1. For every force there is an equal force in the opposite direction on another body, correct?

  2. So when the Suns gravity acts on Earth where is the opposite and equal force?

  3. I also have the same question for centripetal force in a planets orbit.

  • $\begingroup$ Correct, the sun is falling towards the earth, as much as the earth is falling towards the sun. $\endgroup$ – ja72 Dec 5 '10 at 22:38
  • $\begingroup$ Wasn't newton puzzled by the force (action) at a distance. He never grasped how it worked. It turns out we don't quite know either, despite being like 400 years later. $\endgroup$ – ja72 Dec 5 '10 at 22:40
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    $\begingroup$ @Ja72: Since Einstein, we know that the action at a distance in gravity is resolved by local fields, which transmit stress locally. It is not correct that people are puzzled by action at a distance, the field concept resolves this. This was appreciated in Maxwell's time as well. $\endgroup$ – Ron Maimon Dec 15 '11 at 9:58
  • $\begingroup$ Surely forces can act a distance because of the force carrier? (e.g. gluon -> strong) $\endgroup$ – Jonathan. Dec 15 '11 at 19:27

As many others said, the Sun feels the same force towards Earth as the Earth feels towards the sun. That is your equal and opposite force. In practice though the "visible" effects of a force can be deduced through Newton's first law, i.e. ${\bf F} = m{\bf a}$. In other words, you need to divide the force by the mass of the body to determine the net effect on the body itself.


${\bf F_s} = {\bf F_e}$

${\bf F_s} = m_s {\bf a_s}$

${\bf F_e} = m_e {\bf a_e}$


$m_s {\bf a_s} = m_e {\bf a_e}$


${\bf a_s} = {\bf a_s} \frac{m_e}{m_s}$

Now, the last term is $3 \cdot 10^{-6}$! This means that the force that the Earth enacts on the sun is basically doing nothing to the sun.

Another way of seeing this:

$F = \frac{G m_s m_e}{r^2}$

$a_s = \frac{F}{m_s} = \frac{G m_e}{r^2}$

$a_e = \frac{F}{m_e} = \frac{G m_s}{r^2}$

$\frac{a_s}{a_e} = \frac{m_e}{m_s} = 3 \cdot 10^{-6}$

Again, the same big difference in effect.

Regarding the centripetal force, it is still the same force. Gravity provides a centripetal force which is what keeps Earth in orbit.


It's worth pointing out that the mass that acts as the charge for gravity, known as gravitational mass is not, a priori, the same mass that appears in Newtons's law, known as inertial mass. On the other hand it is a fact of nature that they have the same value, and as such we may use a single symbol $m$, instead of two, $m_i$ and $m_g$. This is an underlying, unspoken assumption in the derivation above. This is known as the weak equivalence principle.

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    $\begingroup$ If you point out the difference between inertial and gravitational mass, as is of particular interest when dealing with gravity, you get my up vote. $\endgroup$ – Noldorin Dec 5 '10 at 16:15
  • $\begingroup$ @Noldorin, done. $\endgroup$ – Sklivvz Dec 5 '10 at 16:25
  • $\begingroup$ You get my vote then. It's not absolutely crucial to the argument, but could potentially help avoid much later confusion. $\endgroup$ – Noldorin Dec 6 '10 at 0:15

The opposite is that Earth attracts the sun with the same exact force!
Thinking smaller, the Earth attracts you and you attract the earth with forces that are equal in magnitude and direction, but of opposite sense. Of course, by Newton second's law, that same force will have a much greater effect on you than on the Earth. The same applies to the Earth+Sun combination!

  • $\begingroup$ so, if for example u start doing push up on the ground you are accelerating (hence moving) yourself up and down but at the same time you are also accelerating and moving the heart as well! that helps your motivation, innit? $\endgroup$ – Steve Nov 3 '10 at 12:42
  • $\begingroup$ @Steve, yes, exactly! And if I push something of roughly the same mass as me (another person), we will both lose our equilibrium :) $\endgroup$ – Agos Nov 3 '10 at 17:07
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    $\begingroup$ ok now i'm scared! :P $\endgroup$ – Steve Nov 3 '10 at 17:42

The earth feels a force towards the sun. The sun feels an 'equal and opposite' force towards the earth. In fact, the earth does not rotate around the sun; instead, the sun and the earth (if you are considering only those two bodies) orbit around their center of mass.

  • $\begingroup$ But how can Earth create a force that acts on the Sun equally. Surely if both forces were equal then the barycenter would be exactly half way between the Earth and Sun? $\endgroup$ – Jonathan. Nov 3 '10 at 0:24
  • $\begingroup$ The barycenter comes from mass and geometry, no need for gravity. What I guess is the problem is that you think "force" instead of "field": the field of Earth is weaker, but the resulting forces are equal. (see the edit of my answer) $\endgroup$ – Cedric H. Nov 3 '10 at 0:27
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    $\begingroup$ @Jonathan: that would be true if the Sun and Earth were the same mass. But the Sun is a lot more massive than the Earth, so it moves correspondingly less (i.e. its orbit is smaller). $\endgroup$ – David Z Nov 3 '10 at 5:26
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    $\begingroup$ The barycenter is really close to the center of the sun, which is why it's almost correct to say that the earth orbits around the sun, and depending on the context, you can think of it that way without being too far off. $\endgroup$ – MatrixFrog Nov 3 '10 at 6:32
  • $\begingroup$ This is more apparent with masses that are more equivalent, like the Pluto-Charon system. The difference between the masses of the sun and the earth is like comparing the mass of a grain of sand to a steel ball the size of a beach ball. The earth could fall directly into the sun at a million miles an hour and the sun would barely budge, the difference is so large. $\endgroup$ – Ernie May 17 '11 at 20:37

Sun and Earth interact through gravity. Sun exert a gravitational force on Earth and ... Earth exert and equal and opposite force on Sun (or is it the opposite ?).

Opposite here does not mean "sci-fi anti gravity force repelling objects" but opposite as in "vector of opposite direction and equal magnitude".

The centrifugal force is not a force, but a pseudo force that is introduced because the referential frame is not inertial. In the referential frame where you need to introduce the centifugal force, the "opposite reaction" is the centripedal force itself.

To solve the Earth around the Sun problem, first consider that the sun as an infinite mass and is thus without motion. The sun exert a force on Earth, and in some frame of reference you need to introduce an additional pseudo-force.


The gravitational field "generated" by Earth is weaker, but the forces are equal (in magnitude).

$E_{earth} = \frac{G m_{earth}}{r^2}$

$E_{sun} = \frac{G m_{sun}}{r^2}$

$F = E_{earth} * m_{sun} = E_{sun} * m_{earth}$

Another edit:

Considering the title of your question "is gravity a force": Gravity is one of the four fundamental interaction of nature (strong and weak interactions, and electromagnetism). No complete treatment of this interaction exists at a quantum level, so from a classical point of view, yes gravity is a force, associated with a conservative [gravitational] field, etc.

  • $\begingroup$ Could you format the formulas better, they might be in some universally accepted form but they're simply not easier to read. $\endgroup$ – Jonathan. Nov 3 '10 at 17:04

The mantra, «gravity is not a force»

Gravity is not a force, according to General Relativity (or rather, at least according to Einsteinians' interpretation of GenRel). This seems to have been Einstein's point of view. The other answers are also correct: within a Newtonian framework. But even some physicists have always found Newton's fundamental views on force confusing, and many regard GenRel as having cleared this up. I am a wikipedia editor myself, so I know how little it can be trusted on any subject about which people really care, but here is a reference and I thought this mantra was not controversial...von Hayek's followers' encyclopedia

Force is a classical notion

«At the base of classical mechanics is the notion that a body's motion can be described as a combination of free (or inertial) motion, and deviations from this free motion. Such deviations are caused by external forces acting on a body in accordance with Newton's second law of motion» The point is to have a physical notion of force, not just a mathematical construct.

Contrast with General Relativity

«there is no gravitational force deflecting objects from their natural, straight paths. Instead, gravity corresponds to changes in the properties of space and time, which in turn changes the straightest-possible paths that objects will naturally follow.» Anything physical should have a covariant mathematical formulation...mathematical formulations that are not covariant are not physical. Everything is waves and a local field theory. So there is no need to look for the opposite force...

In summary

All the previous answers are also correct, from the Newtonian point of view which assumes that gravity is a force and then deduces what must be going on, if it is a force. And for many practical purposes such analyses are handy. But reading your question very literally, you are asking whether gravity is a force, and technically, the answer is no.

other links to discussions of this mantra

physics homework help and hindrance

How to get questions answered, if you can filter out the noise

references as the source code of this forum

If something is non-controversial, like the chain rule, references are not necessary. But if something is controversial, or original, it is time to trot out the references. By now, thanks to Ron, I see that this standard mantra is controverted by some people, so I have supplied references. And here is a reference to the idea of supplying references meta-discussion about stack exchange

  • $\begingroup$ This is a bit of a strong statement: yes, you can describe gravity as something other than a force, but you can also describe it as a force in classical physics or in a noninertial reference frame. So it's not correct to definitively state that gravity is not a force, period, in a similar sense to how it's not correct to definitively state that light is made of particles. $\endgroup$ – David Z Dec 15 '11 at 17:14
  • $\begingroup$ @Ron Part of the controversy seems to be fueled by ¿In which direction should grand unification go? If one takes GenRel equally seriously as QM, one insists gravity is not a force. Others take the Heisenberg uncertainty relations and the related notions of quantum mechanical particles more seriously than relativity. But the relative merits of these directions are very speculative, so to me it seems odd to base an answer on personal opinions one way or the other. My impression is that the consensus of citations directly addressing this mantra are as I have indicated. $\endgroup$ – joseph f. johnson Dec 15 '11 at 18:58
  • $\begingroup$ I should, perhaps, make clear that, qua reference, a mere obiter dictum is not what I am thinking of. E.g., a physicist really addressing something else just uses the word «force» without meaning to commit themselves one way or the other. I mean, «directly addressing the mantra». I would be very interested to have a reference to an imporant physicist saying that Einstein's idea that gravity is not a force is wrong. Or to an important scientist or historian who states flatly that although Einstein is usually interpreted as asserting this mantra, this is a misinterpretation of Einstein. $\endgroup$ – joseph f. johnson Dec 15 '11 at 19:04

Just a caveat that is left out of the answers above:

Newton's third law is only approximately true. If a system radiates (either electromagnetically or gravitationally), then the net momentum of the system can change, and you will find that $F_{12}\neq F_{21}$ if the radiation is not uniform in all directions. The reason for this is simple enough: radiation carries energy and momentum away from the system.

For the Earth and the sun, this really doesn't matter much, because the gravitational radiation of the system is so close to zero that we wouldn't even know how to begin to search for the effect.

But for things like colliding black holes, the effect can become very significant. Two orbiting black holes with zero center of mass velocity can end up with more than the escape velocity needed to leave the center of the galaxy, for example.


When you study a two-body problem, which may seem fairly complicated (for the reason that there are two different problems intertwined: knowing the relative position of the two bodies, determine the force acting on each of them; knowing the forces determine the relative position after some time), all is clear once you make the following standard change of coordinates: define the position vector of the center of mass

\begin{equation} \textbf{X}=\frac{1}{m}\sum_i^2 m_i\textbf{x}_i \end{equation}

and the relative position vector

\begin{equation} \textbf{r}=\textbf{x}_2-\textbf{x}_1\end{equation}

This change of coordinates is invertible:

\begin{equation}\textbf{x}_1=\textbf{X}-\frac{m_2}{m}\textbf{r} \end{equation}

\begin{equation} \textbf{x}_2=\textbf{X}+\frac{m_1}{m}\textbf{r} \end{equation}

Now, the magic thing about this change of coordinates is that almost every significant quantity splits into two terms, for example you can check that the total kinetic energy is $T=T_{CM}+T_0$, where the first term is the kinetic energy computed in the center of mass and the second is the kinetic energy of a fictitious particle moving with the center of mass and of mass $m=m_1+m_2$. The kinetic energy in the $CM$ is:

\begin{equation}T_{CM}=\frac{1}{2}m_1(\dot{\textbf{x}}_1-\dot{\textbf{X}})^2+\frac{1}{2}m_2(\dot{\textbf{x}}_2-\dot{\textbf{X}})^2=\frac{1}{2}\mu\dot{\textbf{r}}^2 \end{equation}

where $\mu=m_1m_2/m$ is the reduced mass and it is smaller than both masses. If $m_2>>m_1$, to first order, $\mu=m_1(1-m_1/m_2)$.

The lagrangian of the system is:

\begin{equation}\mathcal{L}=\frac{1}{2}\mu \dot{\textbf{r}}^2+\frac{1}{2}m\dot{\textbf{X}}^2-U(r)=\mathcal{L_0}+\mathcal{L_{CM}} \end{equation}

If you write down the equations of motion, you can see that the $CM$ moves with uniform velocity whereas the second equation is that of a fictitious particle $\mu$ in a central force field. So, in the case of the earth going around the sun, both move in an ellipse around the common $CM$ and since the $CM$ is a weighted mean of the position of the bodies and since the mass of the sun is one million times more than that of the earth, we can say approximately that the sun is static and the earth is spinning around it. Yet, the forces acting on the earth due to the sun and on the sun due to the earth are still equal and opposite, so there's no paradox after all.

Is gravity a force acting at a distance? This idea sounded improbable even to Newton himself. In fact, the field-concept is much more satisfactory and makes the necessity of instantaneous interactions obsolete. With GR, gravity turns into an entirely local, geometric effect.


In this simplified picture, gravity of the Sun is a centripetal force for Earth's orbital motion; centrifugal force that equalizes it comes from the angular momentum of Earth. If Earth didn't have an angular momentum (was, say, created near the sun without relative velocity, at least in the angular direction), it would just fall on the Sun.

  • $\begingroup$ -1: This is completely incorrect--- the centrifugal force is not equal and opposite to the sun's gravity on the Earth, it is just a frame force. It would be there even if the sun was not, if you used a frame rotating with the would-be motion of the Earth. $\endgroup$ – Ron Maimon Dec 15 '11 at 9:59

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