Alright, let's tackle this problem step by step. I'm going to break it down to understand what happens when the force of gravity is proportional to the inverse cube of the distance, and how that affects a planet's orbit when it's suddenly moved to twice its original distance from its star.

### Understanding the Given Information

1. **Gravity's Dependence on Distance**:
   - Normally, in our universe, Newton's law of universal gravitation states that the gravitational force between two objects is inversely proportional to the square of the distance between them. Mathematically, this is:
     \[
     F \propto \frac{1}{r^2}
     \]
   - In this hypothetical universe, gravity is inversely proportional to the cube of the distance:
     \[
     F \propto \frac{1}{r^3}
     \]

2. **Initial Orbit**:
   - The planet is in a stable, perfectly circular orbit around its star. This means that the gravitational force provides the necessary centripetal force to keep the planet moving in a circular path at a constant speed.
   - Let's denote the initial distance from the star to the planet as \( r \), and the initial gravitational force as \( F \).

3. **Change in Orbit**:
   - The planet is suddenly pushed into an orbit exactly twice as far away, so the new distance is \( 2r \).

### Calculating the New Gravitational Force

Given that the gravitational force is inversely proportional to the cube of the distance, we can express the force as:
\[
F = k \cdot \frac{1}{r^3}
\]
where \( k \) is a constant of proportionality.

- **Initial Force (\( F_1 \))**:
  \[
  F_1 = k \cdot \frac{1}{r^3}
  \]

- **New Force (\( F_2 \)) at distance \( 2r \)**:
  \[
  F_2 = k \cdot \frac{1}{(2r)^3} = k \cdot \frac{1}{8r^3} = \frac{1}{8} \cdot k \cdot \frac{1}{r^3} = \frac{F_1}{8}
  \]

So, the new gravitational force is \( \frac{1}{8} \) (or 12.5%) of the original force.

### Determining the Outcome of the New Orbit

Now, let's think about what happens to the planet's orbit after this change.

1. **Initial Circular Orbit**:
   - In a circular orbit, the gravitational force provides the centripetal force needed to keep the planet moving in a circle.
   - The centripetal force is given by:
     \[
     F_{\text{centripetal}} = m \cdot \frac{v^2}{r}
     \]
     where \( m \) is the mass of the planet, and \( v \) is its orbital velocity.
   - Initially, \( F_1 = F_{\text{centripetal}} \).

2. **After the Push**:
   - The planet is suddenly at a distance \( 2r \), and the gravitational force is now \( \frac{F_1}{8} \).
   - However, the planet's velocity hasn't instantaneously changed (since the push is sudden but we're considering the immediate aftermath).
   - Initially, the centripetal force needed at \( 2r \) would be:
     \[
     F_{\text{centripetal, new}} = m \cdot \frac{v^2}{2r}
     \]
     But since the velocity hasn't changed yet, \( v \) is the same as before, so:
     \[
     F_{\text{centripetal, new}} = \frac{1}{2} \cdot m \cdot \frac{v^2}{r} = \frac{F_{\text{centripetal}}}{2} = \frac{F_1}{2}
     \]
     (Because initially, \( F_1 = m \cdot \frac{v^2}{r} \))

   - But the actual gravitational force at \( 2r \) is \( \frac{F_1}{8} \), which is less than \( \frac{F_1}{2} \).

3. **Comparing Forces**:
   - The required centripetal force at \( 2r \) is \( \frac{F_1}{2} \), but the available gravitational force is \( \frac{F_1}{8} \).
   - Since \( \frac{F_1}{8} < \frac{F_1}{2} \), the gravitational force is insufficient to provide the necessary centripetal force for a circular orbit at this speed.

4. **Resulting Motion**:
   - When the centripetal force is less than required, the object cannot maintain a circular orbit. Instead, it will start to move outward because the gravitational pull isn't strong enough to keep it in a circular path.
   - The planet will begin to move away from the star, and its velocity will decrease as it moves to higher orbits (since it's not getting enough inward pull to maintain speed).

5. **Long-term Behavior**:
   - The planet's new orbit will not be stable. It will likely follow a trajectory that moves away from the star, possibly becoming an elliptical orbit with increasing distance or even escaping the star's gravity entirely, depending on the initial push's energy and the specifics of the system.

### Comparing with Inverse-Square Law (For Context)

Just to contrast, in our real universe where gravity follows an inverse-square law:

- If a planet is doubled in distance, the gravitational force becomes \( \frac{F_1}{4} \).
- The required centripetal force at \( 2r \) is \( \frac{F_1}{2} \).
- Here, \( \frac{F_1}{4} < \frac{F_1}{2} \), so similarly, the planet would start to move outward, but the rate of divergence would be different due to the different force scaling.

However, in our inverse-cube scenario, the force drops off much more steeply with distance, leading to a more pronounced outward motion.

### Final Answer

In this hypothetical universe where gravity is inversely proportional to the cube of the distance:

1. **Gravitational Force Change**:
   - When the planet is moved to twice its original distance (\( 2r \)), the new gravitational force is \( \frac{1}{8} \) (or 12.5%) of the original force.

2. **Outcome of the Orbit**:
   - The new gravitational force is insufficient to provide the necessary centripetal force for a circular orbit at the planet's current speed.
   - As a result, the planet will not remain in a stable orbit. Instead, it will start to move away from the star, likely following an outward trajectory that could become increasingly elliptical or even result in the planet escaping the star's gravity entirely.
   - The orbit will not be stable; the planet will tend to move to higher orbits or escape due to the steep decrease in gravitational force with distance.