In a universe where gravity is proportional to the inverse cube of the distance, we analyze the effect of pushing a planet to twice its original orbital distance.

1. **Gravitational Force Calculation**:
   - In our universe, gravity follows the inverse square law: \( F = \frac{G M m}{r^2} \).
   - In this hypothetical universe, gravity follows the inverse cube law: \( F = \frac{G M m}{r^3} \).
   - When the planet is pushed to twice the distance, the new force is \( F' = \frac{G M m}{(2r)^3} = \frac{1}{8} \frac{G M m}{r^3} \).
   - Therefore, the new gravitational force is \( \frac{1}{8} \) of the original force.

2. **Orbital Stability**:
   - For a stable circular orbit, the gravitational force must equal the centripetal force.
   - In this universe, the gravitational acceleration is \( \frac{G M}{r^3} \), while the required centripetal acceleration is \( \frac{G M}{r^2} \).
   - After moving to twice the distance, the gravitational force becomes \( \frac{1}{8} \) of the original, which is weaker than the required centripetal force at twice the distance.
   - This imbalance causes the planet's orbit to become unstable.

3. **Outcome**:
   - The gravitational force is insufficient to maintain a stable circular orbit at the new distance.
   - The planet will likely enter an elliptical orbit or an unstable trajectory, potentially leading to an escape if the velocity is sufficient.

### Final Answer
The new gravitational force is one-eighth of the original force, and the planet's orbit would likely become unstable and elliptical. Therefore, the force decreases by a factor of 8, and the orbit becomes elliptical:

The gravitational force decreases to \boxed{\dfrac{1}{8}} of the original, and the orbit becomes elliptical.