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.
Contents 1 Mass, radius and surface gravity
2 Non-spherically symmetric objects
3
3.1 Schwarzschild solution 3.2 Kerr solution 3.3 Kerr–Newman solution 3.4 Dynamical black holes 4 References 5 External links Mass, radius and surface gravity[edit]
Name Surface gravity Sun 28.02g Mercury 0.38g Venus 0.904g Earth 1g Moon 0.1654g Mars 0.376g Phobos 0.0005814g Deimos 0.000306g Ceres 0.0275g Jupiter 2.53g Io 0.183g Europa 0.134g Ganymede 0.146g Callisto 0.126g Saturn 1.07g Titan 0.14g Enceladus 0.0113g Uranus 0.89g Neptune 1.14g Triton 0.0797g Pluto 0.067g Eris 0.0677g 67P-CG 0.000017g 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.
g = m r 2 displaystyle g= frac 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 (6,371 km).[9] For instance, Mars
has a mass of 6.4185·1023 kg = 0.107
0.107 0.532 2 = 0.38 displaystyle frac 0.107 0.532^ 2 =0.38 times that of Earth. Without using the
g = G M r 2 displaystyle g= frac GM 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 = 4 π 3 G ρ r displaystyle g= frac 4pi 3 Grho r 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 400 km above the
κ displaystyle kappa of a static
k a displaystyle k^ a is a suitably normalized Killing vector, then the surface gravity is defined by k a ∇ a k b = κ k b , displaystyle 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 → − 1 displaystyle k^ a k_ a rightarrow -1 as r → ∞ displaystyle rrightarrow infty , and so that κ ≥ 0 displaystyle kappa geq 0 . For the Schwarzschild solution, we take k a displaystyle k^ a to be the time translation
k a ∂ a = ∂ ∂ t displaystyle k^ a partial _ a = frac partial partial t , and more generally for the
k a ∂ a = ∂ ∂ t + Ω ∂ ∂ φ displaystyle 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 Ω displaystyle Omega is the angular velocity. Schwarzschild solution[edit] Since k a displaystyle k^ a is a
k a ∇ a k b = κ k b displaystyle k^ a ,nabla _ a k^ b =kappa k^ b implies − k a ∇ b k a = κ k b displaystyle -k^ a ,nabla ^ b k_ a =kappa k^ b . In ( t , r , θ , φ ) displaystyle (t,r,theta ,varphi ) coordinates k a = ( 1 , 0 , 0 , 0 ) displaystyle k^ a =(1,0,0,0) . Performing a coordinate change to the advanced Eddington–Finklestein coordinates v = t + r + 2 M ln
r − 2 M
displaystyle v=t+r+2Mln r-2M causes the metric to take the form d s 2 = − ( 1 − 2 M r ) d v 2 + 2 d v d r + r 2 ( d θ 2 + sin 2 θ d φ 2 ) . displaystyle ds^ 2 =-left(1- frac 2M r right),dv^ 2 +2,dv,dr+r^ 2 left(dtheta ^ 2 +sin ^ 2 theta ,dvarphi ^ 2 right). Under a general change of coordinates the
k v = A t v k t displaystyle k^ v =A_ t ^ v k^ t giving the vectors k a ′ = ( 1 , 0 , 0 , 0 ) displaystyle k^ a' =(1,0,0,0) and k a ′ = ( − 1 + 2 M r , 1 , 0 , 0 ) . displaystyle k_ a' =left(-1+ frac 2M r ,1,0,0right). Considering the b = v entry for k a ∇ a k b = κ k b displaystyle k^ a ,nabla _ a k^ b =kappa k^ b gives the differential equation − 1 2 ∂ ∂ r ( − 1 + 2 M r ) = κ . displaystyle - frac 1 2 frac partial partial r left(-1+ frac 2M r right)=kappa . Therefore, the surface gravity for the
M displaystyle M is κ = 1 4 M . displaystyle kappa = frac 1 4M . [13] Kerr solution[edit] The surface gravity for the uncharged, rotating black hole is, simply κ = g − k , displaystyle kappa =g-k, where g = 1 4 M displaystyle g= frac 1 4M is the Schwarzschild surface gravity, and k := M Ω + 2 displaystyle k:=MOmega _ + ^ 2 is the spring constant of the rotating black hole. Ω + displaystyle Omega _ + is the angular velocity at the event horizon. This expression gives a simple Hawking temperature of 2 π T = g − k displaystyle 2pi T=g-k .[14]
Kerr–Newman solution[edit]
The surface gravity for the
κ = r + − r − 2 ( r + 2 + a 2 ) = M 2 − Q 2 − J 2 / M 2 2 M 2 − Q 2 + 2 M M 2 − Q 2 − J 2 / M 2 , displaystyle kappa = frac r_ + -r_ - 2(r_ + ^ 2 +a^ 2 ) = 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 displaystyle Q is the electric charge, J displaystyle J is the angular momentum, we define r ± := M ± M 2 − Q 2 − J 2 / M 2 displaystyle r_ pm :=Mpm sqrt M^ 2 -Q^ 2 -J^ 2 /M^ 2 to be the locations of the two horizons and a := J / M displaystyle a:=J/M .
Dynamical black holes[edit]
^ p. 29, The International System of Units (SI), ed. Barry N. Taylor,
NIST
External links[edit] Newtonian surface gravity Exploratorium – Your Weight on |