Updated spacing between @ latex expressions

astrosander edited
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revision #10329 parent #10071 GitHub c702ebe ← older newer →
en/2.1.38.md +6 −6
@@ -73,17 +73,17 @@
Let's write the equation after a long period of time:
$$ mg=\gamma rv\quad(1) $$
Let's find $m$ through the volume $V$:
−$$ m=\rho V=\frac{4}{3} \rho \pi r^3 $$
+$$ m=\rho V=\frac{4}{3} \rho\pi r^3 $$
And we substitute into $(1)$:
−$$ \frac{4}{3} \rho \pi r^3 g=\gamma rv $$
+$$ \frac{4}{3} \rho\pi r^3 g=\gamma rv $$
From here:
−$$ v = \frac{4}{3} \frac{\rho \pi g}{\gamma } \cdot r^2 =\alpha r^2\quad(2) $$
+$$ v = \frac{4}{3} \frac{\rho\pi g}{\gamma} \cdot r^2 =\alpha r^2\quad(2) $$
−$$ \alpha = \frac{4}{3} \frac{\rho \pi g}{\gamma } =\frac{v}{r^2}=10^8 \,\frac{1}{\text{m}\cdot\text{s}} $$
+$$ \alpha = \frac{4}{3} \frac{\rho\pi g}{\gamma} =\frac{v}{r^2}=10^8 \,\frac{1}{\text{m}\cdot\text{s}} $$
We substitute and find the answer
−$$ v(\frac{r}{2}) = \alpha \frac{r^2}{4}=0.25~\text{m/s} $$
+$$ v(\frac{r}{2}) = \alpha\frac{r^2}{4}=0.25~\text{m/s} $$
−$$ v(\frac{r}{10}) = \alpha \frac{r^2}{100}=0.01~\text{m/s} $$
+$$ v(\frac{r}{10}) = \alpha\frac{r^2}{100}=0.01~\text{m/s} $$
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