<p>I am not an expert in physics and the explanation of the others is excellent. However, I noticed a flaw in your reasoning which they did not address.</p>
<p>You have written:</p>
<blockquote>
<p>Considering the Newton's Law of Gravity equation <span class="math-container">$F = GM/r^2$</span>, if the <strong>radius</strong> of an object becomes super small, then it can technically have immense gravity.</p>
</blockquote>
<p>Hence I deduce that you read the <span class="math-container">$r$</span> in <a href="https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation" rel="noreferrer">the equation</a> as the <em>radius</em> of the object, while in fact it is the <strong>distance between two objects</strong>. So there is no rationale based on this formula that &quot;the smaller the radius is, the stronger the gravity must be&quot;. The proper reading would be &quot;the closer the object is from the black hole, the stronger the gravity is&quot;, but this is valid for all the bodies, not only to black holes.</p>

CC BY-SA 4.0
2021-05-10T19:26:21.793Z
Ahh, right. I sometimes get confused between radius and distance. Thanks for spotting that.
AdiBak
CC BY-SA 4.0
2021-05-11T12:10:22.843Z
There is a certain connection between radius and distance because only a mass with very small spatial (possibly infinitely small) spatial extent creates gravity strong enough that GR effects become noticeable on small scales.
Peter - Reinstate Monica
CC BY-SA 4.0
2021-05-11T19:04:34.647Z
@Peter-ReinstateMonica Well, you might be right, but the Newton's formula is about a **distance of two bodies**, not about the size of one of them. So the reasoning in the OP was based on the incorrect application of the formula.
Honza Zidek
CC BY-SA 4.0
2021-05-12T14:05:08.063Z
@AdiBak: the smaller the radius of the object, the smaller you can be without being inside it.  So the radius is a lower bound on $r$.  (Fun fact which you may have learned: the net gravity *inside* a symmetric spherical shell is 0 everywhere, so burrowing into an object such as the Earth, you can still use $GM/r^2$ but where M only includes the mass of the part below you, not the part you're inside.  (Of course assuming that Newtonian mechanics is a good enough approximation for whatever you're burrowing into, like maybe not a neutron star.)
Peter Cordes
CC BY-SA 4.0
2021-05-10T13:25:06.253Z
2021-05-10T13:25:06.253Z
Honza Zidek
<p>While admiring <a href="https://astronomy.stackexchange.com/a/43738/7982">@ProfRob's answer</a> I'll add some additional perspective/background that may serve as a helpful stepping-stone since not every Astronomy SE reader is prepared to embrace General Relativity in all its glory.</p>
<blockquote>
<p>Can Newton's gravity equation explain why black holes are so strong?</p>
</blockquote>
<ol>
<li>The simple equation<sup>1</sup> <span class="math-container">$F = GMm/r^2$</span> <em>doesn't explain anything</em> but it certainly does give a result that's usually quite useful in Newtonian mechanics where space is &quot;normal&quot; and we can talk about gravitation as a force.</li>
<li>General relativity can be said to <em>offer an explanation</em> of how gravitation works and it gives results that work not only where Newtonian gravity works but in extreme situations as well, where speeds are very high and/or gravity is very very strong (and other situations as well).</li>
<li>Even in situations we are somewhat familliar with, like Mercury's orbit around the Sun or satellites orbiting the Earth, predictions by Newtonian mechanics is measurably off and GR gets it spot-on.</li>
<li>On the flip side, if you want to <em>approximate</em> a trajectory of a star or bit of dust that passes far from a black hole, you certainly can go ahead and use Newton's equations. Trajectories will still be close to Keplerian orbits.</li>
</ol>
<p>I'll make up some numbers for illustration purposes only. If a 20 solar mass star goes supernova and you add up all the ejected mass and energy and it's 12 solar masses, then you'll expect an 8 solar mass black hole left over.</p>
<p>If there is a companion star orbiting it at a large distance, or if you fly past it at a &quot;safe distance&quot; and look at what happens, it will be what you'd expect for a mass of 8 solar masses, whether it's a black hole, a neutron star, a regular star or a (magically self-supporting) ball of concrete.</p>
<p>It's only when you get closer that you need to use GR, and you'd have to use it whether it's a black hole or a more conventional dense object like a neutron star.</p>
<p><sup>1</sup>Remember, for a force there are two masses, for acceleration there's only one <span class="math-container">$a=F/m = GM/r^2$</span></p>

CC BY-SA 4.0
2021-05-10T19:34:17.060Z
I see...thanks for answering!
AdiBak
CC BY-SA 4.0
2021-05-10T05:03:15.157Z
2021-05-10T05:03:15.157Z
uhoh
<p>If Einstein's GR equations are expanded in terms of familiar coordinates (Cartesian, spherical,...), the dominant or leading terms of the expansion (for the acceleration) can be written as the single Newtonian term GM/r^2.  The next terms of the expansion can be considered as GR corrections to this leading term.</p>
<p>Before the publication of GR, 19th-century astronomers noticed that the advance of the perihelion of Mercury was not predicted accurately by Newtonian gravitational physics. This became a major problem.  At some point, astronomer/mathematicians added corrections to the Newtonian Law of Gravitation that produced a new/revised force law that accurately predicted perihelion advance. The results were accurate, but the new gravitation theories were not as rich in their &quot;other&quot; predictions as Einstein's GR.</p>

CC BY-SA 4.0
2021-05-10T12:40:29.753Z
Welcome to the site! I'm not seeing a clear connection to black holes here. Can you edit to make that more explicit?
called2voyage
CC BY-SA 4.0
2021-05-10T14:54:45.787Z
Could you provide some examples of these phenomenological alternatives to general relativity?
Daddy Kropotkin
CC BY-SA 4.0
2021-05-10T19:40:12.037Z
The addition of correction terms were designed to accurately account for the advance of the perihelia of the planets.  The 2 best corrections involved one additional term in the expression for the POTENTIAL:  (1) a 1/(c^2r^2) term, or (2) a 1/(c^2r^3) term.  The latter produced the best and most accurate values for the advance of the perihelia.  An excellent, detailed analysis of these 2 "theories" plus a comparison with GR contributions can be found in the paper by James D. Wells (U Michigan): "When effective theories predict: the inevitability of Mercury’s anomalous perihelion precession.
Ange Purs
CC BY-SA 4.0
2021-05-10T19:53:04.520Z
To address part of the original question, the addition of corrections to the Newtonian potential not only accurately predicted the advance of planetary perihelia, but also predicted the Schwarzschild radius:  This is the radius of a black hole's event horizon.  The Newtonian potential doesnt reveal the Schwarzschild radius/event horizon, but adding a small radially-dependent correction, factored by 1/c^2, does.
Ange Purs
CC BY-SA 4.0
2021-05-11T04:15:47.300Z
Several answers to [How to calculate the planets and moons beyond Newtons's gravitational force?](https://space.stackexchange.com/q/23408/12102) contain the first term (or two?), along with one for velocity.
uhoh
CC BY-SA 4.0
2021-05-10T07:02:28.183Z
2021-05-10T08:18:39.760Z
2021-05-10T08:18:39.760Z
Ange Purs
<p>No you can't and the behaviour of bodies with mass and of light is completely different near a compact, massive object if you use Newtonian physics rather than General Relativity.</p>
<p>In no particular order; features that GR predicts (and which in some cases have now been observationally confirmed) but which Newtonian physics cannot:</p>
<ol>
<li><p>An event horizon. In Newtonian physics there is a misleading numerical coincidence that the escape velocity reaches the speed of light at the Schwarzschild radius. But in Newtonian physics you could still escape by applying a constant thrust. GR predicts that no escape is possible in any circumstances.</p>
</li>
<li><p>Further; this numerical coincidence only applies to light travelling radially. In Newtonian physics the escape &quot;speed&quot; is independent of which direction you fire a body, but in GR light cannot escape from (just above) the Schwarzschild unless it if fired radially outwards. For other directions, the radius at which light can escape is larger.</p>
</li>
<li><p>GR predicts an innermost stable circular orbit. A stable circular orbit is possible at any radius in Newtonian physics.</p>
</li>
<li><p>In GR a particle with some angular momentum and lots of kinetic energy will end up falling into the black hole. In Newtonian physics it will scatter to infinity.</p>
</li>
<li><p>Newtonian physics predicts no precession of a two-body elliptical orbit. GR predicts orbital precession.</p>
</li>
<li><p>Newtonian physics predicts that light travelling close to a massive body has a trajectory that is curved by about half the amount predicted by GR. Even stranger effects are predicted close to the black hole including that light can orbit at 1.5 times the Schwarzschild radius.</p>
</li>
</ol>
<p>The GR approach to gravity is fundamentally and philosophically different to Newtonian gravity. For Newton, gravity is a universal force. In GR, gravity is not a force at all. Freefalling bodies are said to be &quot;inertial&quot;. They accelerate, not because a force acts upon them, but because spacetime is curved by the presence of mass (and energy).</p>
<p>In most cases, where Newtonian gravitational fields are weak, the consequences of this difference are small (but measureable - e.g. the orbit precession of Mercury or gravitational time dilation in GPS clocks), but near large, compact masses, like black holes and neutron stars, the differences become stark and unavoidable.</p>

CC BY-SA 4.0
2021-05-10T07:07:24.813Z
I've just asked [Is there anything different about the gravitation around a black hole and a neutron star of the same mass?](https://astronomy.stackexchange.com/q/43750/7982)
uhoh
CC BY-SA 4.0
2021-05-10T13:24:01.227Z
I am not familiar with the maths - is the relationship between the escape velocity and event horizon really coincidental?
kutschkem
CC BY-SA 4.0
2021-05-10T14:19:16.913Z
@kutschkem The Schwarzchild solution to the GR equations predicts a radius at which it is impossible to escape gravity regardless of your velocity or acceleration. The algebra for computing this radius concludes with an expression that turns out to be identical to the expression obtained when setting the escape velocity equal to $c$ in Newtonian gravity. This is a "coincidence" because the Schwarzchild event horizon has no physical relation to escape velocity at all. Nevertheless, in history, Schwarzchild was aware of this algebra before he achieved the full GR solution.
Ross Presser
CC BY-SA 4.0
2021-05-10T19:33:46.947Z
@ProfRob thanks for the detailed answer!
AdiBak
CC BY-SA 4.0
2021-05-11T11:05:32.287Z
@RossPresser I'd expect a <shrug>coincidence</shrug> if the escape velocity at the Schwarzschild radius were an arbitrary value. Alas, it is c. That is most emphatically *no coincidence.* (That the physics are fundamentally different notwithstanding.)
Peter - Reinstate Monica
CC BY-SA 4.0
2021-05-11T11:41:46.603Z
@Peter-ReinstateMonica that is a point of view; but the fact that $c$ doesn't feature in Newtonian physics at all and isn't the speed limit of anything argues the other way. I think it is an unfortunate coincidence (unfortunate because it leads to fundamental misunderstanding of the nature of the event horizon).
ProfRob
CC BY-SA 4.0
2021-05-11T12:30:15.097Z
@Peter-ReinstateMonica a further point which I've now added is that the Newtonian escape speed argument only works for radial trajectories. For Newton it is an escape speed, independent of trajectory. In GR light sent tangentially will not escape unless $r>1.5r_s$ and the two theories disagree.
ProfRob
CC BY-SA 4.0
2021-05-12T07:27:03.170Z
@ProfRob disregarding the fact that $c$ shows up out of nowhere seems odd to me. $c$ makes no appearance in maxwell's laws but yet $1/\sqrt{\epsilon_0\mu_0}$. 

For clarity, I do not disagree that Newton's gravity does not work near black holes, I just disagree that $c$ showing up in this way is a coincidence. I would suppose this just shows that the constants $G$ and $c$ are somehow connected (a connection not explained by Newton's relativity thus indeed one would still require GR to even find this connection)
Poseidaan
CC BY-SA 4.0
2021-05-09T17:45:09.710Z
2021-05-11T12:26:49.517Z
2021-05-11T12:26:49.517Z
ProfRob
<p>I was just wondering why black hole's gravitational forces are so powerful. I know it's usually explained by Einstein's relativity which states that when an object becomes infinitely dense (a compact mass) it can exert such a force of gravity and warp spacetime. But I also learned about Newton's Law of Gravity equation F = <span class="math-container">$GM/r^2$</span>. Considering this equation, if the radius of an object becomes super small, then it can technically have immense gravity. So, can the gravitational pull of a black hole be explained by Newton's Law of Gravity or am I missing something? Thanks.</p>

CC BY-SA 4.0
2021-05-10T15:12:26.590Z
I don't know what the policy is on cross-site duplicates, but there are some good answers on the Physics Stack Exchange here: https://physics.stackexchange.com/questions/19405/can-a-black-hole-be-explained-by-newtonian-gravity
Toby Bartels
CC BY-SA 4.0
2021-05-11T19:32:18.660Z
"...Einstein's relativity which states that when an object becomes infinitely dense (a compact mass) it can exert such a force of gravity and warp spacetime" - not really correct, all objects exert a force of gravity and warp spacetime.    Black holes just to it to such an extreme level that not even light cannot escape.
Jason Goemaat
CC BY-SA 4.0
2021-05-12T00:45:40.203Z
I almost started to think the question's title started with *Ca**m** Newton's gravity equation ...*
Panzercrisis
CC BY-SA 4.0
2021-05-13T12:10:59.477Z
Very simply, I am even surprised of the answers you got. They are good but made assumption on what you are really asking for. A simple answer would be that as far as you use Newton Law of gravity to describe the motion of the Earth around the Sun then nothing change - > Shall you replace the Sun with a black hole nothing would change. Again, the answer below are great but the question itself does not necessarily call for them. Unless you put your test mass near the various horizons of the black hole. In other words the subject here is unclear, the BH or the gravitating body?
Alchimista
<black-hole><gravity><general-relativity><newtonian-gravity>
Can Newton's gravity equation explain why black holes are so strong?
AdiBak
# Can Newton's gravity equation explain why black holes are so strong?

(Asked by: username_0 on 2021-05-09)

I was just wondering why black hole's gravitational forces are so powerful. I know it's usually explained by Einstein's relativity which states that when an object becomes infinitely dense (a compact mass) it can exert such a force of gravity and warp spacetime. But I also learned about Newton's Law of Gravity equation F = $GM/r^2$. Considering this equation, if the radius of an object becomes super small, then it can technically have immense gravity. So, can the gravitational pull of a black hole be explained by Newton's Law of Gravity or am I missing something? Thanks.

**Tags:** black-hole, gravity, general-relativity, newtonian-gravity
> **Comments:**
> **username_1**: I don't know what the policy is on cross-site duplicates, but there are some good answers on the Physics Stack Exchange here: https://physics.stackexchange.com/questions/19405/can-a-black-hole-be-explained-by-newtonian-gravity  
> **username_2**: "...Einstein's relativity which states that when an object becomes infinitely dense (a compact mass) it can exert such a force of gravity and warp spacetime" - not really correct, all objects exert a force of gravity and warp spacetime.    Black holes just to it to such an extreme level that not even light cannot escape.  
> **username_3**: I almost started to think the question's title started with *Ca**m** Newton's gravity equation ...*  
> **username_4**: Very simply, I am even surprised of the answers you got. They are good but made assumption on what you are really asking for. A simple answer would be that as far as you use Newton Law of gravity to describe the motion of the Earth around the Sun then nothing change - > Shall you replace the Sun with a black hole nothing would change. Again, the answer below are great but the question itself does not necessarily call for them. Unless you put your test mass near the various horizons of the black hole. In other words the subject here is unclear, the BH or the gravitating body?  

---

## 4 answers

### Answer by username_12 on 2021-05-09. Score: 48 | Accepted answer

No you can't and the behaviour of bodies with mass and of light is completely different near a compact, massive object if you use Newtonian physics rather than General Relativity.

In no particular order; features that GR predicts (and which in some cases have now been observationally confirmed) but which Newtonian physics cannot:

1. An event horizon. In Newtonian physics there is a misleading numerical coincidence that the escape velocity reaches the speed of light at the Schwarzschild radius. But in Newtonian physics you could still escape by applying a constant thrust. GR predicts that no escape is possible in any circumstances.
2. Further; this numerical coincidence only applies to light travelling radially. In Newtonian physics the escape "speed" is independent of which direction you fire a body, but in GR light cannot escape from (just above) the Schwarzschild unless it if fired radially outwards. For other directions, the radius at which light can escape is larger.
3. GR predicts an innermost stable circular orbit. A stable circular orbit is possible at any radius in Newtonian physics.
4. In GR a particle with some angular momentum and lots of kinetic energy will end up falling into the black hole. In Newtonian physics it will scatter to infinity.
5. Newtonian physics predicts no precession of a two-body elliptical orbit. GR predicts orbital precession.
6. Newtonian physics predicts that light travelling close to a massive body has a trajectory that is curved by about half the amount predicted by GR. Even stranger effects are predicted close to the black hole including that light can orbit at 1.5 times the Schwarzschild radius.

The GR approach to gravity is fundamentally and philosophically different to Newtonian gravity. For Newton, gravity is a universal force. In GR, gravity is not a force at all. Freefalling bodies are said to be "inertial". They accelerate, not because a force acts upon them, but because spacetime is curved by the presence of mass (and energy).

In most cases, where Newtonian gravitational fields are weak, the consequences of this difference are small (but measureable - e.g. the orbit precession of Mercury or gravitational time dilation in GPS clocks), but near large, compact masses, like black holes and neutron stars, the differences become stark and unavoidable.

> **Comments:**
> **username_8**: I've just asked [Is there anything different about the gravitation around a black hole and a neutron star of the same mass?](https://astronomy.stackexchange.com/q/43750/7982)  
> **username_13**: I am not familiar with the maths - is the relationship between the escape velocity and event horizon really coincidental?  
> **username_14**: @kutschkem The Schwarzchild solution to the GR equations predicts a radius at which it is impossible to escape gravity regardless of your velocity or acceleration. The algebra for computing this radius concludes with an expression that turns out to be identical to the expression obtained when setting the escape velocity equal to $c$ in Newtonian gravity. This is a "coincidence" because the Schwarzchild event horizon has no physical relation to escape velocity at all. Nevertheless, in history, Schwarzchild was aware of this algebra before he achieved the full GR solution.  
> **username_0**: @ProfRob thanks for the detailed answer!  
> **username_6**: @RossPresser I'd expect a <shrug>coincidence</shrug> if the escape velocity at the Schwarzschild radius were an arbitrary value. Alas, it is c. That is most emphatically *no coincidence.* (That the physics are fundamentally different notwithstanding.)  
> **username_12**: @Peter-ReinstateMonica that is a point of view; but the fact that $c$ doesn't feature in Newtonian physics at all and isn't the speed limit of anything argues the other way. I think it is an unfortunate coincidence (unfortunate because it leads to fundamental misunderstanding of the nature of the event horizon).  
> **username_12**: @Peter-ReinstateMonica a further point which I've now added is that the Newtonian escape speed argument only works for radial trajectories. For Newton it is an escape speed, independent of trajectory. In GR light sent tangentially will not escape unless $r>1.5r_s$ and the two theories disagree.  
> **username_15**: @ProfRob disregarding the fact that $c$ shows up out of nowhere seems odd to me. $c$ makes no appearance in maxwell's laws but yet $1/\sqrt{\epsilon_0\mu_0}$.   For clarity, I do not disagree that Newton's gravity does not work near black holes, I just disagree that $c$ showing up in this way is a coincidence. I would suppose this just shows that the constants $G$ and $c$ are somehow connected (a connection not explained by Newton's relativity thus indeed one would still require GR to even find this connection)  

---

### Answer by username_5 on 2021-05-10. Score: 14

I am not an expert in physics and the explanation of the others is excellent. However, I noticed a flaw in your reasoning which they did not address.

You have written:

> Considering the Newton's Law of Gravity equation $F = GM/r^2$, if the **radius** of an object becomes super small, then it can technically have immense gravity.

Hence I deduce that you read the $r$ in [the equation](https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation) as the *radius* of the object, while in fact it is the **distance between two objects**. So there is no rationale based on this formula that "the smaller the radius is, the stronger the gravity must be". The proper reading would be "the closer the object is from the black hole, the stronger the gravity is", but this is valid for all the bodies, not only to black holes.

> **Comments:**
> **username_0**: Ahh, right. I sometimes get confused between radius and distance. Thanks for spotting that.  
> **username_6**: There is a certain connection between radius and distance because only a mass with very small spatial (possibly infinitely small) spatial extent creates gravity strong enough that GR effects become noticeable on small scales.  
> **username_5**: @Peter-ReinstateMonica Well, you might be right, but the Newton's formula is about a **distance of two bodies**, not about the size of one of them. So the reasoning in the OP was based on the incorrect application of the formula.  
> **username_7**: @AdiBak: the smaller the radius of the object, the smaller you can be without being inside it.  So the radius is a lower bound on $r$.  (Fun fact which you may have learned: the net gravity *inside* a symmetric spherical shell is 0 everywhere, so burrowing into an object such as the Earth, you can still use $GM/r^2$ but where M only includes the mass of the part below you, not the part you're inside.  (Of course assuming that Newtonian mechanics is a good enough approximation for whatever you're burrowing into, like maybe not a neutron star.)  

---

### Answer by username_8 on 2021-05-10. Score: 9

While admiring [@ProfRob's answer](https://astronomy.stackexchange.com/a/43738/7982) I'll add some additional perspective/background that may serve as a helpful stepping-stone since not every Astronomy SE reader is prepared to embrace General Relativity in all its glory.

> Can Newton's gravity equation explain why black holes are so strong?

1. The simple equation1 $F = GMm/r^2$ *doesn't explain anything* but it certainly does give a result that's usually quite useful in Newtonian mechanics where space is "normal" and we can talk about gravitation as a force.
2. General relativity can be said to *offer an explanation* of how gravitation works and it gives results that work not only where Newtonian gravity works but in extreme situations as well, where speeds are very high and/or gravity is very very strong (and other situations as well).
3. Even in situations we are somewhat familliar with, like Mercury's orbit around the Sun or satellites orbiting the Earth, predictions by Newtonian mechanics is measurably off and GR gets it spot-on.
4. On the flip side, if you want to *approximate* a trajectory of a star or bit of dust that passes far from a black hole, you certainly can go ahead and use Newton's equations. Trajectories will still be close to Keplerian orbits.

I'll make up some numbers for illustration purposes only. If a 20 solar mass star goes supernova and you add up all the ejected mass and energy and it's 12 solar masses, then you'll expect an 8 solar mass black hole left over.

If there is a companion star orbiting it at a large distance, or if you fly past it at a "safe distance" and look at what happens, it will be what you'd expect for a mass of 8 solar masses, whether it's a black hole, a neutron star, a regular star or a (magically self-supporting) ball of concrete.

It's only when you get closer that you need to use GR, and you'd have to use it whether it's a black hole or a more conventional dense object like a neutron star.

1Remember, for a force there are two masses, for acceleration there's only one $a=F/m = GM/r^2$

> **Comments:**
> **username_0**: I see...thanks for answering!  

---

### Answer by username_9 on 2021-05-10. Score: 5

If Einstein's GR equations are expanded in terms of familiar coordinates (Cartesian, spherical,...), the dominant or leading terms of the expansion (for the acceleration) can be written as the single Newtonian term GM/r^2. The next terms of the expansion can be considered as GR corrections to this leading term.

Before the publication of GR, 19th-century astronomers noticed that the advance of the perihelion of Mercury was not predicted accurately by Newtonian gravitational physics. This became a major problem. At some point, astronomer/mathematicians added corrections to the Newtonian Law of Gravitation that produced a new/revised force law that accurately predicted perihelion advance. The results were accurate, but the new gravitation theories were not as rich in their "other" predictions as Einstein's GR.

> **Comments:**
> **username_10**: Welcome to the site! I'm not seeing a clear connection to black holes here. Can you edit to make that more explicit?  
> **username_11**: Could you provide some examples of these phenomenological alternatives to general relativity?  
> **username_9**: The addition of correction terms were designed to accurately account for the advance of the perihelia of the planets.  The 2 best corrections involved one additional term in the expression for the POTENTIAL:  (1) a 1/(c^2r^2) term, or (2) a 1/(c^2r^3) term.  The latter produced the best and most accurate values for the advance of the perihelia.  An excellent, detailed analysis of these 2 "theories" plus a comparison with GR contributions can be found in the paper by James D. Wells (U Michigan): "When effective theories predict: the inevitability of Mercury’s anomalous perihelion precession.  
> **username_9**: To address part of the original question, the addition of corrections to the Newtonian potential not only accurately predicted the advance of planetary perihelia, but also predicted the Schwarzschild radius:  This is the radius of a black hole's event horizon.  The Newtonian potential doesnt reveal the Schwarzschild radius/event horizon, but adding a small radially-dependent correction, factored by 1/c^2, does.  
> **username_8**: Several answers to [How to calculate the planets and moons beyond Newtons's gravitational force?](https://space.stackexchange.com/q/23408/12102) contain the first term (or two?), along with one for velocity.  

---

