: parabola (escape). $\varepsilon > 0$: hyperbola (unbound).
$a > 0$: ellipse (circle if $e = 0$). $a \to \infty$: parabola ($e = 1$). $a < 0$: hyperbola ($e > 1$).
A spacecraft at altitude 500 km has speed 9000 m/s. What type of orbit is it on, and what is its semi-major axis?
Step 1: $r = 6871$ km $= 6.871 \times 10^6$ m.
Step 2: Compute specific energy.
$$\varepsilon = \frac{9000^2}{2} - \frac{3.986 \times 10^{14}}{6.871 \times 10^6} = 4.05 \times 10^7 - 5.80 \times 10^7 = -1.75 \times 10^7 \text{ J/kg}$$
Step 3: Since $\varepsilon < 0$, this is a bound (elliptical) orbit.
$$a = -\frac{\mu}{2\varepsilon} = \frac{3.986 \times 10^{14}}{2 \times 1.75 \times 10^7} = 1.138 \times 10^7 \text{ m} = 11{,}380 \text{ km}$$
A spacecraft at altitude 300 km fires its engines to reach 10,900 m/s. Is it escaping Earth?
Step 1: $r = 6671$ km. Escape velocity: $v_{\text{esc}} = \sqrt{2\mu/r} = 10{,}929$ m/s.
Step 2: $v = 10{,}900 < 10{,}929$.
Just barely short of escape. The orbit is a very elongated ellipse, not escape.
Step 3: $\varepsilon = 10900^2/2 - \mu/r = 5.94 \times 10^7 - 5.97 \times 10^7 = -3.4 \times 10^5$ J/kg.
$a = \mu/(2 \times 3.4 \times 10^5) = 5.86 \times 10^8$ m $= 586{,}000$ km — a highly elongated ellipse extending past the Moon!
A spacecraft passes Earth at $r = 6671$ km with $v = 12{,}000$ m/s. Find its speed when it's 100,000 km from Earth.
Step 1: Find $a$ from vis-viva at $r_1$.
$$1/a = 2/6.671 \times 10^6 - (12000)^2/3.986 \times 10^{14} = 2.998 \times 10^{-7} - 3.613 \times 10^{-7} = -6.15 \times 10^{-8}$$
$a = -1.626 \times 10^7$ m (negative $\to$ hyperbola, unbound).
Step 2: At $r_2 = 10^8$ m, use vis-viva:
$$v_2^2 = \mu(2/r_2 - 1/a) = 3.986 \times 10^{14}(2/10^8 + 1/1.626 \times 10^7) = 3.24 \times 10^7$$
$$v_2 = 5693 \text{ m/s}$$
Practice Problems
Show Answer Key
1. $r = 6971$ km. $\varepsilon = 8500^2/2 - \mu/r = 3.612 \times 10^7 - 5.718 \times 10^7 = -2.106 \times 10^7$. $a = \mu/(2 \times 2.106 \times 10^7) = 9461$ km. Elliptical.
2. $v_{\text{esc}} = \sqrt{2\mu/(6871000)} = 10{,}768$ m/s.
3. $v_{\text{esc}} = \sqrt{2\mu/r} = 8929$ m/s. $v_{\infty} = \sqrt{14000^2 - 8929^2} = \sqrt{1.163 \times 10^8} = 10{,}785$ m/s.
4. Both have $E = -\mu m/(2a)$ and the same $a$, so same energy. ✓
5. $v_{\text{new}} = 7668 + 500 = 8168$ m/s. $1/a = 2/r - v^2/\mu = 2/6.771 \times 10^6 - 8168^2/3.986 \times 10^{14} = 2.954 \times 10^{-7} - 1.673 \times 10^{-7} = 1.281 \times 10^{-7}$. $a = 7808$ km.
6. At periapsis: $v_p^2 = \mu(2/r_p - 1/a)$. At apoapsis: $v_a^2 = \mu(2/r_a - 1/a)$. Angular momentum $L = mrv\sin\theta$, at apsides $\theta = 90°$ so $L = mv_p r_p = mv_a r_a$. ✓
7. $v = \sqrt{\mu(2/6671000 - 1/25000000)} = \sqrt{3.986 \times 10^{14} \times 2.396 \times 10^{-7}} = 9771$ m/s.
8. $a = -\mu/(2\varepsilon) = 3.986 \times 10^{14}/(2 \times 2.94 \times 10^7) = 6779$ km. ✓
9. $a = -\mu/v_{\infty}^2 = -3.986 \times 10^{14}/9 \times 10^6 = -4.43 \times 10^7$ m. $v_p = \sqrt{\mu(2/r_p - 1/a)} = \sqrt{3.986 \times 10^{14}(2/7 \times 10^6 + 1/4.43 \times 10^7)} = 11{,}095$ m/s.
10. If $1/a = 0$: $v^2 = \mu \cdot 2/r \Rightarrow v = \sqrt{2\mu/r}$. ✓