In a universe where the force of gravity is proportional to the inverse cube of the distance between two objects, we need to determine the new gravitational force and the outcome of the planet's orbit after being pushed to twice its original distance.

1. **Gravitational Force Comparison**:
   - Original distance: \( r \)
   - Original gravitational force: \( F_1 \propto \frac{1}{r^3} \)
   - New distance: \( 2r \)
   - New gravitational force: \( F_2 \propto \frac{1}{(2r)^3} = \frac{1}{8r^3} \)
   - Therefore, the new gravitational force is \( \frac{1}{8} \) times the original force.

2. **Orbital Dynamics**:
   - For a stable circular orbit, the gravitational force provides the necessary centripetal force.
   - Orbital velocity in this universe: \( v \propto \frac{1}{r} \)
   - When the planet is pushed to twice the distance, the required orbital velocity for a circular orbit at \( 2r \) is halved.
   - However, the planet retains its original velocity, which is higher than the required velocity for a circular orbit at \( 2r \).

3. **Energy Considerations**:
   - Total mechanical energy in the original circular orbit is zero (derived from kinetic and potential energies).
   - After being pushed to \( 2r \), the planet's total mechanical energy becomes positive, indicating an unbound orbit.
   - Escape velocity in this universe is equal to the circular orbit velocity at any radius. The planet's velocity at \( 2r \) is twice the escape velocity at \( 2r \), leading to an unbound trajectory.

4. **Conclusion**:
   - The new gravitational force is \( \frac{1}{8} \) times the old force.
   - The planet's velocity is too high for a circular orbit at \( 2r \), and its positive total energy results in an unbound orbit, leading to escape from the star.

\[
\boxed{\dfrac{1}{8}}
\]

The planet will likely escape the star on an unbound orbit.