### Understanding the Problem

First, let's restate the problem to ensure we understand it correctly.

In our universe, gravity follows an inverse-square law: the force between two masses is proportional to \( \frac{1}{r^2} \), where \( r \) is the distance between them. Here, we're considering a hypothetical universe where gravity follows an inverse-cube law: \( F \propto \frac{1}{r^3} \).

We have a planet in a stable, perfectly circular orbit around its star. Suddenly, the planet is pushed to a new orbit that is exactly twice as far away from the star. We need to:

1. Determine how the gravitational force changes when the distance is doubled.
2. Predict the likely outcome for the planet's new orbit.

### Step 1: Calculating the Change in Gravitational Force

Original distance: \( r \)

New distance: \( 2r \)

Original force: \( F \propto \frac{1}{r^3} \)

New force: \( F' \propto \frac{1}{(2r)^3} = \frac{1}{8r^3} \)

So, \( F' = \frac{1}{8} F \).

Thus, the new gravitational force is one-eighth of the original force.

### Step 2: Analyzing the Orbit After the Push

Now, let's think about the planet's orbit after it's pushed to \( 2r \).

In our universe (inverse-square gravity), when a planet is moved to a higher orbit, it's typically in an elliptical orbit unless it's given the exact correct velocity for a circular orbit at the new radius. Here, the planet is "pushed" to twice the distance, implying that its velocity might not be adjusted to maintain a circular orbit at \( 2r \).

But first, let's recall how circular orbits work in both inverse-square and inverse-cube gravity.

#### Circular Orbits in Inverse-Square Gravity (Our Universe)

For a circular orbit, the centripetal force is provided by gravity:

\( \frac{GMm}{r^2} = \frac{mv^2}{r} \)

Solving for \( v \):

\( v = \sqrt{\frac{GM}{r}} \)

The orbital velocity decreases with \( \sqrt{r} \).

#### Circular Orbits in Inverse-Cube Gravity

Similarly, for inverse-cube gravity:

\( \frac{GMm}{r^3} = \frac{mv^2}{r} \)

Solving for \( v \):

\( v = \sqrt{\frac{GM}{r^2}} = \frac{\sqrt{GM}}{r} \)

So, the orbital velocity decreases with \( \frac{1}{r} \).

This is a key difference: in inverse-square gravity, velocity decreases as \( \frac{1}{\sqrt{r}} \), but in inverse-cube gravity, it decreases as \( \frac{1}{r} \).

#### Initial Circular Orbit at \( r \)

At distance \( r \), the planet has velocity \( v = \frac{\sqrt{GM}}{r} \).

#### After Being Pushed to \( 2r \)

The planet is now at \( 2r \). For a circular orbit at \( 2r \), the required velocity would be:

\( v_{\text{new circular}} = \frac{\sqrt{GM}}{2r} \)

But the planet was originally moving at \( \frac{\sqrt{GM}}{r} \). When it's pushed to \( 2r \), its velocity might not change instantaneously (assuming the push is purely radial). So, its velocity at \( 2r \) is still \( \frac{\sqrt{GM}}{r} \).

But the required velocity for a circular orbit at \( 2r \) is \( \frac{\sqrt{GM}}{2r} \), which is half of its current velocity.

This means the planet is moving too fast to maintain a circular orbit at \( 2r \). 

#### Determining the New Orbit

With gravity being inverse-cube, let's see what happens when the planet is at \( 2r \) with velocity \( \frac{\sqrt{GM}}{r} \).

The gravitational force provides the centripetal acceleration:

\( \frac{GM}{(2r)^3} = \frac{v^2}{2r} \)

Plugging \( v = \frac{\sqrt{GM}}{r} \):

\( \frac{GM}{8r^3} = \frac{GM}{r^2 \cdot 2r} = \frac{GM}{2r^3} \)

But \( \frac{GM}{8r^3} \neq \frac{GM}{2r^3} \), so the planet is not in equilibrium; it's moving too fast for the gravitational force at \( 2r \).

This excess velocity will cause the planet to move outward, but since gravity is stronger at closer distances (due to the \( \frac{1}{r^3} \) dependence), the planet may not escape to infinity. Instead, it might enter an elliptical orbit where it oscillates between \( r \) and \( 3r \) or some other bounds.

However, in an inverse-cube force law, stable orbits are not possible except for very specific conditions (like zero angular momentum, which isn't the case here). Typically, orbits under inverse-cube gravity are not stable; they tend to precess or spiral.

Given that, the planet, when pushed to \( 2r \) with its original velocity, would likely not have a stable orbit. It would probably spiral outward or inward depending on the exact dynamics, but generally, orbits under inverse-cube gravity are not stable circles or ellipses as we know them.

But let's think more carefully about the energy and angular momentum.

#### Energy Considerations

Total mechanical energy \( E \) is the sum of kinetic and potential energy.

Potential energy \( U \) in inverse-cube gravity can be found by integrating the force:

\( F = -\frac{k}{r^3} \), so \( U = -\int F \, dr = \frac{k}{2r^2} \) (plus a constant, which we can set to zero at infinity).

Kinetic energy \( K = \frac{1}{2}mv^2 \).

At \( r \), \( v = \frac{\sqrt{GM}}{r} \), so \( K = \frac{GMm}{2r^2} \), \( U = \frac{GMm}{2r^2} \), so \( E = 0 \).

At \( 2r \), if the planet has the same speed (which is unlikely unless the push is instantaneous and doesn't change velocity), \( K = \frac{GMm}{8r^2} \), \( U = \frac{GMm}{8r^2} \), so \( E = 0 \) again.

But this suggests the total energy is the same, which might imply the planet could escape, but this contradicts the idea that inverse-cube gravity is stronger at close distances, making escape harder.

Alternatively, if the push is radial and instantaneous, the velocity perpendicular to the radial direction (which determines angular momentum) might remain the same, but the speed could change.

Actually, the exact change in velocity depends on how the push is applied. If it's purely radial, the tangential velocity remains \( \frac{\sqrt{GM}}{r} \), but the radial velocity is added.

This makes the orbit more complex, likely leading to an unstable path.

#### More Accurate Approach: Effective Potential

In central force problems, we can consider the effective potential which includes the angular momentum.

The effective potential \( U_{\text{eff}} \) is given by:

\( U_{\text{eff}}(r) = U(r) + \frac{L^2}{2mr^2} \)

For inverse-cube gravity, \( U(r) = \frac{k}{2r^2} \), and \( L \) is angular momentum.

At \( r \), \( v = \frac{\sqrt{GM}}{r} \), so \( L = mvr = m \frac{\sqrt{GM}}{r} r = m \sqrt{GM} \).

At \( 2r \), the angular momentum is the same if no torque is applied (i.e., the push is radial), so \( L = m \sqrt{GM} \).

Then, \( U_{\text{eff}}(2r) = \frac{k}{8r^2} + \frac{GM}{8r^2} = \frac{GM}{4r^2} \) (since \( k = GMm \)).

The kinetic energy is \( \frac{L^2}{2m(2r)^2} = \frac{GM}{8r^2} \), but this seems inconsistent.

This suggests that the planet may not have enough energy to maintain a circular orbit at \( 2r \), leading to an unstable situation where it would likely spiral inward or outward, but given the nature of inverse-cube gravity, it's more likely to spiral inward and eventually crash into the star.

### Conclusion

1. **Change in Gravitational Force**: When the planet is moved to twice the distance, the gravitational force becomes one-eighth of the original force.
   
2. **Outcome of the New Orbit**: The planet's new orbit is unlikely to be stable. Given that the original circular orbit required a specific velocity, and moving to twice the distance without adjusting the velocity means the planet is moving too fast for a circular orbit at the new distance, it will not maintain a circular path. Instead, it will likely enter an unstable orbit and could spiral inward toward the star due to the nature of the inverse-cube gravitational force. 

In summary, the gravitational force decreases to one-eighth, and the planet's new orbit will be unstable, likely leading to a spiral inward toward the star.