Question

The gravitational force on $m_{1}$ due to $m_{2}$ is

• A. $\frac{G m_{1} m_{2}}{\left|\vec{r}_{2}-\vec{r}_{1}\right|^{3}}\left(\vec{r}_{2}-\vec{r}_{1}\right)$ $\left|r_{2}-r_{1}\right|$
• B. $\frac{G m_{1} m_{2}}{\left|\bar{r}_{2}-\bar{r}_{1}\right|^{2}}\left(\vec{r}_{2}-\vec{r}_{1}\right)$
• C. $\frac{G m_{1} m_{2}\left(\vec{r}_{1}-\vec{r}_{2}\right)}{\left|\vec{r}_{1}-\vec{r}_{2}\right|^{3}}$
• D. $\frac{G m_{1} m_{2}}{\left|\vec{r}_{1}-\vec{r}_{2}\right|^{2}}\left(\vec{r}_{1}-\vec{r}_{2}\right)$

Answer 1

Answer: (A)

$m_{2}$ attract $m_{1}$ towards itself.
$\vec{F}_{12}=\frac{G m_{1} m_{2}}{|\bar{r}|^{2}} \hat{r}$ where $\vec{r}=\vec{r}_{2}-\vec{r}_{1}$ and $\hat{r}=\frac{\vec{r}_{2}-\vec{r}_{1}}{\left|\vec{r}_{2}-\vec{r}_{1}\right|}$
$\Rightarrow \vec{F}_{12}=\frac{G m_{1} m_{2}}{\left|\vec{r}_{2}-\vec{r}_{1}\right|^{2}} \cdot \frac{\left(\vec{r}_{2}-\vec{r}_{1}\right)}{\left|\vec{r}_{2}-\vec{r}_{1}\right|}$