Shrinking the Solar System

For a simplistic analysis, assume that the Earth is in uniform circular motion about the Sun:

$\large{\frac{G M_{Sun}}{R_{orbit}^2} = \frac{v_{orbit}^2}{R_{orbit}} = R_{orbit} \omega_{orbit}^2 \\ G M_{Sun} = R_{orbit}^3 \omega_{orbit}^2 \\ G \frac{4}{3} \pi R_{Sun}^3 \rho} = R_{orbit}^3 \omega_{orbit}^2$

If the solar system shrinks uniformly, both $R_{Sun}$ and $R_{orbit}$ will shrink by the same proportion, and each side of the above equation is proportional to the cubic power of its respective "R" value. Therefore, $\omega_{orbit}$ will remain unchanged after the solar system shrinks. Assuming that the change in the solar system scaling has no effect on the speed of the Earth's rotation about it's own axis, there will still be 365 days in a year.

Equate the forces of attraction between the Sun and the Earth to the centripetal force that keeps the Earth in its approximately circular orbit, and express the angular velocity $\omega$ in terms of $T$ , the period of revolution. This gives - $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$ $\frac { GmM }{ { r }^{ 2 } } =\quad mr{ \omega }^{ 2 }\quad =\quad mr\frac { 4{ \pi }^{ 2 } }{ { T }^{ 2 } }$ $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$ where $m$ and $M$ are the respective masses of the Earth and Sun, and $r$ is the average distance between them. Divide by $m$ and express $M$ in terms of the average density $\rho$ and radius $R$ of the Sun as follows: $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$

$G\frac { \frac { 4 }{ 3 } \pi { R }^{ 3 }\rho }{ { r }^{ 2 } } \quad =\quad \frac { 4{ \pi }^{ 2 } }{ { T }^{ 2 } } r\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad This\quad yields,\quad T=\sqrt { \frac { 3\pi }{ G\rho } { \left( \frac { r }{ R } \right) }^{ 3 } } \\ for\quad the\quad period\quad of\quad revolution.$

It can be seen that the Earth's rotation period only depends on universal gravitational constant $G$ , the average density of the Sun $\rho$ and the ratio $\frac { r }{ R }$ . Therefore if the density of matter remains constant, any scaling of the solar system leaves the length of the year unchanged. It can also be seen that only the density and the size of the Sun are relevant. Any body that is small in size relative to the sun would have the same period and follow the same orbit.

This gives the answer to be $\boxed { 365 }$ days