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Categories: Astrodynamics | Celestial mechanics

Specific orbital energy

In astrodynamics the specific orbital energy $\epsilon\,\!$ (or vis-viva energy) of an orbiting body traveling through space under standard assumptions is the sum of its potential energy ( $\epsilon_p\,\!$ ) and kinetic energy ( $\epsilon_k\,\!$ ) per unit mass. According to the orbital energy conservation equation (also referred to as vis-viva equation) it is the same at all points of the trajectory:

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

where:

$v\,\!$ is the orbital speed of the orbiting body
$r\,\!$ is the orbital position of the orbiting body
$\mu\,\!$ is the standard gravitational parameter
$h\,\!$ is the specific relative angular momentum of the orbiting body
$e\,\!$ is the orbit eccentricity

It is expressed in J/kg = m²s^-2 or MJ/kg = km²s^-2.

Contents

1 Equation forms for different orbits

Equation forms for different orbits

For an elliptical orbit specific orbital energy equation simplifies to:

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

where:

$\mu\,\!$ is the standard gravitational parameter
$a\,\!$ is semi-major axis of the orbiting body

For a parabolic orbit this equation simplifies to:

$\epsilon=0\,\!$

For a hyperbolic trajectory this specific orbital energy equation takes form:

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

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 an escape orbit (parabolic orbit).

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

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

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:

$\mu\,\!$ is the standard gravitational parameter
$a\,\!$ is semi-major axis of the orbiting body

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.

Additional 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 this is $(2 a - R) g$ ; $2 a - R$ 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

The International Space Station has an orbital period of 91.74 minutes, hence the semi-major axis is 6738 km [1].

The energy is −29.6 MJ/kg [2]: 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 [3].

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

The energy is −30.8 MJ/kg [4]: 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 [5], the net delta-v to reach this orbit is 8.0 km/s [6].

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).

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:

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

Categories: Astrodynamics | Celestial mechanics

Science Fair Project Encyclopedia Contents Page

03-10-2013 05:06:04
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