2.1.2 Vis Viva Equation

2.1.2 Vis Viva Equation:

The Vis Viva equation is one of the most useful equations in orbital mechanics is given by:

$\ GM(\frac{2}{r_{}} - \frac{1}{a}) = v^{2}$

2.1.2.1 What You Can Find With It

1. Velocity at any point in an orbit (given the orbit shape)

Know the orbit's semi-major axis a
Want velocity at distance r
Example: "How fast is a satellite moving at 400 km altitude in an orbit with a = 7000 km?"

2. Required velocity for a specific orbit

Want to achieve a certain orbit (apoapsis/periapsis)
Need to know burn velocity
Example: "What velocity do I need at 200 km altitude to reach 35,786 km apoapsis?" (GTO transfer)

3. Orbit shape from current state

Know current position r and velocity v
Solve for semi-major axis a
Determine if orbit is circular, elliptical, parabolic, or hyperbolic
Example: "My rocket is at 300 km going 9 km/s --- what orbit am I in?"

4. Apoapsis/Periapsis calculations

Combined with energy/angular momentum
Find the highest/lowest points of orbit
Example: "After a burn at periapsis, what's my new apoapsis?"

2.1.2.2 When to Use It

USE vis-viva when:

✅ You need velocity at a specific radius

✅ You know the orbit geometry (a, or r_p and r_a)

✅ You're doing Hohmann transfers or orbit changes

✅ You need Δv calculations between two circular orbits

✅ You're checking if you have escape velocity

✅ Quick sanity checks on orbital velocities

DON'T use vis-viva when:

❌ You need time or position along orbit → use Kepler's equation instead

❌ You need the shape of the trajectory in space → use orbital elements or position vectors

❌ You're dealing with atmospheric drag or thrust → vis-viva assumes pure Keplerian motion

❌ You need angular position (true anomaly, eccentric anomaly) → different tools needed

❌ Multiple gravitational bodies significantly affect motion → vis-viva is two-body only

Derivation:

We're going to start using the conservation of mechanical energy in a system

$ME = \ KE\ + \ PE$

Where KE is kinetic energy and PE is potential energy

$KE = \frac{mv^{2}}{2}$

$PE = \frac{GMm}{r}$ (orbital energy)

That leaves us with

$ME = \ \frac{mv^{2}}{2}\ + \ \frac{GMm}{r}$

The beauty of this equation is that at any point in an orbit the mechanical energy should be equal, so we're going to pick two different points in the orbit.

Apoapsis(Highest Point in Orbit)

Velocity at Apoapsis = ${v}_{{{a}}}$

Distance at Apoapsis = ${r}_{{{a}}}$

Periapsis(Lowest Point in Orbit)

Velocity at Periapsis = ${v}_{{{p}}}$

Distance at Periapsis = ${r}_{{{p}}}$

We can then equal the equations

$\frac{m{v_{a}}^{2}}{2}\ + \ \frac{GMm}{r_{a}} = \frac{m{v_{p}}^{2}}{2}\ + \ \frac{GMm}{r_{p}}$

Now we have a problem that has 4 variables and we want to reduce it down to do this we need to use one more conservation law. Conservation of angular momentum

${L}={m}{r}{v}$

So in this case

${L}={m}{r}_{{{p}}}{v}_{{{p}}}={m}{r}_{{{a}}}{v}_{{{a}}}$

$\frac{mr_{p}v_{p}}{mr_{a}} = v_{a}$

$\frac{r_{p}v_{p}}{r_{a}} = v_{a}$

Substituting

$\frac{\frac{r_{p}v_{p}}{r_{a}}^{2}}{2}\ + \ \frac{GM}{r_{a}} = \frac{{v_{p}}^{2}}{2}\ + \ \frac{GM}{r_{p}}$

$\frac{(r_{p}v_{p})^{2}}{{{2r}_{a}}^{2}}^{}\ + \ \frac{GM}{r_{a}} = \frac{{v_{p}}^{2}}{2}\ + \ \frac{GM}{r_{p}}$

$\frac{{r_{p}}^{2}{v_{p}}^{2}}{{2r_{a}}^{2}}^{}\ + \ \frac{GM}{r_{a}} = \frac{{v_{p}}^{2}}{2}\ + \ \frac{GM}{r_{p}}$

$\frac{GM}{r_{a}} - \ \frac{GM}{r_{p}} = \frac{{v_{p}}^{2}}{2} - \frac{{r_{p}}^{2}{v_{p}}^{2}}{{{2r}_{a}}^{2}}^{}$

$GM(\frac{1}{r_{a}} - \ \frac{1}{r_{p}}) = \frac{1}{2}{v_{p}}^{2}(1 - {\frac{{r_{p}}^{2}}{{r_{a}}^{2}})}^{}$

$GM(\frac{r_{p} - \ r_{a}}{r_{a}r_{p}}) = \frac{1}{2}{v_{p}}^{2}(1 - {\frac{{r_{p}}^{2}}{{r_{a}}^{2}})}^{}$

$GM(\frac{r_{p} - \ r_{a}}{r_{a}r_{p}}) = \frac{1}{2}{v_{p}}^{2}(1 - {\frac{{r_{p}}^{}}{{r_{a}}^{}})}^{}(1 + {\frac{{r_{p}}^{}}{{r_{a}}^{}})}^{}$

$GM(\frac{r_{p} - \ r_{a}}{r_{a}r_{p}}) = \frac{1}{2}{v_{p}}^{2}({\frac{{r_{a}}^{} - \ {r_{p}}^{}}{{r_{a}}^{}})}^{}({\frac{{{r_{a}}^{} + {\ r}_{p}}^{}}{{r_{a}}^{}})}^{}$: Cancel$({r_{a}}^{} - \ {r_{p}}^{})$

$(\frac{GM}{r_{a}r_{p}}) = \frac{1}{2}{v_{p}}^{2}({\frac{{{r_{a}}^{} + {\ r}_{p}}^{}}{{{r_{a}}^{2}}^{}})}^{}$

$(\frac{{{r_{a}}^{2}}^{}GM}{r_{a}r_{p}({{r_{a}}^{} + {\ r}_{p}}^{})}) = \frac{1}{2}{v_{p}}^{2}$

$2(\frac{{{r_{a}}^{}}^{}*\ GM}{r_{p}({{r_{a}}^{} + {\ r}_{p}}^{})}) = {v_{p}}^{2}$

Now all we need to do is include eccentricity then we have our equation

Eccentricity is defined as: $a = \frac{r_{a} + \ r_{p}}{2}$

$2a = r_{a} + \ r_{p}$

$2a - \ r_{p} = r_{a}$

$2(\frac{r_{a}*\ GM}{{2a*r}_{p}}) = {v_{p}}^{2}$

$(\frac{r_{a}*\ GM}{{a*r}_{p}}) = {v_{p}}^{2}$

$(\frac{(2a - \ r_{p})*\ GM}{{a*r}_{p}}) = {v_{p}}^{2}$

$\ GM(\frac{(2a - \ r_{p})}{{a*r}_{p}}) = {v_{p}}^{2}$

$\ GM(\frac{2}{r_{p}} - \frac{1}{a}) = {v_{p}}^{2}$

This can be generalized to work at any point in orbit to

$\ GM(\frac{2}{r_{}} - \frac{1}{a}) = v^{2}$

2.1.2.3 Solving Problems:

Problems are provided by Claude AI

Basic Applications

A satellite orbits Earth at an altitude of 400 km. Calculate its orbital velocity. (Earth's mass ${M}={5.5972}\cdot{10}^{{{24}}}$

Earth's radius ${r}={6371}{k}{m}$)

Starting from our initial equation and assuming its a circular orbit

Here our radius will be that of the earth plus the height It's orbiting at so ${r}={6371}{k}{m}+{400}{k}{m}$

$\ GM(\frac{2}{r_{}} - \frac{1}{a}) = v^{2}$

$\ (6.6743*10^{- 11})(5.5972*10^{24})(\frac{2}{(6771000)} - \frac{1}{6771000}) = v^{2}$

$\ (3.7357391*10^{14})(\frac{1}{(6771000)}) = v^{2}$

$\ (55172634.77) = v^{2}$

$\ = v^{2}$

${\left\lbrace{7},{427.45}={v}\right\rbrace}^{{null}}$

At what distance from Earth's center does a spacecraft need to be for its orbital velocity to equal 5 km/s?

Starting from the Vis-Viva equation

$\ GM(\frac{2}{r_{}} - \frac{1}{a}) = v^{2}$ we know that ${v}={5000}$ and we want to find ${r}$ so

$(6.6743*10^{- 11})(5.5972*10^{24})(\frac{2}{r_{}} - \frac{1}{a}) = 5000^{2}$

But remember the height will be the radius of earth plus the distance off the floor so

${r}={6371}+{h}$

It's worth noting here that because it's a circular orbit $a = \frac{d}{2} = r + h$ so our semi-major axis is just our radius

$(6.6743*10^{- 11})(5.5972*10^{24})(\frac{2}{{6371 + h}_{}} - \frac{1}{6371 + h}) = 5000^{2}$

$(6.6743*10^{- 11})(5.5972*10^{24})(\frac{1}{6371 + h}) = 5000^{2}$ Solving for r

$\frac{(6.6743*10^{- 11})(5.5972*10^{24})}{5000^{2}} = 6371 + h$

$\frac{(3.7357391*10^{14})}{5000^{2}} = 6371 + h$

$\frac{(3.7357391*10^{14})}{25000000^{}} = 6371 + h$

${14942956.4}{m}={6371}{k}{m}+{h}$ converting units from km to m

${14942956.4}{m}={6371000}{m}+{h}$

${14942956.4}{m}-{6371000}{m}={h}$

${8571956}{m}={h}$ or a height of

${8571.956}{k}{m}={h}$

A comet has a perihelion distance of 0.6 AU and aphelion distance of 35 AU from the Sun. Find its velocity at both perihelion and aphelion. (Sun's mass = 1.989 × 10³⁰ kg)

Starting from the vis-viva equation:

$\ GM(\frac{2}{r_{}} - \frac{1}{a}) = v^{2}$

Here we are going to have 2 r's 0.6AU and 35 AU

Our semi major axis is going to be given by$\frac{(35 + 0.6)AU}{2}$ so $\frac{(35.6)AU}{2} = 17.8AU$

For our perihelion

$\ GM(\frac{2}{{0.6AU}_{}} - \frac{1}{17.8AU}) = v^{2}$

$\ GM(3.2771\frac{1}{AU}) = v^{2}$ ${1}{A}{U}={1.496}\cdot{10}^{{{11}}}{m}$ so $\frac{1}{AU} = \frac{1}{1.496*10^{11}m}$ so $\frac{3.2771}{AU} = 2.80715*10^{- 12}\frac{1}{m}$

$\ GM(2.80715*10^{- 12}\frac{1}{m}) = v^{2}$

Putting in our constants

$\ (6.6743*10^{- 11}\frac{m^{3}}{kgs^{2}})(1.989*10^{30}kg)(2.80715*10^{- 12}\frac{1}{m}) = v^{2}$

$\ 372634378\frac{m^{2}}{s^{2}} = v^{2}$

$\ = v$

$\ 19303.7\frac{m}{s} = v$

Doing the same thing but now for our Aphelion with 35 AU in place for 0.6AU

$\ GM(\frac{2}{{35AU}_{}} - \frac{1}{17.8AU}) = v^{2}$

$\ GM(9.6308*10^{- 4}\frac{1}{AU}) = v^{2}$

$9.6308*10^{- 4}\frac{1}{AU} = 1.4407*10^{- 14}\frac{1}{m}$

$\ GM(1.4407*10^{- 14}\frac{1}{m}) = v^{2}$ replacing constants

$(6.6743*10^{- 11}\frac{m^{3}}{kgs^{2}})(1.989*10^{30}kg)\ (1.4407*10^{- 14}\frac{1}{m}) = v^{2}$

$1912645.4\frac{m^{2}}{s^{2}} = v^{2}$

$={v}^{{{2}}}$

$1382.98\frac{m}{s} = v$