Specific orbital energy

In thegravitational two-body problem,thespecific orbital energy$\varepsilon$(orvis-viva energy) of twoorbiting bodiesis the constant sum of their mutualpotential energy($\varepsilon _{p}$) and theirkinetic energy($\varepsilon _{k}$), divided by thereduced mass.According to theorbital energy conservation equation(also referred to as vis-viva equation), it does not vary with time: $${\begin{aligned}\varepsilon &=\varepsilon _{k}+\varepsilon _{p}\\&={\frac {v^{2}}{2}}-{\frac {\mu }{r}}=-{\frac {1}{2}}{\frac {\mu ^{2}}{h^{2}}}\left(1-e^{2}\right)=-{\frac {\mu }{2a}}\end{aligned}}$$ where

• $v$is the relativeorbital speed;
• $r$is theorbital distancebetween the bodies;
• $\mu ={G}(m_{1}+m_{2})$is the sum of thestandard gravitational parametersof the bodies;
• $h$is thespecific relative angular momentumin the sense ofrelative angular momentumdivided by the reduced mass;
• $e$is theorbital eccentricity;
• $a$is thesemi-major axis.

It is typically expressed in${\frac {\text{MJ}}{\text{kg}}}$(megajouleper kilogram) or${\frac {{\text{km}}^{2}}{{\text{s}}^{2}}}$(squared kilometer per squared second). For anelliptic orbitthe specific orbital energy is the negative of the additional energy required to accelerate a mass of one kilogram toescape velocity(parabolic orbit). For ahyperbolic 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 ascharacteristic energy.

Equation forms for different orbits

For anelliptic orbit,the specific orbital energy equation, when combined withconservation of specific angular momentumat one of the orbit'sapsides,simplifies to:

$$\varepsilon =-{\frac {\mu }{2a}}$$ where

• $\mu =G\left(m_{1}+m_{2}\right)$is thestandard gravitational parameter;
• $a$issemi-major axisof the orbit.

Proof

For an elliptic orbit withspecific angular momentumhgiven by $$h^{2}=\mu p=\mu a\left(1-e^{2}\right)$$ we use the general form of the specific orbital energy equation, $$\varepsilon ={\frac {v^{2}}{2}}-{\frac {\mu }{r}}$$ with the relation that the relative velocity atperiapsisis $$v_{p}^{2}={h^{2} \over r_{p}^{2}}={h^{2} \over a^{2}(1-e)^{2}}={\mu a\left(1-e^{2}\right) \over a^{2}(1-e)^{2}}={\mu \left(1-e^{2}\right) \over a(1-e)^{2}}$$ Thus our specific orbital energy equation becomes $$\varepsilon ={\frac {\mu }{a}}{\left[{1-e^{2} \over 2(1-e)^{2}}-{1 \over 1-e}\right]}={\frac {\mu }{a}}{\left[{(1-e)(1+e) \over 2(1-e)^{2}}-{1 \over 1-e}\right]}={\frac {\mu }{a}}{\left[{1+e \over 2(1-e)}-{2 \over 2(1-e)}\right]}={\frac {\mu }{a}}{\left[{e-1 \over 2(1-e)}\right]}$$ and finally with the last simplification we obtain: $$\varepsilon =-{\mu \over 2a}$$

For aparabolic orbitthis equation simplifies to $$\varepsilon =0.$$

For ahyperbolic trajectorythis specific orbital energy is either given by $$\varepsilon ={\mu \over 2a}.$$

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

In this case the specific orbital energy is also referred to ascharacteristic energy(or$C_{3}$) and is equal to the excess specific energy compared to that for a parabolic orbit.

It is related to thehyperbolic excess velocity$v_{\infty }$(theorbital velocityat infinity) by $$2\varepsilon =C_{3}=v_{\infty }^{2}.$$

It is relevant for interplanetary missions.

Thus, iforbital position vector($\mathbf {r}$) andorbital 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 elliptic 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

• $\mu ={G}(m_{1}+m_{2})$is thestandard gravitational parameter;
• $a\,\!$issemi-major axisof the orbit.

In the case of circular orbits, this rate is one half of the gravitation 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.

Additional energy

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

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

The quantity$2a-R$is the height the ellipse extends above the surface, plus theperiapsis distance(the distance the ellipse extends beyond the center of the Earth). For the Earth and$a$just little more than$R$the additional specific energy is$(gR/2)$;which is the kinetic energy of the horizontal component of the velocity, i.e.${\textstyle {\frac {1}{2}}V^{2}={\frac {1}{2}}gR}$,$V={\sqrt {gR}}$.

Examples

ISS

TheInternational Space Stationhas anorbital periodof 91.74 minutes (5504s), hence byKepler's Third Lawthe semi-major axis of its orbit is 6,738km.

The specific orbital energy associated with this orbit is −29.6MJ/kg: the potential energy is −59.2MJ/kg, and the kinetic energy 29.6MJ/kg. Compared with the potential energy at the surface, which is −62.6MJ/kg., the extra potential energy is 3.4MJ/kg, and the total extra energy is 33.0MJ/kg. The average speed is 7.7km/s, the netdelta-vto reach this orbit is 8.1km/s (the actual delta-v is typically 1.5–2.0km/s more foratmospheric dragandgravity drag).

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

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

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

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

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

Voyager 1

ForVoyager 1,with respect to the Sun:

• $\mu =GM$= 132,712,440,018 km³⋅s⁻²is thestandard gravitational parameterof the Sun
• r= 17billionkilometers
• v= 17.1 km/s

Hence: $$\varepsilon =\varepsilon _{k}+\varepsilon _{p}={\frac {v^{2}}{2}}-{\frac {\mu }{r}}=\mathrm {146\,km^{2}s^{-2}} -\mathrm {8\,km^{2}s^{-2}} =\mathrm {138\,km^{2}s^{-2}}$$

Thus the hyperbolic excess velocity (the theoreticalorbital velocityat infinity) is given by $$v_{\infty }=\mathrm {16.6\,km/s}$$

However,Voyager 1does not have enough velocity to leave theMilky 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:

• ais the acceleration due tothrust(the time-rate at whichdelta-vis spent)
• gis the gravitational field strength
• vis 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 betweenvanda.

Thus, when applying delta-v to increase specific orbital energy, this is done most efficiently ifais applied in the direction ofv,and when |v| is large. If the angle betweenvandgis 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 alsogravity 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. Such maneuver is called anOberth maneuveror powered flyby.

When applying delta-v todecreasespecific orbital energy, this is done most efficiently ifais applied in the direction opposite to that ofv,and again when |v| is large. If the angle betweenvandgis 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.

Ifais in the direction ofv: $$\Delta \varepsilon =\int v\,d(\Delta v)=\int v\,adt$$

Tangential velocities at altitude

Orbit	Center-to-center distance	Altitude above the Earth's surface	Speed	Orbital period	Specific orbital energy
Earth's own rotation at surface (for comparison— not an orbit)	6,378km	0km	465.1m/s(1,674km/h or 1,040mph)	23h 56min 4.09sec	−62.6MJ/kg
Orbiting at Earth's surface (equator) theoretical	6,378km	0km	7.9km/s (28,440km/h or 17,672mph)	1h 24min 18sec	−31.2MJ/kg
Low Earth orbit	6,600–8,400km	200–2,000km	Circular orbit: 7.7–6.9km/s (27,772–24,840km/h or 17,224–15,435mph) respectively Elliptic orbit: 10.07–8.7km/s respectively	1h 29min – 2h 8min	−29.8MJ/kg
Molniya orbit	6,900–46,300km	500–39,900km	1.5–10.0km/s (5,400–36,000km/h or 3,335–22,370mph) respectively	11h 58min	−4.7MJ/kg
Geostationary	42,000km	35,786km	3.1km/s (11,600km/h or 6,935mph)	23h 56min 4.09sec	−4.6MJ/kg
Orbit of the Moon	363,000–406,000km	357,000–399,000km	0.97–1.08km/s (3,492–3,888km/h or 2,170–2,416mph) respectively	27.27days	−0.5MJ/kg

The lower axis gives orbital speeds of some orbits.

See also

• Specific energy change of rockets
• Characteristic energyC3 (Double the specific orbital energy)