Question
Suppose the gravitational potential due to a small system is $\frac{\text{k}}{\text{r}^2}$ at a distance r from it. What will be the gravitational field? Can you think of any such system? What happens if there were negative masses?
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Answer
The gravitational potential due to the system is given as $\text{V}=\frac{\text{k}}{\text{r}^2}.$
Gravitational field due to the system:$\text{E}=-\frac{\text{dv}}{\text{dr}}$
$\Rightarrow\text{E}=-\frac{\text{d}}{\text{dr}}\Big(\frac{\text{k}}{\text{r}^3}\Big)=-\Big(-\frac{2\text{k}}{\text{r}^3}\Big)$
$\Rightarrow\text{E}=\frac{2\text{k}}{\text{r}^3}$
We can see that for this system, $\text{E}\propto\frac{1}{\text{r}^3}$ This type of system is not possible because $F_g$ is always proportional to inverse of square of distance(experimental fact). If there were negative masses, then this type of system is possible. This system is a dipole of two masses, i.e., two masses, one positive and the other negative, separated by a small distance. In this case, the gradational field due to the dipole is proportional to $\frac{1}{\text{r}^3}. $
Gravitational field due to the system:$\text{E}=-\frac{\text{dv}}{\text{dr}}$
$\Rightarrow\text{E}=-\frac{\text{d}}{\text{dr}}\Big(\frac{\text{k}}{\text{r}^3}\Big)=-\Big(-\frac{2\text{k}}{\text{r}^3}\Big)$
$\Rightarrow\text{E}=\frac{2\text{k}}{\text{r}^3}$
We can see that for this system, $\text{E}\propto\frac{1}{\text{r}^3}$ This type of system is not possible because $F_g$ is always proportional to inverse of square of distance(experimental fact). If there were negative masses, then this type of system is possible. This system is a dipole of two masses, i.e., two masses, one positive and the other negative, separated by a small distance. In this case, the gradational field due to the dipole is proportional to $\frac{1}{\text{r}^3}. $
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