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I don't know if I should post this here, but I feel like the problem I'm having is in the mathematics of this problem, so here goes:-
I had this problem on Newton's Gravitational Theory and it is as follows.
The magnitude of the gravitational force experienced by a small spaceship of mass $M$ inside an inter-galactic dust cloud (assumed to be spherically symmetric but not necessarily uniform) when it is at a distance from of $r$ from the centre of the cloud is found to be $$F(r) = \alpha r + \frac{\beta}{r}$$ Find the density of the dust cloud ($G$ is Newton's gravitational constant).
So, I did the usual, starting with defining the elemental mass $dm$ :-
$$dm = \rho (r) dV,$$ $$dV = 4\pi r^2dr$$
where, $\rho(r)$ is the density function of the inter-galactic cloud which varies on $r$, and $dV$ is the volume taken by the elemental mass $dm$ (which I imagined to be a spherical shell of radius $r$ and thickness $dr$).
So, continuing with the following assumptions, we get:-
$$F(r) = \alpha r + \frac{\beta}{r}$$ $$\implies \int_{0}^{r} \frac{GMdm}{r^2} = \alpha r + \frac{\beta}{r}$$ $$\implies GM \int_{0}^{r} \frac{\rho(r) (4\pi \cancel{r^2})dr}{\cancel{r^2}} = \alpha r + \frac{\beta}{r}$$ $$\implies 4\pi GM \int_{0}^{r} \rho (r) dr = \alpha r + \frac{\beta}{r}$$
Now, differentiating both sides of the equation w.r.t $r$:-
$$\implies 4 \pi GM \left(\frac{d}{dr} (r) \right)\rho(r) = \alpha - \frac{\beta}{r^2}$$ $$\implies \rho (r) = \frac{1}{4 \pi GM} (\alpha - \frac{\beta}{r^2})$$
This would be one of my most joyful moments of my life if this answer was on the list of options that were given, and sadly it wasn't.
I thought up of another way, but I think it is fundamentally wrong:-
$$F = \frac{GMm}{r^2}$$ $$\implies \alpha r + \frac{\beta}{r} = \frac{GM\frac{4}{3}\pi \rho \cancel{r^3}^r}{\cancel{r^2}}$$ $$\implies \rho = \frac{3\alpha}{4\pi GM}\left( 1 + \frac{\beta}{\alpha r^2}\right)$$
But surprisingly, this is one of the options. I do not have any answer key for the question, so I don't exactly know which is supposed to be correct.
Any help would be appreciated. Thanks in advance.