### Comparison of the New Gravitational Force to the Old One

In this universe, the gravitational force is proportional to the inverse cube of the distance (\(F \propto 1/r^3\)). For the original orbit at distance \(r\), let the force be \(F_\text{old} \propto 1/r^3\).

When the planet is pushed to a new distance of \(2r\), the new force is \(F_\text{new} \propto 1/(2r)^3 = 1/(8r^3)\). Therefore,

\[
\frac{F_\text{new}}{F_\text{old}} = \frac{1/(8r^3)}{1/r^3} = \frac{1}{8}.
\]

The new gravitational force is **one-eighth (1/8)** as strong as the old one. This is purely due to the increased distance; the masses and gravitational constant are unchanged.

### Likely Outcome for the Planet's New Orbit

To determine the outcome, we must consider the planet's velocity after the "sudden push." I interpret this as an instantaneous radial displacement to \(2r\) while keeping the planet's original tangential velocity unchanged (a common assumption in such hypothetical scenarios, as the push is radial and does not alter the tangential speed). Here's the step-by-step reasoning:

1. **Orbital Requirements in This Universe**:
   - The gravitational force law is \(F = GMm / r^3\), where \(G\) is the gravitational constant, \(M\) is the star's mass, and \(m\) is the planet's mass.
   - For a stable circular orbit at radius \(R\), the gravitational force provides the centripetal force: \(GMm / R^3 = m v^2 / R\).
   - Solving gives the required circular orbital speed: \(v_\text{circ} = \sqrt{GM} / R\) (i.e., speed is inversely proportional to radius).
   - In the original orbit at \(r\), the speed is \(v_\text{old} = \sqrt{GM} / r\).
   - At \(2r\), the speed required for a new circular orbit would be \(v_\text{required} = \sqrt{GM} / (2r) = (1/2) v_\text{old}\).

2. **The Planet's Actual Speed After the Push**:
   - The planet retains its original tangential speed \(v_\text{old}\).
   - This is **twice** the speed needed for a circular orbit at \(2r\) (\(v_\text{old} = 2 \times v_\text{required}\)).

3. **Energy Analysis**:
   - In this universe, the gravitational potential energy is \(U = - (GMm) / (2 r^2)\) (derived by integrating the force law).
   - Total mechanical energy \(E =\) kinetic energy \(K + U\).
   - For circular orbits, \(E = 0\) (they are marginally bound).
   - Original energy (at \(r\)): \(E_\text{old} = 0\).
   - After the push (at \(2r\), with unchanged \(v_\text{old}\)):
     - \(K_\text{new} = (1/2) m v_\text{old}^2 = (1/2) m (GM / r^2) = m (GM / (2 r^2))\) (unchanged kinetic energy).
     - \(U_\text{new} = - (GMm) / (2 (2r)^2) = - (GMm) / (8 r^2) = - m (GM / (8 r^2))\).
     - \(E_\text{new} = m (GM / (2 r^2)) - m (GM / (8 r^2)) = m (GM / r^2) (1/2 - 1/8) = m (GM / r^2) (3/8) > 0\).
   - Since \(E > 0\) and the potential approaches 0 as \(r \to \infty\), the planet has enough energy to escape to infinity.

4. **Interpretation**:
   - The planet's speed is too high relative to the weaker gravity at \(2r\) (which is only 1/8 as strong).
   - Gravity cannot provide enough inward acceleration to keep the planet bound; the "centripetal acceleration" implied by the planet's speed and new radius (\(v_\text{old}^2 / (2r)\)) is 4 times greater than what gravity can supply.
   - As a result, the planet will follow an **unbound (hyperbolic-like) trajectory** and escape the star's gravitational influence entirely, never returning to a closed orbit.

This outcome assumes the push does not adjust the velocity (if it did, and the speed was reduced to \(v_\text{required}\), the new orbit could be circular and stable—but the problem's wording suggests a disruptive "sudden" push, not a carefully adjusted one). This is analogous to similar scenarios in our inverse-square universe, where suddenly doubling the radius while keeping speed constant also leads to escape (though with \(E = 0\) there).