# What are the scalar equations for velocity and displacement if acceleration obeys the inverse-square law?

In basic high school physics/calculus you learn that you can formulate equations for velocity and displacement under constant acceleration as:

$$a(t) = a_0$$

$$v(t) = a_0t + v_0$$

$$x(t) = \frac{1}{2}a_0t^2 + v_0t + x_0$$

My question is how would you formulate a similar equation when acceleration is dependent on the inverse-square of distance from a point, such as Coulomb's Law or Isaac Newton's inverse-square law of universal gravitation?

Where:

$$a(x) = k/x^2$$

$$v(t) = ? + v_0$$

$$x(t) = ? + x_0$$

Also I imagine that there may be no solution as $$x$$ approaches zero and acceleration goes to infinity. But let's say that we have two positive charges sitting at rest some distance $$r$$ away from each other in endless space. What will their displacement be at some later time $$t$$?

• This is an interesting question, that can be thoroughly discussed. The answer is not a simple one (in the sense that you don't have an easy expression for x(t)), and furthermore requires at least an operative knowledge of calculus. How much do you know calculus? – pp.ch.te Nov 14 '18 at 15:02
• For something like a planet, x does NOT approach zero. "x" is measured from the center of the planet, so the surface of the planet limits how low "x" can go. – David White Nov 14 '18 at 16:35
• @DavidWhite Right for a planet there is a different simple harmonic equation that would describe what would happen if you drilled a hole through the center of the earth. But what about for a very small charged particle with no radius, like a point charge? – Nick Sotiros Nov 17 '18 at 7:52
• For point charges, the inverse square law applies until you get very close to the charge. At that point, repulsive or attractive forces approach infinity as "x" approaches zero. Of course, in the real world, when this happens you switch from Newtonian mechanics (the inverse square law) to quantum mechanics. – David White Nov 17 '18 at 18:52

Since $$a=\frac{dv}{dt}=\frac{dx}{dt}\frac{dv}{dx}=v\frac{dv}{dx}=\frac{d}{dx}(\frac{1}{2}v^2)$$, integration gives $$v^2=v_\infty^2-\frac{2k}{x}$$. Then $$t=\int\frac{dx}{v}=\int\frac{dx}{\sqrt{v_\infty^2-\frac{2k}{x}}}$$. If you evaluate that integral (hint: substitute $$x=\frac{2k}{v_\infty^2}\csc^2\theta$$), you'll have $$t$$ as a function of $$x$$, not the other way round.
• @NickSotiros SInce we can get $t$ as a function of $x$, we can also get $v=1/(dt/dx)$ as a function of $x$, then get $a=d(v^2/2)/dx$ as a function of $x$. I think with some effort you can also get $x$, hence $a$, as a function of $t$. – J.G. Nov 17 '18 at 16:12