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We know that gravity is a force. But what is it's direction? Can it be expressed by vector and how can we do that? This question can also be asked for Coulomb's Law.

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  • $\begingroup$ "We know that gravity is a force." - Newtonian gravity is a force, but General Relativity is the currently accepted theory of gravitation. $\endgroup$ – Alfred Centauri Jan 22 '18 at 13:27
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If we use the centre of the earth as origin, we have

$$\mathbf{F}=-\frac{GM_{\oplus}m}{r^3}\mathbf{r} \tag{$r>R_{\oplus}$}$$

where $\mathbf{r}=(x,y,z)$ and $\displaystyle \left| \frac{\mathbf{r}}{r^3} \right|=\frac{1}{r^2}$.

At the surface of the earth,

$$g=\frac{GM_{\oplus}}{R_{\oplus}^2} \approx 9.8 \text{ m s}^{-2}$$

where $M_{\oplus}$ and $R_{\oplus}$ is the mass and radius of the earth respectively.

We assume the earth and the test mass have spherical symmetry in their densities.

The Columb's law version is

$$\mathbf{F}=\frac{Q_1 Q_2}{4\pi \epsilon_0 r^3}\mathbf{r}$$

assuming point charges or negligible electrostatic induction.

See another answer with electrostatic induction here.

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  • $\begingroup$ Nice answer. But you're missing a factor of $R_{\oplus}$ in your definition of $g$, which should read: $g=\frac{GM_{\oplus}}{R_{\oplus}^2} \approx 9.8 \text{ m s}^{-2}$ $\endgroup$ – Ben51 Jan 22 '18 at 18:11
  • $\begingroup$ You might want to spell out what r is. $\endgroup$ – M. Enns Jan 22 '18 at 18:18
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As a first statement, I like to begin by stating that gravity is always towards the mass (e.g. always attractive). In other words, if mass A pulls on mass B, I would state that the direction of the gravitational force is towards mass A. If you set up coordinate system, you may then put this in by hand. (The same line of reasoning applies to the coulomb force, except now the direction of the force may be either toward or away depending on the signs of the charges.)

If we wish to be more formal, we can set up a spherical coordinate system centred on mass A. The direction of the gravitational force on mass B will always be in the $-\hat{r}$ direction, where $\vec{r}$ is a vector that describes the location of mass B.

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  • $\begingroup$ Chung Tak....why is there a negative sign in Your formula for gravitation?and an extra r?I could not understand this. $\endgroup$ – Theoretical Feb 4 '18 at 6:37
  • $\begingroup$ @AsifIqubal --> in my formula for gravity? $\endgroup$ – Bob Feb 6 '18 at 1:19
  • $\begingroup$ Yes whoever wrote the answer.I didn't understand why you did that? $\endgroup$ – Theoretical Feb 6 '18 at 4:48
  • $\begingroup$ I didn't explicitly write a formula for gravity (there are 3 separate answers to this question). I wrote a negative sign because if we work in spherical coordinates and place a large mass at the origin, the force of gravity on a (small) test particle due to the large mass will always be pointing toward the origin (hence in the $-\vec{r}$ direction). $\endgroup$ – Bob Feb 6 '18 at 12:29
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Imagine you have only two masses in the universe $M$ and $m$, then the gravitational force that $m$ feels due to $M$ is indeed a vector that points towards $M$. This is called a central force, and as you point out Coulomb's force also behaves like that.

If you add another mass $M'$ into the picture the problem becomes more complex, in the sense that $m$ will feel now two forces: one pointing towards $M$ (${\bf F}$) and the other one pointing to $M'$ (${\bf F}'$). The resulting force ${\bf F} + {\bf F}'$ is also a vector but not necessarily point in any particular direction

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  • $\begingroup$ How can a vector "not point in any particular direction?" Having a particular magnitude and a particular direction is pretty much what defines a vector. $\endgroup$ – Solomon Slow Jan 22 '18 at 17:41
  • $\begingroup$ @jameslarge Sure, what I meant is that it does not point in either the direction of $M$ or $M'$ $\endgroup$ – caverac Jan 22 '18 at 17:43

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