How is $$g=-\nabla V$$ where $V$ is gravitational potential and $g$ is acceleration due to gravity. I am new to calculus.
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$\begingroup$ Here V varies with distance from (0,0,0). $\endgroup$– protectgoodlivingbeingaskCommented Dec 14, 2020 at 13:44
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$\begingroup$ The gravitational force is a conservative vector field so it can be written as the gradient of a potential function. $\endgroup$– CharlieCommented Dec 14, 2020 at 14:01
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$\begingroup$ As I am new to calculus could you please elaborate why I cannot just divide v by r=$\sqrt{x^2+y^2+z^2}$ and instead $\nabla V to g$. $\endgroup$– protectgoodlivingbeingaskCommented Dec 14, 2020 at 14:04
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$\begingroup$ I believe OP is asking this having seen that given a spherically potential: $V(r) = -\frac{GM}{r}$, the gravitational field strength is $\frac{GM}{r^2}$ - OP, please note that this is just a special case of $\nabla V$ = $\frac{\partial}{\partial r}V(r) =\frac{GM}{r^2}$, the former formula does not hold for general $V$ $\endgroup$– Nihar KarveCommented Dec 14, 2020 at 14:52
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$\begingroup$ are you aware how potential energy is defined for a general conservative force? $\endgroup$– Vulgar MechanickCommented Dec 14, 2020 at 15:22
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3 Answers
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Every conservative force (e.g. gravity) can be presented as a gradient of some scalar potential (let it be $-\nabla U$, because work of conservative force depends only on initial and final position, $$ W_\mathrm{conservative} = \int\limits_{\mathbf{r}_1}^{\mathbf{r}_2} \mathbf{F}\cdot\mathrm{d}\mathbf{r} = \int\limits_{\mathbf{r}_1}^{\mathbf{r}_2}-(\nabla U)\cdot\mathrm{d}\mathbf{r} = -(U(\mathbf{r}_2)-U(\mathbf{r}_1)). $$ If we define the change of potential energy as a $\Delta U = -W_\mathrm{conservative}$, then $U$ is the desired potential energy. Since $g = F/m$ and $V=U/m$ once finds the relation $g = -\nabla V$.
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$\begingroup$ As I am new to calculus could you please elaborate why I cannot just divide v by r=$\sqrt{x^2+y^2+z^2}$ to get it? $\endgroup$ Commented Dec 14, 2020 at 14:05
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$\begingroup$ Unfortunately, for the force $\propto 1/r^2$ it works (more or less, because the sign is wrong) but when you consider other conservative forces, e.g. Hooke's Law $F=-kx$ what shall you do? Potential energy is then $-\int (-kx) dx = kx^2/2$ $\endgroup$– KarolCommented Dec 14, 2020 at 17:49
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This is really a definition of a potential function. The way a potential function works is that we want to conceive of the force as the result of the object trying to minimize its available energy. Hence we assign an energy to each point in space, which is what the potential field does, and the force experienced must tend to take it toward places of lower energy.
Hence the negative gradient, which gives the direction at which the energy decreases most readily.
answered Dec 14, 2020 at 16:19
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$\begingroup$ "..object trying to minimize its available energy" Why? $\endgroup$ Commented Dec 14, 2020 at 17:31
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If you put your mass M at the origin of an xyz system, then its gravitational potential (with zero at infinity) can be written as V = -GM$(x^2 + y^2 + z^2)^{(-1/2)}$. The gradient is (dV/dx)i + (dV/dy)j + (dV/dz)k. In this case (dV/dx) = [-GM(-1/2)$(x^2 + y^2 + z^2)^{(-3/2)}$][(2x)]. The y and z components are similar. Adding these three gives the negative of the gradient as: [-GM/($r^3$)][xi + yj + zk] which gives g (as a vector). Or,in polar coordinates: V = -GM$r^{-1}$ and the gradient is GM/$r^2$.
