Question

Two metal spheres of equal radius $\text{r}$ and equal densities are touching each other. The force of attraction $\text{F}$ between them is

Answer 1

Step 1: Given: We have two metal spheres of equal radius $\text{r}$ and equal densities $\text{ρ}$.

Step 2: Relations between mass and densities are:

$\text{m=ρv}$

So the density and volume of the first sphere are:

${\text{ρ}}_1=\frac{{\text{m}}_1}{{\text{v}}_1}$, ${\text{v}}_1=\frac{4}{3}{\text{πr}}^3$

Similarly, the density and volume of the second sphere are:

${\text{ρ}}_2=\frac{{\text{m}}_2}{{\text{v}}_2}$, ${\text{v}}_2=\frac{4}{3}{\text{πr}}^3$

Step 3: Finding the force of attraction between them:

According to Newton's Gravitational force for both spheres are ${\text{F}}_1=\frac{{\text{Gm}}_1{\text{m}}_2}{{\text{r}}^2}$, ${\text{F}}_2=\frac{{\text{Gm}}_1{\text{m}}_2}{{\text{r}}^2}$

So, the force of attraction between them.

${\text{F}}_{}=\frac{{\text{Gm}}_1{\text{m}}_2}{{\text{r}}^2}\left(1\right)$

Now, putting values of ${\text{m}}_{1,}{m}_2$ in the equation $\left(1\right)$, we get

${\text{F}}_{}=\frac{\text{G(ρv)(ρv)}}{{\text{r}}^2}$, $(\text{G,ρ}\text{are constant)}$

Now, we can say that $\text{F∝}\frac{{\text{v}}^2}{{\text{r}}^2}\text{(2)}$

By putting values of $v$ in the equation $\text{(2)}$, we get

$\text{F∝}\frac{{\text{(}{{\displaystyle \frac{4}{3}}}^{}{\text{πr}}^3\text{)}}^2}{{\text{r}}^2}$, $(\frac{4}{3}\mathrm{\pi }\mathrm{is}\mathrm{constant})$

$\text{F∝}\frac{{{\text{(r}}^3\text{)}}^2}{{\text{r}}^2}$

${\text{F∝ r}}^{6-2}$

${\text{F∝ r}}^4$

So, the force of attraction between two spheres are ${\text{F∝ r}}^4$: