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Rabu, 11 Juni 2014

SpaceX Perkenalkan Kapsul Antariksa Baru Pengganti Soyuz TMA

Elon Musk berfoto bersama dengan kapsul Dragon V2. Image credit: SpaceX
Baru-baru ini perusahaan antariksa swasta terkenal asal Amerika, SpaceX, memperkenalkan kapsul antariksa baru yang akan digunakan untuk membawa astronot ke Stasiun Luar Angkasa Internasional (ISS). Kapsul yang diberinama Dragon V2 itu akan menggantikan fungsi dari pesawat ulang alik (Space Shuttle) yang sudah dipensiunkan NASA pada 2011 lalu. Kapsul Dragon V2 ini sekaligus akan menggantikan peran dari kapsul Soyuz yang selama 3 tahun ini digunakan oleh astronot untuk terbang ke ISS, dimana biaya penerbangan (pergi-pulang) untuk satu orang astronot sekira 70 juta dollar.

Pelan Tapi Pasti Bulan Pergi Meninggalkan Bumi

Pelan Tapi Pasti Bulan Pergi Meninggalkan Bumi

Bulan. Image credit: wikiversity
Bumi dan Bulan bisa dibilang sebagai sahabat yang tumbuh, berkembang dan menjalani hidup bersama-sama untuk mengelilingi Matahari dan galaksi. Berdasarkan penelitian, Bulan lahir sebagai akibat dari tabrakan antara Bumi muda dengan obyek seukuran planet Mars yang disebut Theia sekira 4,5 miliar tahun yang lalu. Kemudian, kedua obyek baik Bumi dan Bulan bersama-sama menjalani hidup berdua.

Rabu, 30 April 2014

Parameters - Longitude Of The Periapsis

Longitude of the periapsis


In celestial mechanics, the longitude of the periapsis (symbolized ϖ) of an orbiting body is the longitude (measured from the point of the vernal equinox) at which the periapsis (closest approach to the central body) would occur if the body's inclination were zero. For motion of a planet around the Sun, this position could be called longitude of perihelion. The longitude of periapsis is a compound angle, with part of it being measured in the plane of reference and the rest being measured in the plane of the orbit. Likewise, any angle derived from the longitude of periapsis (e.g. mean longitude and true longitude) will also be compound.
Sometimes, the term longitude of periapsis is used to refer to ω, the angle between the ascending node and the periapsis. That usage of the term is especially common in discussions of binary stars and exoplanets.[1] However, the angle ω is less ambiguously known as the argument of periapsis.

Parameters - Argument Of Periapsis

Argument of periapsis


The argument of periapsis (also called argument of perifocus or argument of pericenter), symbolized as ω, is one of the orbital elements of an orbiting body. Specifically, ω is the angle between the orbit's periapsis (the point of closest approach to the central point) and the orbit's ascending node (the point where the body crosses the plane of reference from South to North). The angle is measured in the orbital plane and in the direction of motion. For specific types of orbits, words such as "perihelion" (for Sun-centered orbits), "perigee" (for Earth-centered orbits), "periastron" (for orbits around stars) and so on may replace the word "periapsis". See apsis for more information.

An argument of periapsis of 0° means that the orbiting body will be at its closest approach to the central body at the same moment that it crosses the plane of reference from South to North. An argument of periapsis of 90° means that the orbiting body will reach periapsis at its northmost distance from the plane of reference.
Adding the argument of periapsis to the longitude of the ascending node gives the longitude of the periapsis. However, especially in discussions of binary stars and exoplanets, the terms "longitude of periapsis" or "longitude of periastron" are often used synonymously with "argument of periapsis".

Parameters - Longitude Of The Asending Node

Longitude of the ascending node


The longitude of the ascending node.
The longitude of the ascending node (☊ or Ω) is one of the orbital elements used to specify the orbit of an object in space. It is the angle from a reference direction, called the origin of longitude, to the direction of the ascending node, measured in a reference plane.[1] Commonly used reference planes and origins of longitude include:

Paramaters - Azimuth

Azimuth


The azimuth is the angle formed between a reference direction (North) and a line from the observer to a point of interest projected on the same plane as the reference direction
An azimuth (Listeni/ˈæzɪməθ/) is an angular measurement in a spherical coordinate system. The vector from an observer (origin) to a point of interest is projected perpendicularly onto a reference plane; the angle between the projected vector and a reference vector on the reference plane is called the azimuth.
An example is the position of a star in the sky. The star is the point of interest, the reference plane is the horizon or the surface of the sea, and the reference vector points north. The azimuth is the angle between the north vector and the perpendicular projection of the star down onto the horizon.[1]
Azimuth is usually measured in degrees (°). The concept is used in navigation, astronomy, engineering, mapping, mining and artillery.

Parameters - Horizontal Coordinate System

Horizontal coordinate system


Azimuth is measured from the north point (sometimes from the south point) of the horizon around to the east; altitude is the angle above the horizon.
The horizontal coordinate system is a celestial coordinate system that uses the observer's local horizon as the fundamental plane. This coordinate system divides the sky into the upper hemisphere where objects are visible, and the lower hemisphere where objects cannot be seen since the earth is in the way. The great circle separating hemispheres is called celestial horizon or rational horizon. The pole of the upper hemisphere is called the zenith. The pole of the lower hemisphere is called the nadir. [1]
The horizontal coordinates are:

Parameters - Orbital inclination Change

Orbital inclination change


Orbital inclination change is an orbital maneuver aimed at changing the inclination of an orbiting body's orbit. This maneuver is also known as an orbital plane change as the plane of the orbit is tipped. This maneuver requires a change in the orbital velocity vector (delta v) at the orbital nodes (i.e. the point where the initial and desired orbits intersect, the line of orbital nodes is defined by the intersection of the two orbital planes).
In general, inclination changes can take a very large amount of delta v to perform, and most mission planners try to avoid them whenever possible to conserve fuel. This is typically achieved by launching a spacecraft directly into the desired inclination, or as close to it as possible so as to minimize any inclination change required over the duration of the spacecraft life. Planetary flybys are the most efficient way to achieve large inclination changes, but they are only effective for interplanetary missions.

Parameters - Orbital Inclination

Orbital inclination



Fig. 1: One view of inclination i (green) and other orbital parameters
Inclination in general is the angle between a reference plane and another plane or axis of direction.

 

Parameters - Eccentricity

Orbital eccentricity



An elliptic Kepler orbit with an eccentricity of 0.7 (red ellipse), a parabolic Kepler orbit (green) and a hyperbolic Kepler orbit with an eccentricity of 1.3 (blue outer line)
The orbital eccentricity of an astronomical object is a parameter that determines the amount by which its orbit around another body deviates from a perfect circle. A value of 0 is a circular orbit, values between 0 and 1 form an elliptical orbit, 1 is a parabolic escape orbit, and greater than 1 is a hyperbola. The term derives its name from the parameters of conic sections, as every Kepler orbit is a conic section. It is normally used for the isolated two-body problem, but extensions exist for objects following a rosette orbit through the galaxy.

Parameters - Apsis

Apsis


"Apogee", "Aphelion", "Perigee" and "Perihelion" redirect here. For the literary journal, see Perigee: Publication for the Arts. For Edenbridge's Album, see Aphelion (album). For the architectural term, see Apse. For other uses, see Apogee (disambiguation) and Perihelion (disambiguation).

Apsides 1) Apoapsis; 2) Periapsis; 3)
Focus
An apsis (Greek ἁψίς, gen. ἁψίδος), plural apsides (/ˈæpsɨdz/; Greek: ἁψίδες), is a point of greatest or least distance of a body in an elliptic orbit about a larger body. For a body orbiting the Sun the greatest and least distance points are called respectively aphelion and perihelion (/æpˈhliən/, /ˌpɛrɨˈhliən/), whereas for any satellite of Earth including the Moon the corresponding points are apogee and perigee (/ˈpɛrɨ/). The generic suffix, independent of the particular central body, can be either apsis or centre, hence apoapsis, apocentre or apapsis (from ἀπ(ό) (ap(ó)), meaning "from"), and periapsis or pericentre (from περί (peri), meaning "around"). During the Apollo program, the terms pericynthion and apocynthion (referencing Cynthia, an alternative name for the Greek Moon goddess Artemis) were used when referring to the Moon.[1]
A straight line connecting the periapsis and apoapsis is the line of apsides. This is the major axis of the ellipse, its greatest diameter. For a two-body system the center of mass of the system lies on this line at one of the two foci of the ellipse. When one body is sufficiently larger than the other it may be taken to be at this focus. However whether or not this is the case, both bodies are in similar elliptical orbits each having one focus at the system's center of mass, with their respective lines of apsides being of length inversely proportional to their masses. Historically, in geocentric systems, apsides were measured from the center of the Earth. However in the case of the Moon, the center of mass of the Earth-Moon system or Earth-Moon barycenter, as the common focus of both the Moon's and Earth's orbits about each other, is about 75% of the way from Earth's center to its surface.
In orbital mechanics, the apsis technically refers to the distance measured between the centers of mass of the central and orbiting body. However, in the case of spacecraft, the family of terms are commonly used to describe the orbital altitude of the spacecraft from the surface of the central body (assuming a constant, standard reference radius)

Parameters - Semi-minor Axis

Semi-minor axis



The semi-major (in red*) and semi-minor axis (in blue*) of an ellipse.
(* on some browsers)
In geometry, the semi-minor axis (also semiminor axis) is a line segment associated with most conic sections (that is, with ellipses and hyperbolas) that is at right angles with the semi-major axis and has one end at the center of the conic section. It is one of the axes of symmetry for the curve: in an ellipse, the shorter one; in a hyperbola, the one that does not intersect the hyperbola.

Parameters - Semi-major Axis

Semi-major axis



The semi-major and semi-minor axis of an ellipse
In geometry, the major axis of an ellipse is the longest diameter: a line (line segment) that runs through the center and both foci, with ends at the widest points of the shape. The semi-major axis is one half of the major axis, and thus runs from the centre, through a focus, and to the edge of the ellipse; essentially, it is the radius of an orbit at the orbit's two most distant points. For the special case of a circle, the semi-major axis is the radius. One can think of the semi-major axis as an ellipse's long radius.
The length of the semi-major axis a of an ellipse is related to the semi-minor axis' length b through the eccentricity e and the semi-latus rectum , as follows:

Orbit - Two-line Element Set

Two-line element set

Orbit - Tundra Orbit

Tundra orbit

Orbit - Polar Orbit

Polar orbit

Orbit - Orbit Of The Moon

Orbit of the Moon

Orbit - Molniya Orbit

Orbit - High Earth Orbit

High Earth orbit

Orbit - Medium Earth Orbit

Medium Earth orbit

Orbit - Low Earth Orbit

Orbit - Geostationary Orbit

Geostationary orbit


Geostationary orbits (top view). To an observer on the rotating Earth, both satellites appear stationary in the sky at their respective locations.
Geostationary orbits (side view)
A 5 × 6 degree view of a part of the geostationary belt, showing several geostationary satellites. Those with inclination 0 degrees form a diagonal belt across the image; a few objects with small inclinations to the Equator are visible above this line. The satellites are pinpoint, while stars have created small trails due to the Earth's rotation.
A geostationary orbit, geostationary Earth orbit or geosynchronous equatorial orbit[1] (GEO), is an orbit whose position in sky remains the same for a stationary observer on earth. This effect is achieved with a circular orbit 35,786 kilometres (22,236 mi) above the Earth's equator and following the direction of the Earth's rotation.[2] An object in such an orbit has an orbital period equal to the Earth's rotational period (one sidereal day), and thus appears motionless, at a fixed position in the sky, to ground observers. Communications satellites and weather satellites are often placed in geostationary orbits, so that the satellite antennas which communicate with them do not have to move to track them, but can be pointed permanently at the position in the sky where they stay. Using this characteristic, ocean color satellites with visible sensors (e.g. the Geostationary Ocean Color Imager (GOCI)) can also be operated in geostationary orbit in order to monitor sensitive changes of ocean environments.
A geostationary orbit is a particular type of geosynchronous orbit, the distinction being that while an object in geosynchronous orbit returns to the same point in the sky at the same time each day, an object in geostationary orbit never leaves that position.
The notion of a geosynchronous satellite for communication purposes was first published in 1928 (but not widely so) by Herman Potočnik.[3] The first appearance of a geostationary orbit in popular literature was in the first Venus Equilateral story by George O. Smith,[4] but Smith did not go into details. British science fiction author Arthur C. Clarke disseminated the idea widely, with more details on how it would work, in a 1945 paper entitled "Extra-Terrestrial Relays — Can Rocket Stations Give Worldwide Radio Coverage?", published in Wireless World magazine. Clarke acknowledged the connection in his introduction to The Complete Venus Equilateral.[5] The orbit, which Clarke first described as useful for broadcast and relay communications satellites,[6] is sometimes called the Clarke Orbit.[7] Similarly, the Clarke Belt is the part of space about 35,786 km (22,236 mi) above sea level, in the plane of the Equator, where near-geostationary orbits may be implemented. The Clarke Orbit is about 265,000 km (165,000 mi) long.

Practical uses

Most commercial communications satellites, broadcast satellites and SBAS satellites operate in geostationary orbits. A geostationary transfer orbit is used to move a satellite from low Earth orbit (LEO) into a geostationary orbit. (Russian television satellites have used elliptical Molniya and Tundra orbits due to the high latitudes of the receiving audience.) The first satellite placed into a geostationary orbit was the Syncom-3, launched by a Delta-D rocket in 1964.
A worldwide network of operational geostationary meteorological satellites is used to provide visible and infrared images of Earth's surface and atmosphere. These satellite systems include:
A statite, a hypothetical satellite that uses a solar sail to modify its orbit, could theoretically hold itself in a geostationary "orbit" with different altitude and/or inclination from the "traditional" equatorial geostationary orbit.

Orbital stability

A geostationary orbit can only be achieved at an altitude very close to 35,786 km (22,236 mi), and directly above the Equator. This equates to an orbital velocity of 3.07 km/s (1.91 mi/s) or a period of 1,436 minutes, which equates to almost exactly one sidereal day or 23.934461223 hours. This ensures that the satellite is locked to the Earth's rotational period and has a stationary footprint on the ground. All geostationary satellites have to be located on this ring.
A combination of lunar gravity, solar gravity, and the flattening of the Earth at its poles causes a precession motion of the orbital plane of any geostationary object, with a period of about 53 years and an initial inclination gradient of about 0.85 degrees per year, achieving a maximum inclination of 15 degrees after 26.5 years. To correct for this orbital perturbation, regular orbital stationkeeping manoeuvres are necessary, amounting to a delta-v of approximately 50 m/s per year.
A second effect to be taken into account is the longitude drift, caused by the asymmetry of the Earth – the Equator is slightly elliptical. There are two stable (at 75.3°E, and at 104.7°W) and two unstable (at 165.3°E, and at 14.7°W) equilibrium points. Any geostationary object placed between the equilibrium points would (without any action) be slowly accelerated towards the stable equilibrium position, causing a periodic longitude variation. The correction of this effect requires orbit control manoeuvres with a maximum delta-v of about 2 m/s per year, depending on the desired longitude.
Solar wind and radiation pressure also exert small forces on satellites which, over time, cause them to slowly drift away from their prescribed orbits.
In the absence of servicing missions from the Earth or a renewable propulsion method, the consumption of thruster propellant for station-keeping places a limitation on the lifetime of the satellite.

Communications

Satellites in geostationary orbits are far enough away from Earth that communication latency becomes significant — about a quarter of a second for a trip from one ground-based transmitter to the satellite and back to another ground-based transmitter; close to half a second for a round-trip communication from one Earth station to another and then back to the first.
For example, for ground stations at latitudes of φ = ±45° on the same meridian as the satellite, the time taken for a signal to pass from Earth to the satellite and back again can be computed using the cosine rule, given the geostationary orbital radius r (derived below), the Earth's radius R and the speed of light c, as
\Delta t = \frac{2}{c} \sqrt{R^2 + r^2 - 2 R r \cos\varphi} \approx253\,\mathrm{ms}
(Note that r is the orbital radius, the distance from the centre of the Earth, not the height above the Equator.)
This delay presents problems for latency-sensitive applications such as voice communication.[8]
Geostationary satellites are directly overhead at the Equator, and become lower in the sky the further north or south one travels. As the observer's latitude increases, communication becomes more difficult due to factors such as atmospheric refraction, Earth's thermal emission, line-of-sight obstructions, and signal reflections from the ground or nearby structures. At latitudes above about 81°, geostationary satellites are below the horizon and cannot be seen at all.[9]

Orbit allocation

Satellites in geostationary orbit must all occupy a single ring above the Equator. The requirement to space these satellites apart to avoid harmful radio-frequency interference during operations means that there are a limited number of orbital "slots" available, thus only a limited number of satellites can be operated in geostationary orbit. This has led to conflict between different countries wishing access to the same orbital slots (countries near the same longitude but differing latitudes) and radio frequencies. These disputes are addressed through the International Telecommunication Union's allocation mechanism.[10][11] In the 1976 Bogotá Declaration, eight countries located on the Earth's equator claimed sovereignty over the geostationary orbits above their territory, but the claims gained no international recognition.[12]

Limitations to usable life of geostationary satellites

When they run out of thruster fuel, the satellites are at the end of their service life as they are no longer able to keep in their allocated orbital position. The transponders and other onboard systems generally outlive the thruster fuel and, by stopping N-S station keeping, some satellites can continue to be used in inclined orbits (where the orbital track appears to follow a figure-eight loop centred on the Equator),[13][14] or else be elevated to a "graveyard" disposal orbit.

Derivation of geostationary altitude

Comparison of Geostationary Earth Orbit with GPS, GLONASS, Galileo and Compass (medium earth orbit) satellite navigation system orbits with the International Space Station, Hubble Space Telescope and Iridium constellation orbits, and the nominal size of the Earth.[a] The Moon's orbit is around 9 times larger (in radius and length) than geostationary orbit.[b]
In any circular orbit, the centripetal force required to maintain the orbit (Fc) is provided by the gravitational force on the satellite (Fg). To calculate the geostationary orbit altitude, one begins with this equivalence:
\mathbf{F}_\text{c} = \mathbf{F}_\text{g}
By Newton's second law of motion,[15] we can replace the forces F with the mass m of the object multiplied by the acceleration felt by the object due to that force:
m \mathbf{a}_\text{c} = m \mathbf{g}
We note that the mass of the satellite m appears on both sides — geostationary orbit is independent of the mass of the satellite.[c] So calculating the altitude simplifies into calculating the point where the magnitudes of the centripetal acceleration required for orbital motion and the gravitational acceleration provided by Earth's gravity are equal.
The centripetal acceleration's magnitude is:
|\mathbf{a}_\text{c}| = \omega^2 r
where ω is the angular speed, and r is the orbital radius as measured from the Earth's center of mass.
The magnitude of the gravitational acceleration is:
|\mathbf{g}| = \frac{G M}{r^2}
where M is the mass of Earth, 5.9736 × 1024 kg, and G is the gravitational constant, 6.67428 ± 0.00067 × 10−11 m3 kg−1 s−2.
Equating the two accelerations gives:
r^3 = \frac{G M}{\omega^2} \to r = \sqrt[3]{\frac{G M}{\omega^2}}
The product GM is known with much greater precision than either factor alone; it is known as the geocentric gravitational constant μ = 398,600.4418 ± 0.0008 km3 s−2:
r = \sqrt[3]{\frac\mu{\omega^2}}
The angular speed ω is found by dividing the angle travelled in one revolution (360° = 2π rad) by the orbital period (the time it takes to make one full revolution). In the case of a geostationary orbit, the orbital period is one sidereal day, or 86,164.09054 seconds).[16] This gives:
\omega \approx \frac{2 \mathrm\pi~\mathrm{rad}} {86\,164~\mathrm{s}} \approx 7.2921 \times 10^{-5}~\mathrm{rad} / \mathrm{s}
The resulting orbital radius is 42,164 kilometres (26,199 mi). Subtracting the Earth's equatorial radius, 6,378 kilometres (3,963 mi), gives the altitude of 35,786 kilometres (22,236 mi).
Orbital speed (how fast the satellite is moving through space) is calculated by multiplying the angular speed by the orbital radius:
v = \omega r \approx 3.0746~\mathrm{km}/\mathrm{s} \approx 11\,068~\mathrm{km}/\mathrm{h} \approx 6877.8~\mathrm{mph}\text{.}
By the same formula we can find the geostationary-type orbit of an object in relation to Mars (this type of orbit above is referred to as an areostationary orbit if it is above Mars). The geocentric gravitational constant GM (which is μ) for Mars has the value of 42,828 km3s−2, and the known rotational period (T) of Mars is 88,642.66 seconds. Since ω = 2π/T, using the formula above, the value of ω is found to be approx 7.088218×10−5 s−1. Thus, r3 = 8.5243×1012 km3, whose cube root is 20,427 km; subtracting the equatorial radius of Mars (3396.2 km) we have 17,031 km.

Orbit - Geosynchronous Orbit

Geosynchronous orbit

Orbit - Sun-synchronous Orbit

Sun-synchronous orbit

Orbit - Heliocentric Orbit

Heliocentric orbit

Orbit - Lunar Orbit

Lunar orbit

Orbit - Lissajous Orbit

Lissajous orbit

Orbit - Halo Orbit

Halo orbit

Orbit - Aerostationary Orbit

Areostationary orbit

Orbit - Aerosynchronous Orbit

Areosynchronous orbit


Areosynchronous orbits are class of synchronous orbits for artificial satellites around the planet Mars. As with all synchronous orbits, an areosynchronous orbit has an orbital period equal in length to Mars's sidereal day. A satellite in areosynchronous orbit does not necessarily maintain a fixed position in the sky as seen by an observer on the surface of Mars, however such a satellite will return to the same apparent position every Martian day.
The orbital altitude required to maintain an areosynchonous orbit is approximately 17,000 kilometres (11,000 mi). If a satellite in areosynchonous orbit were to be used as a communication relay link, it "would experience communications ranges of 17,000 to 20,000 kilometres (11,000 to 12,000 mi)" to various points on the visible Martian surface.[1]
An areosynchronous orbit that is equatorial (in the same plane as the equator of Mars), circular, and prograde (rotating about Mars's axis in the same direction as the planet's surface) is known as an areostationary orbit (ASO). To an observer on the surface of Mars, the position of a satellite in ASO would appear to be fixed in a constant position in the sky. The ASO is analogous to a geostationary orbit (GSO) about Earth.[citation needed]
Although no satellites currently occupy areosynchronous or areostationary orbits, some scientists[who?] foresee a future telecommunications network for the exploration of Mars.[citation needed]

Orbit - Synchronous Orbit

Synchronous orbit

Orbit - Parking Orbit

Orbit - Osculating Orbit

Orbit - Inclined Orbit

Orbit - Hyperbolic Trajectory

Orbit - Graveyard Orbit

Orbit - Parabolic Trajectory

Orbit - Highly Elliptical Orbit

Orbit - Elliptic Orbit

Elliptic orbit

Orbit - Circular Orbit

Circular orbit


A circular orbit is depicted in the top-left quadrant of this diagram, where the gravitational potential well of the central mass shows potential energy, and the kinetic energy of the orbital speed is shown in red. The height of the kinetic energy remains constant throughout the constant speed circular orbit.
For other meanings of the term "orbit", see orbit (disambiguation)
A circular orbit is the orbit at a fixed distance around any point by an object rotating around a fixed axis.
Below we consider a circular orbit in astrodynamics or celestial mechanics under standard assumptions. Here the centripetal force is the gravitational force, and the axis mentioned above is the line through the center of the central mass perpendicular to the plane of motion.
In this case not only the distance, but also the speed, angular speed, potential and kinetic energy are constant. There is no periapsis or apoapsis. This orbit has no radial version.

Circular acceleration

Transverse acceleration (perpendicular to velocity) causes change in direction. If it is constant in magnitude and changing in direction with the velocity, we get a circular motion. For this centripetal acceleration we have
 \mathbf{a} = - \frac{v^2}{r} \frac{\mathbf{r}}{r} = - \omega^2 \mathbf{r}
where:

Velocity

The relative velocity is constant:
 v = \sqrt{ G(M\!+\!m) \over{r}} = \sqrt{\mu\over{r}}
where:

Equation of motion

The orbit equation in polar coordinates, which in general gives r in terms of θ, reduces to:
r={{h^2}\over{\mu}}
where:
This is just another way of writing \mu=rv^2 again.

Angular speed and orbital period

\omega^2 r^3=\mu
Hence the orbital period (T\,\!) can be computed as:
T=2\pi\sqrt{r^3\over{\mu}}
Compare two proportional quantities, the free-fall time (time to fall to a point mass from rest)
T_{ff}=\frac{\pi}{2\sqrt{2}}\sqrt{r^3\over{\mu}} (17.7 % of the orbital period in a circular orbit)
and the time to fall to a point mass in a radial parabolic orbit
T_{par}=\frac{\sqrt{2}}{3}\sqrt{r^3\over{\mu}} (7.5 % of the orbital period in a circular orbit)
The fact that the formulas only differ by a constant factor is a priori clear from dimensional analysis.

Energy

The specific orbital energy (\epsilon\,) is negative, and
{v^2\over{2}}=-\epsilon
-{\mu\over{r}}=2\epsilon
Thus the virial theorem applies even without taking a time-average:
  • the kinetic energy of the system is equal to the absolute value of the total energy
  • the potential energy of the system is equal to twice the total energy
The escape velocity from any distance is √2 times the speed in a circular orbit at that distance: the kinetic energy is twice as much, hence the total energy is zero.

Delta-v to reach a circular orbit

Maneuvering into a large circular orbit, e.g. a geostationary orbit, requires a larger delta-v than an escape orbit, although the latter implies getting arbitrarily far away and having more energy than needed for the orbital speed of the circular orbit. It is also a matter of maneuvering into the orbit. See also Hohmann transfer orbit.

Orbital velocity in general relativity

In Schwarzschild metric, the orbital velocity for a circular orbit with radius R is given by the following formula:
v = \sqrt{\frac{GM}{r-r_S}}
where \scriptstyle r_S = \frac{2GM}{c^2} is the Schwarzschild radius of the central body.

Derivation

For the sake of convenience, the derivation will be written in units in which \scriptstyle c=G=1.
The four-velocity of a body on a circular orbit is given by:
u^\mu = (\dot{t}, 0, 0, \dot{\phi})
(\scriptstyle r is constant on a circular orbit, and the coordinates can be chosen so that \scriptstyle \theta=\frac{\pi}{2}). The dot above a variable denotes derivation with respect to proper time \scriptstyle \tau.
For a massive particle, the components of the four-velocity satisfy the following equation:
\left(1-\frac{2M}{r}\right) \dot{t}^2 - r^2 \dot{\phi}^2 = 1
We use the geodesic equation:
\ddot{x}^\mu + \Gamma^\mu_{\nu\sigma}\dot{x}^\nu\dot{x}^\sigma = 0
The only nontrivial equation is the one for \scriptstyle \mu = r. It gives:
\frac{M}{r^2}\left(1-\frac{2M}{r}\right)\dot{t}^2 - r\left(1-\frac{2M}{r}\right)\dot{\phi}^2 = 0
From this, we get:
\dot{\phi}^2 = \frac{M}{r^3}\dot{t}^2
Substituting this into the equation for a massive particle gives:
\left(1-\frac{2M}{r}\right) \dot{t}^2 - \frac{M}{r} \dot{t}^2 = 1
Hence:
\dot{t}^2 = \frac{r}{r-3M}
Assume we have an observer at radius \scriptstyle r, who is not moving with respect to the central body, that is, his four-velocity is proportional to the vector \scriptstyle \partial_t. The normalization condition implies that it is equal to:
v^\mu = \left(\sqrt{\frac{r}{r-2M}},0,0,0\right)
The dot product of the four-velocities of the observer and the orbiting body equals the gamma factor for the orbiting body relative to the observer, hence:
\gamma = g_{\mu\nu}u^\mu v^\nu = \left(1-\frac{2M}{r}\right) \sqrt{\frac{r}{r-3M}} \sqrt{\frac{r}{r-2M}} = \sqrt{\frac{r-2M}{r-3M}}
This gives the velocity:
v = \sqrt{\frac{M}{r-2M}}
Or, in SI units:
v = \sqrt{\frac{GM}{r-r_S}}