Many questions can be answered using the vis-viva equation:
$$v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right)$$
which gives the velocity of an object in a Keplerian orbit at distance $r$ from a body of mass $M$ and with a semi-major axis $a$. $G$ is the gravitational constant. And for convenience and accuracy, the product $GM$ or standard gravitational parameter for the Sun and for the Earth are 1.327×1020 and 3.986×1014 m3/s2.
For circular orbits set $r=a$ and get
$$v^2 = GM/a.$$
The orbit's circumference $C=2\pi a$ and the time for one orbit (period) is $T=C/v = C=2\pi a / v$ so
$$T^2 = 4 \pi^2 \frac{a^3}{GM}$$
The mass of a sphere is
$$M = \frac{4}{3} \pi R^3 \rho$$
and we're going to keep the density of the Sun fixed at 1408 kg/m3 which is only 41% higher than water! (see At what depth below the Sun's surface does the density reach that of water?) So:
$$T^2 = 4 \pi^2 \frac{a^3}{GM}$$
$$T^2 = \frac{3 \pi}{G \rho} \left(\frac{a}{R}\right)^3$$
or
$$T = \sqrt{\frac{3 \pi}{G \rho}} \ \ \left(\frac{a}{R}\right)^{3/2}$$
punch line: So the period is going to be one year, i.e. about 365 days whether we use the current values for $a$ and $R$ or scale them up or down by any factor!!
In other words, while:
in plain English, in fact a sun-density sized soccer ball will keep a grape-sized object (of the same density as the planet it represents) in the same scale orbit with the same orbital period. This does in fact all scale down.
...is nearly correct. If the Sun were a 22 cm diameter ball with the same 1.4 g/cm^3 average density, and the sesame seed-sized Earth was 47.4 meters away with a diameter of 2 millimeters and the same average density of 5.5 g/cm^2, then it would orbit the soccer ball-sized Sun once a year, unless there were external forces pulling on it from other astronomical object.
Alternatively you could keep the Sun and Earth and all the planets the same sizes and distances but make them a hundred times less dense, and the orbital periods would be $\sqrt{\text{100}} = $ 10 times longer.
This is actually variant of the rule of thumb that the period of a low orbit around a spherical body is just inversely related to the square root of density. So a dust particle orbiting a 1 meter diameter spherical chunk of "average Earth" will orbit in about 90 minutes just like the ISS orbits around the whole Earth in about 90 minutes.
But you can always replace a spherically symmetric mass distribution with a smaller spherically symmetric mass distribution, (even a point).
Not the same, but similar to what's discussed in this answer to Delta-V required for lift-off from a planet/ asteroid
