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Forgive me if my question has an obvious answer, but I need to know the answer. I always thought that more massive/energetic objects had a stronger force of gravity than less massive ones; that is, the Sun would create a stronger gravitational pull than the Earth does.

Upon looking into Newton's law of universal gravitation, I began to wonder whether I misunderstood this. Does the gravity between the Sun and Earth have an equal force on both? As in, does the Earth pull the Sun with as much force as the Sun does on Earth? If so, would the Sun just resist accelerating to Earth because it is extremely massive? Mass is just an object's resistance toward accelerating from a force, right?

If this is the case, would it apply to GR? Sorry for hypothesizing, I'm honestly trying to get a grasp on it.

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The magnitude of the force of gravity between two bodies is proportional to the product of their masses: $$F=G\frac{m_1m_2}{r^2}$$ This doesn't change depending on which body you're applying the force to, i.e. if you interchange the masses. The magnitude is the same.

What does change is the direction of the force. Force is a vector quantity, denoted as $\vec{F}$ or $\mathbf{F}$. If we write the equation for gravity using proper vector notation, we have $$\mathbf{F}=G\frac{m_1m_2}{|\mathbf{r}_1-\mathbf{r_2}|^2}\frac{\mathbf{r_1}-\mathbf{r_2}}{|\mathbf{r_1}-\mathbf{r_2}|}$$ Here, the positions of the objects are represented by vectors, $\mathbf{r}_1$ and $\mathbf{r_2}$. Additionally, $|\mathbf{x}|$ denotes the norm of a vector $\mathbf{x}$ - its magnitude.

Now, if you interchange the masses, the direction of the force changes, although $|\mathbf{r_1}-\mathbf{r_2}|=|\mathbf{r_2}-\mathbf{r_1}|$, because this refers to the magnitude of the vectors. So the force applied on one object is the opposite of the force applied on the other object. This is Newton's third law.

The acceleration is more interesting. The force on object $1$ due to gravity is $$F_1=m_1g_1$$ Here, $$g_1=\frac{Gm_2}{r^2}$$ where $m_2$ is the other mass. This should tell you that $g_1\neq g_2$, except when $m_1=m_2$.

I'm not an expert in general relativity, but I do know that it describes how spacetime curves due to the presence of one body. The solution to the Einstein Field Equations, the metric, is different for different bodies, because one piece of it, the stress-energy tensor, is different for objects of different mass/energy/etc.

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  • $\begingroup$ I thought gravity is a tensor field, not vector? $\endgroup$ – Sir Cumference Dec 8 '15 at 22:33
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    $\begingroup$ @SirCumference-Pies It's a vector field in Newtonian gravity. It's described via tensors in general relativity. It's important to be careful when using the term "gravity", though, the way you used it. Gravity can refer to (in Newtonian gravity) a force or the acceleration caused by that force. In general relativity, it refers to the bending of spacetime, not a quantity - in other words, "gravity" doesn't have a unit or specific quantity. $\endgroup$ – HDE 226868 Dec 8 '15 at 22:34

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