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The surface gravity, g, of an astronomical object is the gravitational acceleration experienced at its surface at the equator, including the effects of rotation. The surface gravity may be thought of as the 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. 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 the system, are meters per second squared. It may also be expressed as a multiple of the 's , which is equal to

In , the surface gravity may be expressed as , which is obtained by first expressing the gravity in , where the unit of acceleration and surface gravity is per second squared (cm/s2), and then taking the base-10 of the cgs value of the surface gravity. Therefore, the surface gravity of Earth could be expressed in cgs units as , and then taking the base-10 ("log  g") of 980.665, giving 2.992 as "log  g".

The surface gravity of a is very high, and of a even higher. A white dwarf's surface gravity is around 100,000 g () whilst the neutron star's compactness gives it a surface gravity of up to with typical values of order (that is more than 1011 times that of Earth). One measure of such immense gravity is that neutron stars have an of around 100,000 km/s, about a third of the speed of light. Since black holes do not have a surface, the surface gravity is not defined.


Relationship of surface gravity to mass and radius
+ Surface gravity of various
Solar System bodies
(1978). 9780552108843, Corgi.

(1  g = 9.80665 m/s2, the average surface gravitational acceleration on Earth)
28.02 g
0.377 g
0.905 g
1 g (midlatitudes)
0.165 7 g (average)
0.379 g (midlatitudes)
0.000 581 g
0.000 306 g
0.022 g (equator)
0.025 g (equator)
0.029 g
2.528 g (midlatitudes)
0.183 g
0.134 g
0.146 g
0.126 g
1.065 g (midlatitudes)
0.006 48 g
0.011 5 g
0.014 9 g
0.023 7 g
0.026 9 g
0.138 g
0.022 8 g
0.003 9–0.005 1 g
0.886 g (equator)
0.007 9 g
0.025 4 g
0.023 g
0.037 2 g
0.036 1 g
1.137 g (midlatitudes)
0.007 g
0.079 4 g
0.063 g
0.029 4 g
0.084 g
0.0247 g (equator)
0.000 017 g
In the Newtonian theory of , 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 , 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 a or , will usually be approximately round, approaching hydrostatic equilibrium (where all points on the surface have the same amount of gravitational 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-perfect when the rotation rate is low. However, for young, massive stars, the equatorial velocity can be quite high—up to 200 km/s or more—causing a significant amount of . Examples of such include , , and .

The fact that many large celestial objects are approximately spheres makes it easier to calculate their surface gravity. According to the , 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, §XII, pp. 218–226, 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 with a given mass will be approximately inversely proportional to the square of its , 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 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, 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. 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} where is the surface gravity of an object, expressed as a multiple of the Earth's, is its mass, expressed as a multiple of the 's mass () and its radius, expressed as a multiple of the Earth's (mean) radius (6,371 km). 2.7.4 Physical properties of the Earth, web page, accessed on line May 27, 2007. For instance, has a mass of  = 0.107 Earth masses and a mean radius of 3,390 km = 0.532 Earth radii. Mars Fact Sheet, web page at NASA NSSDC, accessed May 27, 2007. The surface gravity of Mars is 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 from Newton's law of universal gravitation, which gives the formula g = \frac{GM}{r^2} where is the mass of the object, is its radius, and is the gravitational constant. If denote the mean of the object, this can also be written as g = \frac{4\pi}{3} G \rho r so that, for fixed mean density, the surface gravity is proportional to the radius . Solving for mass, this equation can be written as g = G \left ( \frac{4\pi \rho}{3} \right ) ^{2/3} M^{1/3} But density is not constant, but increases as the planet grows in size, as they are not incompressible bodies. That is why the experimental relationship between surface gravity and mass does not grow as 1/3 but as 1/2: g = M^{1/2} here with in times Earth's surface gravity and in times Earth's mass. In fact, the exoplanets found fulfilling the former relationship have been found to be rocky planets. Thus, for rocky planets, density grows with mass as \rho \propto M^{1/4}.


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. It has been found that for giant planets with masses in the range up to 100 times Earth's mass, their surface gravity is nevertheless very similar and close to 1, a region named the gravity plateau.


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 of and centrifugal force. This causes stars and planets to be , which means that their surface gravity is smaller at the equator than at the poles. This effect was exploited by 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, the measured surface gravity may be used to deduce things about the object's internal structure. This fact has been put to practical use since 1915–1916, when Roland Eötvös's was used to prospect for near the city of (now , .) Prediction by Eötvös' torsion balance data in Hungary , Gyula Tóth, Periodica Polytechnica Ser. Civ. Eng. 46, #2 (2002), pp. 221–229. In 1924, the torsion balance was used to locate the oil fields in .

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 the gravitational time dilation factor (which goes to zero at the event horizon). For the Schwarzschild case, this value is mathematically well behaved for all non-zero values of and .

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 is the acceleration, as exerted at infinity, needed to keep an object at the horizon. Mathematically, if k^a is a suitably normalized , 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, take k^a to be the k^a \partial_a = \frac \partial {\partial t}, and more generally for the Kerr–Newman solution take 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 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^{a'}_{v} =(1,0,0,0) and k_{a'} = g_{a'v} = \left(-1+\frac{2M} r ,1,0,0\right).

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

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

(2025). 9781848163829, Imperial College Press. .
Extract of page 44


Kerr solution
The surface gravity for the uncharged, rotating black hole is, simply \kappa = g - k , where 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 the Kerr–Newman solution is \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, define 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.
(1984). 9780226870335, University Of Chicago Press. .
Recently there has been a shift towards defining the surface gravity of dynamical black holes whose spacetime does not admit a timelike Killing 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. Semiclassical results indicate that the peeling surface gravity is ill-defined for transient objects formed in finite time of a distant observer.


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