Surface gravity

Thesurface gravity,g,of anastronomical objectis thegravitational accelerationexperienced at its surface at the equator, including the effects of rotation. The surface gravity may be thought of as theaccelerationdue 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. For objects where the surface is deep in the atmosphere and the radius not known, the surface gravity is given at the 1 bar pressure level in the atmosphere.

Surface gravity is measured in units of acceleration, which, in theSIsystem, aremeters per second squared.It may also be expressed as a multiple of theEarth'sstandard surface gravity,which is equal to

g=9.80665m/s²

Inastrophysics,the surface gravity may be expressed aslogg,which is obtained by first expressing the gravity incgs units,where the unit of acceleration and surface gravity iscentimetersper second squared (cm/s²), and then taking the base-10logarithmof the cgs value of the surface gravity.Therefore, the surface gravity of Earth could be expressed in cgs units as980.665 cm/s²,and then taking the base-10logarithm( "logg") of 980.665, and we get 2.992 as" logg".

The surface gravity of awhite dwarfis very high, and of aneutron stareven higher. A white dwarf's surface gravity is around 100,000g(10⁶m/s²) whilst the neutron star's compactness gives it a surface gravity of up to7×10¹²m/s²with typical values of order10¹²m/s²(that is more than 10¹¹times that of Earth). One measure of such immense gravity is that neutron stars have anescape velocityof around100,000 km/s,about a third of thespeed of light.For black holes, the surface gravity must be calculated relativistically.

Relationship of surface gravity to mass and radius

Surface gravity of various
Solar System bodies

(1g= 9.80665 m/s²,the average surface gravitational acceleration on Earth)

Name	Surface gravity
Sun	28.02g
Mercury	00.377g
Venus	00.905g
Earth	01g(midlatitudes)
Moon	00.165 7g(average)
Mars	00.379g(midlatitudes)
Phobos	00.000 581g
Deimos	00.000 306g
Pallas	00.022g(equator)
Vesta	00.025g(equator)
Ceres	00.029g
Jupiter	02.528g(midlatitudes)
Io	00.183g
Europa	00.134g
Ganymede	00.146g
Callisto	00.126g
Saturn	01.065g(midlatitudes)
Mimas	00.006 48g
Enceladus	00.011 5g
Tethys	00.014 9g
Dione	00.023 7g
Rhea	00.026 9g
Titan	00.138g
Iapetus	00.022 8g
Phoebe	00.003 9–0.005 1g
Uranus	00.886g(equator)
Miranda	00.007 9g
Ariel	00.025 4g
Umbriel	00.023g
Titania	00.037 2g
Oberon	00.036 1g
Neptune	01.137g(midlatitudes)
Proteus	00.007g
Triton	00.079 4g
Pluto	00.063g
Charon	00.029 4g
Eris	00.084g
Haumea	00.0247g(equator)
67P-CG	00.000 017g

In theNewtoniantheory ofgravity,thegravitational forceexerted by an object is proportional to its mass: an object with twice the mass produces twice as much force. Newtonian gravity also follows aninverse 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 oflight,which also follows an inverse square law: with relation to distance, light becomes less visible. Generally speaking, this can be understood as geometric dilution corresponding to point-source radiation into three-dimensional space.

A large object, such as aplanetorstar,will usually be approximately round, approachinghydrostatic equilibrium(where all points on the surface have the same amount ofgravitational potential energy). On a small scale, higher parts of the terrain are eroded, with eroded material deposited in lower parts of the terrain. On a large scale, the planet or star itself deforms until equilibrium is reached.For most celestial objects, the result is that the planet or star in question can be treated as a near-perfectspherewhen the rotation rate is low. However, for young, massive stars, the equatorialazimuthalvelocity can be quite high—up to 200 km/s or more—causing a significant amount ofequatorial bulge.Examples of suchrapidly rotating starsincludeAchernar,Altair,Regulus AandVega.

The fact that many large celestial objects are approximately spheres makes it easier to calculate their surface gravity. According to theshell theorem,the gravitational force outside a spherically symmetric body is the same as if its entire mass were concentrated in the center, as was established bySir Isaac Newton.Therefore, the surface gravity of aplanetorstarwith a given mass will be approximately inversely proportional to the square of itsradius,and the surface gravity of a planet or star with a given average density will be approximately proportional to its radius. For example, the recently discoveredplanet,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,and if it is a rocky planet with a large iron core, it should have a radius approximately 50% larger than that of Earth.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: $$g\propto {\frac {m}{r^{2}}}$$ wheregis the surface gravity of an object, expressed as a multiple of theEarth's,mis its mass, expressed as a multiple of theEarth's mass (5.976×10²⁴kg) andrits radius, expressed as a multiple of the Earth's (mean) radius (6,371 km).For instance,Marshas a mass of6.4185×10²³kg= 0.107 Earth masses and a mean radius of 3,390 km = 0.532 Earth radii.The surfacegravity of Marsis therefore approximately $${\frac {0.107}{0.532^{2}}}=0.38$$ times that of Earth. Without using the Earth as a reference body, the surface gravity may also be calculated directly fromNewton's law of universal gravitation,which gives the formula $$g={\frac {GM}{r^{2}}}$$ whereMis the mass of the object,ris its radius, andGis thegravitational constant.If we letρ=M/Vdenote the meandensityof the object, we can also write this as $$g={\frac {4\pi }{3}}G\rho r$$ so that, for fixed mean density, the surface gravitygis proportional to the radiusr.

Gas giants

For gas giant planets such as Jupiter, Saturn, Uranus, and Neptune, the surface gravity is given at the 1 bar pressure level in the atmosphere.

Non-spherically symmetric objects

Most real astronomical objects are not perfectly spherically symmetric. One reason for this is that they are often rotating, which means that they are affected by the combined effects ofgravitational forceandcentrifugal force.This causes stars and planets to beoblate,which means that their surface gravity is smaller at the equator than at the poles. This effect was exploited byHal Clementin his SF novelMission 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 1915–1916, whenRoland Eötvös'storsion balancewas used to prospect foroilnear the city ofEgbell(nowGbely,Slovakia.)In 1924, the torsion balance was used to locate theNash Domeoil fields inTexas.

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.

Black holes

In relativity, the Newtonian concept of acceleration turns out not to be clear cut. For a black hole, which must be treated relativistically, one cannot define a surface gravity as the acceleration experienced by a test body at the object's surface because there is no surface, although the event horizon is a natural alternative candidate, but this still presents a problem because the acceleration of a test body at the event horizon of a black hole turns out to be infinite in relativity. Because of this, a renormalized value is used that corresponds to the Newtonian value in the non-relativistic limit. The value used is generally the local proper acceleration (which diverges at the event horizon) multiplied by thegravitational time dilationfactor (which goes to zero at the event horizon). For the Schwarzschild case, this value is mathematically well behaved for all non-zero values ofrandM.

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 staticKilling horizonis the acceleration, as exerted at infinity, needed to keep an object at the horizon. Mathematically, if$k^{a}$is a suitably normalizedKilling vector,then the surface gravity is defined by $$k^{a}\,\nabla _{a}k^{b}=\kappa k^{b},$$ where the equation is evaluated at the horizon. For a static and asymptotically flat spacetime, the normalization should be chosen so that$k^{a}k_{a}\to -1$as$r\to \infty$,and so that$\kappa \geq 0$.For the Schwarzschild solution, we take$k^{a}$to be thetime translationKilling vector${\textstyle k^{a}\partial _{a}={\frac {\partial }{\partial t}}}$,and more generally for theKerr–Newman solutionwe take${\textstyle k^{a}\partial _{a}={\frac {\partial }{\partial t}}+\Omega {\frac {\partial }{\partial \varphi }}}$,the linear combination of the time translation and axisymmetry Killing vectors which is null at the horizon, where$\Omega$is the angular velocity.

Schwarzschild solution

Since$k^{a}$is a Killing vector$k^{a}\,\nabla _{a}k^{b}=\kappa k^{b}$implies$-k^{a}\,\nabla ^{b}k_{a}=\kappa k^{b}$.In$(t,r,\theta,\varphi )$coordinates$k^{a}=(1,0,0,0)$.Performing a coordinate change to the advanced Eddington–Finklestein coordinates${\textstyle v=t+r+2M\ln |r-2M|}$causes the metric to take the form $$ds^{2}=-\left(1-{\frac {2M}{r}}\right)\,dv^{2}+\left(dv\,dr+\,dr\,dv\right)+r^{2}\left(d\theta ^{2}+\sin ^{2}\theta \,d\varphi ^{2}\right).$$

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

Considering theb=$v$entry for$k^{a}\,\nabla _{a}k^{b}=\kappa k^{b}$gives the differential equation${\textstyle -{\frac {1}{2}}{\frac {\partial }{\partial r}}\left(-1+{\frac {2M}{r}}\right)=\kappa.}$

Therefore, the surface gravity for theSchwarzschild solutionwith mass$M$is$\kappa ={\frac {1}{4M}}$($\kappa ={c^{4}}/{4GM}$in SI units).

Kerr solution

The surface gravity for the uncharged, rotating black hole is, simply $$\kappa =g-k,$$ where${\textstyle g={\frac {1}{4M}}}$is the Schwarzschild surface gravity, and$k:=M\Omega _{+}^{2}$is the spring constant of the rotating black hole.$\Omega _{+}$is the angular velocity at the event horizon. This expression gives a simple Hawking temperature of$2\pi T=g-k$.

Kerr–Newman solution

The surface gravity for theKerr–Newman solutionis $$\kappa ={\frac {r_{+}-r_{-}}{2\left(r_{+}^{2}+a^{2}\right)}}={\frac {\sqrt {M^{2}-Q^{2}-J^{2}/M^{2}}}{2M^{2}-Q^{2}+2M{\sqrt {M^{2}-Q^{2}-J^{2}/M^{2}}}}},$$ where$Q$is the electric charge,$J$is the angular momentum, we define${\textstyle r_{\pm }:=M\pm {\sqrt {M^{2}-Q^{2}-J^{2}/M^{2}}}}$to be the locations of the two horizons and$a:=J/M$.

Dynamical black holes

Surface gravity for stationary black holes is well defined. This is because all stationary black holes have a horizon that is Killing.Recently there has been a shift towards defining the surface gravity of dynamical black holes whose spacetime does not admit a timelikeKilling vector (field).Several definitions have been proposed over the years by various authors, such as peeling surface gravity and Kodama surface gravity.As of current, there is no consensus or agreement on which definition, if any, is correct.Semiclassicalresults indicate that the peeling surface gravity is ill-defined for transient objects formed in finite time of a distant observer.

External links

• Newtonian surface gravity
• Exploratorium – Your Weight on Other Worlds