## Step 1: Understand the relationship between gravity and distance in the given universe.
In this universe, the force of gravity is proportional to the inverse cube of the distance between two objects. This means if the distance between two objects is doubled, the gravitational force between them becomes 1/8 of its original value (since 2^3 = 8).

## Step 2: Calculate the change in gravitational force when the planet's orbit is doubled.
Given that the planet is pushed into an orbit exactly twice as far away from its star, we apply the inverse cube law. The new gravitational force (F_new) compared to the old force (F_old) is calculated as F_new = F_old / 2^3, because the distance is doubled. Thus, F_new = F_old / 8.

## Step 3: Determine the outcome for the planet's new orbit.
In a stable, perfectly circular orbit, the gravitational force provides the necessary centripetal force to keep the planet in orbit. If the gravitational force decreases to 1/8 of its original value, the centripetal force available to keep the planet in a circular orbit at the new distance also decreases. However, the planet's velocity and the requirements for a stable orbit at the new distance must be considered. For a circular orbit, the centripetal force (F_c) is given by F_c = (m * v^2) / r, where m is the mass of the planet, v is its velocity, and r is the radius of the orbit. Since the gravitational force decreases more rapidly with distance than the centripetal force requirement for a circular orbit (which is inversely proportional to radius), the planet would not be able to maintain a circular orbit at the new distance with its current velocity. The planet would likely enter an elliptical orbit, with the point of closest approach (periapsis) being the original orbit and the point of farthest distance (aphelion) being beyond the new orbit, due to conservation of angular momentum and energy.

The final answer is: $\boxed{\frac{1}{8}}$