In a universe where the gravitational force follows an inverse cube law, pushing a planet to twice its original orbital radius results in the gravitational force decreasing to 1/8 of its original value. Here's the breakdown:

1. **Force Change**: The gravitational force becomes $$ \frac{1}{8} $$ of the original force because $$ F \propto \frac{1}{r^3} $$, so doubling $$ r $$ results in $$ F' = \frac{F}{8} $$.

2. **Orbital Outcome**: Upon being pushed to $$ 2r $$, the planet's required centripetal force for a circular orbit at that distance exceeds the actual gravitational force. This disparity causes the planet to lack sufficient centripetal force, leading it to move outward. Since the planet's total mechanical energy becomes positive after the push, it is no longer gravitationally bound and escapes the star's gravitational pull.

**Answer**:

When the planet is moved to twice its original orbital distance, the gravitational force exerted on it decreases to **1/8** of its original value. Due to the insufficient centripetal force at the new distance, the planet's orbit becomes unstable, and it escapes the star's gravity, resulting in an unbound trajectory away from the star.