# Surface gravity

Surface gravity

The surface gravity, "g", of an astronomical or other object is the gravitational acceleration experienced at its surface. The surface gravity may be thought of as the acceleration due to gravity experienced by a hypothetical test particle which is very close to the object's surface and which, in order not to disturb the system, has negligible mass.

Surface gravity is measured in units of acceleration, which, in the SI system, are meters per second squared. It may also be expressed as a multiple of the Earth's standard surface gravity, "g" = 9.80665 m/s2. [p. 29, [http://physics.nist.gov/Document/sp330.pdf The International System of Units (SI)] , ed. Barry N. Taylor, NIST Special Publication 330, 2001.] In astrophysics, the surface gravity may be expressed as log "g", which is obtained by first expressing the gravity in cgs units, where the unit of acceleration is centimeters per second squared, and then taking the base 10 logarithm. [cite web
last = Smalley | first = B. | date = 2006-07-13
url =http://www.astro.keele.ac.uk/~bs/publs/review_text.html
title =The Determination of T"eff" and log "g" for B to G stars
publisher = Keele University | accessdate = 2007-05-31
]

In the Newtonian theory of gravity, the gravitational force exerted by an object is proportional to its mass: an object with twice the mass produces twice as much force. Newtonian gravity also follows an inverse square law, so that moving an object twice as far away divides its gravitational force by four, and moving it ten times as far away divides it by 100. This is similar to the intensity of light, which also follows an inverse square law: lights far away give the observer less light.

A large object, such as a planet or star, will usually be approximately round, because large mountains will be squashed down, and large valleys filled in, by the object's own gravity. This hydrostatic equilibrium condition usually also applies to changes of density inside the object, so that the density at a place inside the object will depend only on how far it is from the object's center. If this is so, the object is called "spherically symmetric". [ [http://www.newton.dep.anl.gov/askasci/gen99/gen99251.htm Why is the Earth round?] , at Ask A Scientist, accessed online May 27, 2007.] The gravitational force outside a spherically symmetric body is the same as if its entire mass were concentrated in the center, as was established by Sir Isaac Newton. [Book I, &sect;XII, pp. 218&ndash;226, "Newton's Principia: The Mathematical Principles of Natural Philosophy", Sir Isaac Newton, tr. Andrew Motte, ed. N. W. Chittenden. New York: Daniel Adee, 1848. First American edition.] Therefore, the surface gravity of a planet or star will be approximately inversely proportional to the square of its radius. For example, the recently-discovered planet, Gliese 581 c, has at least 5 times the mass of Earth, but is unlikely to have 5 times its surface gravity. If its mass is no more than 5 times that of the Earth, as is expected, [ [http://www.eso.org/public/outreach/press-rel/pr-2007/pr-22-07.html Astronomers Find First Earth-like Planet in Habitable Zone] , ESO 22/07, press release from the European Southern Observatory, April 25, 2007] and if it is a rocky planet with a large iron core, it should have a radius approximately 50% larger than that of Earth. [ [http://adsabs.harvard.edu/abs/2007arXiv0704.3841U The HARPS search for southern extra-solar planets XI. Super-Earths (5 & 8 M_Earth) in a 3-planet system] , S. Udry, X. Bonfils), X. Delfosse, T. Forveille, M. Mayor, C. Perrier, F. Bouchy, C. Lovis, F. Pepe, D. Queloz, and J.-L. Bertaux. arXiv:astro-ph/0704.3841.] [http://adsabs.harvard.edu/abs/2007arXiv0704.3454V Detailed Models of super-Earths: How well can we infer bulk properties?] , Diana Valencia, Dimitar D. Sasselov, and Richard J. O'Connell, arXiv:astro-ph/0704.3454.] Gravity on such a planet's surface would be approximately 2.2 times as strong as on Earth. If it is an icy or watery planet, its radius might be as large as twice the Earth's, in which case its surface gravity might be no more than 1.25 times as strong as the Earth's.

These proportionalities may be expressed by the formula "g" = "m"/"r"2, where "g" is the surface gravity of an object, expressed as a multiple of the Earth's, "m" is its mass, expressed as a multiple of the Earth's mass (5.976·1024 kg) and "r" its radius, expressed as a multiple of the Earth's (mean) radius (6371 km). [ [http://www.kayelaby.npl.co.uk/general_physics/2_7/2_7_4.html 2.7.4 Physical properties of the Earth] , web page, accessed on line May 27, 2007.] For instance, Mars has a mass of 6.4185·1023 kg = 0.107 Earth masses and a mean radius of 3390 km = 0.532 Earth radii. [ [http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html Mars Fact Sheet] , web page at NASA NSSDC, accessed May 27, 2007.] The surface gravity of Mars is therefore approximately:$frac\left\{0.107\right\}\left\{0.532^2\right\} = 0.38$ times that of Earth. Without using the Earth as a reference body, the surface gravity may also be calculated directly from Newton's Law of Gravitation, which gives the formula:$g = \left\{frac\left\{GM\right\}\left\{r^2$where "M" is the mass of the object, "r" is its radius, and "G" is the gravitational constant.If we let ρ = "m"/"V" denote the mean density of the object, we can also write this as:$g = \left\{frac\left\{4pi\right\}\left\{3\right\} G ho r\right\}$so that, for fixed mean density, the surface gravity "g" is proportional to the radius "r".

Since gravity is inversely proportional to the square of the distance, a space station 100 miles above the Earth feels almost the same gravitational force as we do on the Earth's surface. The reason a space station does not plummet to the ground is not that it is not subject to gravity, but that it is in a free-fall orbit.

Non-spherically symmetric objects

Most real astronomical objects are not absolutely spherically symmetric. One reason for this is that they are often rotating, which means that they are affected by the combined effects of gravitational force and centrifugal force. This causes stars and planets to be oblate, which means that their surface gravity is smaller at the equator than at the poles. This effect was exploited by Hal Clement in his SF novel "Mission of Gravity", dealing with a massive, fast-spinning planet where gravity was much higher at the poles than at the equator.

To the extent that an object's internal distribution of mass differs from a symmetric model, we may use the measured surface gravity to deduce things about the object's internal structure. This fact has been put to practical use since 1915&ndash;1916, when Roland Eötvös's torsion balance was used to prospect for oil near the city of Egbell (now Gbely, Slovakia.) [ [http://www.lct.com/technical-pages/pdf/Li_G_Tut.pdf Ellipsoid, geoid, gravity, geodesy, and geophysics] , Xiong Li and Hans-J&uuml;rgen G&ouml;tze, "Geophysics", 66, #6 (November&ndash;December 2001), pp. 1660&ndash;1668. DOI [http://dx.doi.org/10.1190/1.1487109 10.1190/1.1487109] .] , p. 1663; [http://www.pp.bme.hu/ci/2002_2/pdf/ci2002_2_09.pdf Prediction by E&ouml;tv&ouml;s' torsion balance data in Hungary] , Gyula T&oacute;th, "Periodica Polytechnica Ser. Civ. Eng." 46, #2 (2002), pp. 221&ndash;229.] , p. 223. In 1924, the torsion balance was used to locate the Nash Dome oil fields in Texas., p. 223. It is sometimes useful to calculate the surface gravity of simple hypothetical objects which are not found in nature. The surface gravity of infinite planes, tubes, lines, hollow shells, cones, and even more unrealistic structures may be used to provide insights into the behavior of real structures.

urface gravity of a black hole

In relativity, the Newtonian concept of acceleration turns out not to be clear cut. For a black hole, which can only be truly treated relativistically, one cannot define a surface gravity as the acceleration experienced by a test body at the object's surface. This is because the acceleration of a test body at the event horizon of a black hole turns out to be infinite in relativity.

Therefore, when one talks about the surface gravity of a black hole, one is defining a notion that behaves analogously to the Newtonian surface gravity, but is not the same thing. In fact, the surface gravity of a general black hole is not well defined. However, one can define the surface gravity for a black hole whose event horizon is a Killing horizon.

The surface gravity $kappa$ of a static Killing horizon is the acceleration, as exerted at infinity, needed to keep an object at the horizon. Mathematically, if $k^a$ is a suitably normalized Killing vector, then the surface gravity is defined by

:$k^a abla_a k^b = kappa k^b$,

where the equation is evaluated at the horizon. For a static and asymptotically flat spacetime, the normalization should be chosen so that $k^a k_a ightarrow -1$ as $r ightarrowinfty$, and so that $kappa geq 0$. For the Schwarzschild solution, we take $k^a$ to be the time translation Killing vector $k^apartial_a = frac\left\{partial\right\}\left\{partial t\right\}$, and more generally for the Kerr-Newman solution we take $k^apartial_a = frac\left\{partial\right\}\left\{partial t\right\}+Omegafrac\left\{partial\right\}\left\{partialphi\right\}$, the linear combination of the time translation and axisymmetry Killing vectors which is null at the horizon, where $Omega$ is the angular velocity.

The Schwarzschild solution

Since $k^a$ is a Killing vector $k^a abla_a k^b = kappa k^b$ implies $-k^a abla^b k_a = kappa k^b$. In $\left(t,r, heta,phi\right)$ coordinates $k^\left\{a\right\}=\left(1,0,0,0\right)$. Performing a coordinate change to the advanced Eddington-Finklestein coordinates $v = t+r+2Mln |r-2M|$ causes the metric to take the form $ds^\left\{2\right\} = -left\left(1-frac\left\{2M\right\}\left\{r\right\} ight\right)dv^\left\{2\right\}+2dvdr+r^\left\{2\right\}\left(d heta^\left\{2\right\}+sin^\left\{2\right\} heta dphi^\left\{2\right\}\right)$.

Under a general change of coordinates the Killing vector transforms as $k^\left\{v\right\} = A_\left\{t\right\}^\left\{v\right\}k^\left\{t\right\}$ giving the vectors $k^\left\{a\text{'}\right\}=\left(1,0,0,0\right)$ and $k_\left\{a\text{'}\right\} = left\left(-1+frac\left\{2M\right\}\left\{r\right\},1,0,0 ight\right)$

Considering the b=v entry for $k^a abla_a k^b = kappa k^b$ gives the differential equation $-frac\left\{1\right\}\left\{2\right\}frac\left\{partial\right\}\left\{partial r\right\}left\left(-1+frac\left\{2M\right\}\left\{r\right\} ight\right) = kappa$

Therefore the surface gravity for the Schwarzschild solution with mass $M$ is $kappa = frac\left\{1\right\}\left\{4M\right\}$.

The Kerr-Newman solution

The surface gravity for the Kerr-Newman solution is

:$kappa = frac\left\{r_+-r_-\right\}\left\{2\left(r_+^2+a^2\right)\right\} = frac\left\{sqrt\left\{M^2-Q^2-J^2/M^2\left\{2M^2-Q^2+2Msqrt\left\{M^2-Q^2-J^2/M^2$,

where $Q$ is the electric charge, $J$ is the angular velocity, we define $r_pm := M pm sqrt\left\{M^2-Q^2-J^2/M^2\right\}$ to be the locations of the two horizons and $a := J/M$.

References

* [http://farside.ph.utexas.edu/teaching/301/lectures/node152.html Newtonian surface gravity]

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