In this hypothetical universe, the gravitational force \( F \) is proportional to the inverse cube of the distance \( r \), expressed as \( F = \frac{k}{r^3} \), where \( k \) is a constant. Consider a planet in a stable circular orbit at an initial distance \( r \) from its star. After being pushed to a new distance exactly twice as far, \( R = 2r \), we analyze the changes.

### Change in Gravitational Force
- At the initial distance \( r \), the gravitational force is \( F_{\text{initial}} = \frac{k}{r^3} \).
- At the new distance \( R = 2r \), the gravitational force is:
  \[
  F_{\text{new}} = \frac{k}{(2r)^3} = \frac{k}{8r^3} = \frac{1}{8} F_{\text{initial}}.
  \]
Thus, the new gravitational force is **one-eighth** of the original force.

### Outcome for the Planet's New Orbit
For a stable circular orbit in an inverse-cube gravity universe:
- The centripetal force is provided by gravity: \( F = \frac{mv^2}{r} \), where \( m \) is the planet's mass and \( v \) is its orbital speed.
- At the initial orbit:
  \[
  \frac{k}{r^3} = \frac{mv_{\text{initial}}^2}{r} \implies v_{\text{initial}} = \sqrt{\frac{k}{m r^2}}.
  \]
- At the new distance \( R = 2r \), the required orbital speed for a circular orbit is:
  \[
  F_{\text{new}} = \frac{k}{8r^3} \implies \frac{k}{8r^3} = \frac{mv_{\text{circ}}^2}{2r} \implies v_{\text{circ}} = \sqrt{\frac{k}{4m r^2}} = \frac{1}{2} \sqrt{\frac{k}{m r^2}} = \frac{v_{\text{initial}}}{2}.
  \]
- However, when the planet is pushed radially to \( 2r \) (with no immediate change in tangential velocity due to angular momentum conservation), its tangential speed at the new distance remains \( v_{\text{initial}} \):
  \[
  v_{\text{tangent}} = v_{\text{initial}} = \sqrt{\frac{k}{m r^2}}.
  \]
- Comparing this speed to \( v_{\text{circ}} \):
  \[
  v_{\text{tangent}} = \sqrt{\frac{k}{m r^2}} > \frac{1}{2} \sqrt{\frac{k}{m r^2}} = v_{\text{circ}}.
  \]
  Specifically, \( v_{\text{tangent}} = 2v_{\text{circ}} \), meaning the planet is moving too fast for a circular orbit at \( R = 2r \).

### Orbital Stability and Escape
- The total energy at the new position determines the orbit's nature. The gravitational potential energy for inverse-cube gravity is \( U = -\frac{k}{2r^2} \) (derived from \( F = -\frac{dU}{dr} \)).
- Initial total energy at \( r \) (stable circular orbit): Kinetic energy \( K_{\text{initial}} = \frac{1}{2} m v_{\text{initial}}^2 = \frac{1}{2} m \left( \frac{k}{m r^2} \right) = \frac{k}{2r^2} \), and potential energy \( U_{\text{initial}} = -\frac{k}{2r^2} \). Thus:
  \[
  E_{\text{initial}} = K_{\text{initial}} + U_{\text{initial}} = \frac{k}{2r^2} - \frac{k}{2r^2} = 0.
  \]
- At the new distance \( R = 2r \), with tangential velocity still \( v_{\text{initial}} \):
  - Kinetic energy: \( K_{\text{new}} = \frac{1}{2} m v_{\text{tangent}}^2 = \frac{1}{2} m \left( \frac{k}{m r^2} \right) = \frac{k}{2r^2} \).
  - Potential energy: \( U_{\text{new}} = -\frac{k}{2 (2r)^2} = -\frac{k}{8r^2} \).
  - Total energy:
    \[
    E_{\text{new}} = K_{\text{new}} + U_{\text{new}} = \frac{k}{2r^2} - \frac{k}{8r^2} = \frac{3k}{8r^2} > 0.
    \]
- A **positive total energy** implies the orbit is **unbound**. The planet has sufficient kinetic energy to escape the gravitational pull of the star and will follow a hyperbolic trajectory, moving away indefinitely without returning.

### Conclusion
- The new gravitational force is **one-eighth** of the original force.
- The planet will **escape the star's gravitational influence** due to its increased tangential velocity at the new distance, leading to a positive total energy and an unbounded orbit.

This outcome contrasts with an inverse-square law universe, where similar conditions might lead to an elliptical or escape trajectory depending on the energy. Here, the inverse-cube force ensures that the excess speed causes gravitational escape.