Hint 3

In the presence of a uniform dust cloud, there will be two contributions to the gravitational field at a distance $r$ from the Sun: the usual $1/r^2$ piece due to the Sun plus a piece from the cloud. The latter can be found from the gravitational field at a distance $r$ from the centre of a spherical distribution of matter,

$${\bf g}=-{GM(r)\over r^2}{\bf e}_r,$$

where is $M(r)$ is the total mass inside the radius $r$. In the case of uniform density this mass is

$$M(r)={4\pi\over 3}\rho r^3,$$

which gives a field proportional to $r$.

Equating the combined gravitational field at $r$ to the radial acceleration of the Earth gives the modified relation between the orbital period $T$ and $r$:

$${4\pi^2\over T^2}={GM_{\rm sun}\over r^3}+{4\pi G\over 3}\rho.$$

The first term on the right-hand side is just equal to $4\pi^2/T_0^2$ where $T_0$ is one year (the period if $\rho=0$). Hence we have

$${1\over T^2}-{1\over T_0^2}={G\over 3\pi}\rho.$$

Expanding this to first order in $\Delta T=T-T_0$ we get

$$\rho\simeq-{6\pi\over G}{\Delta T\over T_0^3}.$$

For $\Delta T=-1$ day (the period must get shorter) the corresponding density would be $\rho=7.8\times 10^{-7}\ \hbox{\rm kg m}^{-3}$.