Specific orbital energy

In the gravitational two-body problem, the specific orbital energy

For an elliptic 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).

where For an elliptic orbit with specific angular momentum h given by

we use the general form of the specific orbital energy equation,

For a parabolic orbit this equation simplifies to

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

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

) and is equal to the excess specific energy compared to that for a parabolic orbit.

It is related to the hyperbolic excess velocity

is known, then the energy can be computed and from that, for any other position, the orbital speed.

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

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

is the height the ellipse extends above the surface, plus the periapsis distance (the distance the ellipse extends beyond the center of the Earth).

; which is the kinetic energy of the horizontal component of the velocity, i.e.

The International Space Station has an orbital period of 91.74 minutes (5504 s), hence by Kepler's Third Law the semi-major axis of its orbit is 6,738 km.

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 1.5–2.0 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.

Compare with the potential energy at the surface, which is −62.6 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).

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.

Assume: Then the time-rate of change of the specific energy of the rocket is

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.

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 gradually making an elliptic orbit smaller, it means applying thrust each time when near the periapsis.