In the gravitational two-body problem, the specific orbital energy $\epsilon\,\!$ (or vis-viva energy) of two orbiting bodies is the constant sum of their mutual potential energy ($\epsilon_p\,\!$) and their total kinetic energy ($\epsilon_k\,\!$), divided by the reduced mass. According to the orbital energy conservation equation (also referred to as vis-viva equation), it does not vary with time:

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In the gravitational two-body problem, the specific orbital energy $\epsilon\,\!$ (or vis-viva energy) of two orbiting bodies is the constant sum of their mutual potential energy ($\epsilon_p\,\!$) and their total kinetic energy ($\epsilon_k\,\!$), divided by the reduced mass. According to the orbital energy conservation equation (also referred to as vis-viva equation), it does not vary with time:

$\epsilon = \epsilon_k+\epsilon_p \!$
$\epsilon = {v^2\over{2}}-{\mu\over{r}} = -{1\over{2}}{ \mu^2\over{h^2}}\left(1-e^2\right) = -\frac{\mu }{2a}$

where

It is expressed in J/kg = m2s−2 or MJ/kg = km2s−2. For an elliptical orbit the specific orbital energy is the negative of the additional energy required to accelerate a mass of one kilogram to escape velocity (parabolic orbit). For a hyperbolic orbit, it is equal to the excess energy compared to that of a parabolic orbit. In this case the specific orbital energy is also referred to as characteristic energy.

## Equation forms for different orbits

For an elliptical orbit, the specific orbital energy equation, when combined with conservation of specific angular momentum at one of the orbit's apsides, simplifies to:[1]

$\epsilon = -{\mu \over{2a}}\,\!$

where

Proof:

For an elliptical orbit with specific angular momentum h given by
${h^2} = \mu p = \mu a (1-e^2)\,\!$
we use the general form of the specific orbital energy equation,
$\epsilon={v^2\over{2}}-{\mu\over{r}}$
with the relation that the relative velocity at periapsis is
${v_p^2} = {h^2 \over{{r_p}}^2} = {h^2 \over{{a^2(1-e)}}^2} = {\mu a (1-e^2) \over{{a^2(1-e)}}^2} = {\mu (1-e^2) \over{{a(1-e)}}^2}\,\!$
Thus our specific orbital energy equation becomes
$\epsilon = {\mu \over{a}} {\left [ {(1-e^2) \over{{2(1-e)}}^2} - {1 \over{(1-e)}} \right ] } = {\mu \over{a}} {\left [ {{(1-e)(1+e)} \over{{2(1-e)}}^2} - {1 \over{(1-e)}} \right ] } = {\mu \over{a}} {\left [ {(1+e) \over{{2(1-e)}}} - {2 \over{2(1-e)}} \right ] } = {\mu \over{a}} {\left [ {{e-1} \over{2(1-e)}} \right ] }\,\!$
and finally with the last simplification we obtain:
$\epsilon = -{\mu \over{2a}}\,\!$

For a parabolic orbit this equation simplifies to

$\epsilon=0\,\!.$

For a hyperbolic trajectory this specific orbital energy is either given by

$\epsilon = {\mu \over{2a}}\,\!.$

or the same as for an ellipse, depending on the convention for the sign of a.

In this case the specific orbital energy is also referred to as characteristic energy (or $C_3\,\!$) and is equal to the excess specific energy compared to that for a parabolic orbit.

It is related to the hyperbolic excess velocity $v_{\infty} \,\!$ (the orbital velocity at infinity) by

$2\epsilon=C_3=v_{\infty}^2\,\!.$

It is relevant for interplanetary missions.

Thus, if orbital position vector ($\mathbf{r}\,\!$) and orbital velocity vector ($\mathbf{v}\,\!$) are known at one position, and $\mu\,\!$ is known, then the energy can be computed and from that, for any other position, the orbital speed.

## Rate of change

For an elliptical orbit the rate of change of the specific orbital energy with respect to a change in the semi-major axis is

$\frac{\mu}{2a^2}\,\!$

where

In the case of circular orbits, this rate is one half of the gravity at the orbit. This corresponds to the fact that for such orbits the total energy is one half of the potential energy, because the kinetic energy is minus one half of the potential energy.

If the central body has radius R, then the additional energy of an elliptic orbit compared to being stationary at the surface is

$\ -\frac{\mu}{2a}+\frac{\mu}{R} = \frac{\mu (2a-R)}{2aR}.$

• For the Earth and a just little more than $R/2$ this is $(2a-R)g$ ; the quantity $2a-R$ is the height the ellipse extends above the surface, plus the periapsis distance (the distance the ellipse extends beyond the center of the Earth); the latter times g is the kinetic energy of the horizontal component of the velocity.

## Examples

### ISS

The International Space Station has an orbital period of 91.74 minutes, hence the semi-major axis is 6,738 km.

The energy is −29.6 MJ/kg: the potential energy is −59.2 MJ/kg, and the kinetic energy 29.6 MJ/kg. Compare with the potential energy at the surface, which is −62.6 MJ/kg. The extra potential energy is 3.4 MJ/kg, the total extra energy is 33.0 MJ/kg. The average speed is 7.7 km/s, the net delta-v to reach this orbit is 8.1 km/s (the actual delta-v is typically 1.5–2 km/s more for atmospheric drag and gravity drag).

The increase per meter would be 4.4 J/kg; this rate corresponds to one half of the local gravity of 8.8 m/s².

For an altitude of 100 km (radius is 6471 km):

The energy is −30.8 MJ/kg: the potential energy is −61.6 MJ/kg, and the kinetic energy 30.8 MJ/kg. Compare with the potential energy at the surface, which is −62.6 MJ/kg. The extra potential energy is 1.0 MJ/kg, the total extra energy is 31.8 MJ/kg.

The increase per meter would be 4.8 J/kg; this rate corresponds to one half of the local gravity of 9.5 m/s². The speed is 7.8 km/s, the net delta-v to reach this orbit is 8.0 km/s.

Taking into account the rotation of the Earth, the delta-v is up to 0.46 km/s less (starting at the equator and going east) or more (if going west).

### Voyager 1

For Voyager 1, with respect to the Sun:

Hence:

• $\epsilon=\epsilon_k+\epsilon_p={v^2\over{2}}-{\mu\over{r}} =$ 146 km2s−2 - 8 km2s−2 = 138 km2s−2

Thus the hyperbolic excess velocity (the theoretical orbital velocity at infinity) is given by

$v_{\infty}=\,\!$ 16.6 km/s

However, Voyager 1 does not have enough velocity to leave the Milky Way. The computed speed applies far away from the Sun, but at such a position that the potential energy with respect to the Milky Way as a whole has changed negligibly, and only if there is no strong interaction with celestial bodies other than the Sun.

## Applying thrust

Assume:

• a is the acceleration due to thrust (the time-rate at which delta-v is spent)
• g is the gravitational field strength
• v is the velocity of the rocket

Then the time-rate of change of the specific energy of the rocket is $\mathbf{v} \cdot \mathbf{a}$: an amount $\mathbf{v} \cdot (\mathbf{a}-\mathbf{g})$ for the kinetic energy and an amount $\mathbf{v} \cdot \mathbf{g}$ for the potential energy.

The change of the specific energy of the rocket per unit change of delta-v is

$\frac{\mathbf{v \cdot a}}{|\mathbf{a}|}$

which is |v| times the cosine of the angle between v and a.

Thus, when applying delta-v to increase specific orbital energy, this is done most efficiently if a is applied in the direction of v, and when |v| is large. If the angle between v and g is obtuse, for example in a launch and in a transfer to a higher orbit, this means applying the delta-v as early as possible and at full capacity. See also gravity drag. When passing by a celestial body it means applying thrust when nearest to the body. When gradually making an elliptic orbit larger, it means applying thrust each time when near the periapsis.

When applying delta-v to decrease specific orbital energy, this is done most efficiently if a is applied in the direction opposite to that of v, and again when |v| is large. If the angle between v and g is acute, for example in a landing (on a celestial body without atmosphere) and in a transfer to a circular orbit around a celestial body when arriving from outside, this means applying the delta-v as late as possible. When passing by a planet it means applying thrust when nearest to the planet. When gradually making an elliptic orbit smaller, it means applying thrust each time when near the periapsis.

If a is in the direction of v:

$\Delta \epsilon = \int v\, d (\Delta v) = \int v\, a dt$

## Earth orbits

orbit center-to-center
distance
altitude above
the Earth's surface
speed Orbital period specific orbital energy
Earth's surface (for comparison) 6,400 km 0 km 7.89 km/s (17,650 mph) 85 minutes -62.6 MJ/kg
Low Earth orbit 6,600 to 8,400 km 200 to 2,000 km circular orbit: 7.8 to 6.9 km/s (17,450 mph to 15,430 mph) respectively
elliptic orbit: 8.2 to 6.5 km/s respectively
89 to 128 min -29.8 MJ/kg
Molniya orbit 6,900 to 46,300 km 500 to 39,900 km 10.0 to 1.5 km/s (22,370 mph to 3,335 mph) respectively 11 h 58 min -4.7 MJ/kg
GEO 42,000 km 35,786 km 3.1 km/s (6,935 mph) 23 h 56 min -4.6 MJ/kg
Orbit of the Moon 363,000 to 406,000 km 357,000 to 399,000 km 1.08 to 0.97 km/s (2,416 to 2,170 mph) respectively 27.3 days -0.5 MJ/kg