The vector form of universal law of gravitation

The vector form of universal law of gravitation
  1. $\overrightarrow{\mathrm{F}}=\mathrm{G} \frac{\mathrm{m}_1 \mathrm{~m}_2}{\mathrm{r}} \overrightarrow{\mathrm{r}}$
  2. $\overrightarrow{\mathrm{F}}=\mathrm{G} \frac{\mathrm{m}_1 \mathrm{~m}_2}{\mathrm{r}^3} \hat{\mathrm{r}}$
  3. $\overrightarrow{\mathrm{F}}=\mathrm{G} \frac{\mathrm{m}_1 \mathrm{~m}_2}{\mathrm{r}^2} \overrightarrow{\mathrm{r}}$
  4. $\overrightarrow{\mathrm{F}}=\mathrm{G} \frac{\mathrm{m}_1 \mathrm{~m}_2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}}$

Solution

Vector form of law of gravitation is given as $ \overrightarrow{\mathrm{F}}=\mathrm{G} \frac{\mathrm{m}_1 \mathrm{~m}_2}{\mathrm{r}^2} \hat{\mathrm{r}} $ As, $\hat{r}=\frac{\vec{r}}{r}$ $ \therefore \overrightarrow{\mathrm{F}}=\mathrm{G} \frac{\mathrm{m}_1 \mathrm{~m}_2}{\mathrm{r}^3} \overrightarrow{\mathrm{r}} $

Asked in: AP EAMCET 2023 (18 May Shift 2)

Practice more Gravitation questions on Aicharya