Excellent question! This scenario forces us to abandon our intuition based on the real world and think from first principles. Here’s a breakdown of what would happen.

### Part 1: The Change in Gravitational Force

In our universe, the force of gravity follows an inverse-square law ($F \propto 1/r^2$). In this hypothetical universe, it follows an **inverse-cube law ($F \propto 1/r^3$)**.

Let's compare the force at the original distance ($r_{old}$) with the force at the new distance ($r_{new}$).

*   **Original Force ($F_{old}$):** Proportional to $1 / (r_{old})^3$
*   **New Distance ($r_{new}$):** The problem states this is $2 \times r_{old}$
*   **New Force ($F_{new}$):** Proportional to $1 / (r_{new})^3 = 1 / (2 \times r_{old})^3$

Now, let's calculate the effect of doubling the distance:

$F_{new} \propto 1 / (2r_{old})^3 = 1 / (8 \times (r_{old})^3)$

So, the new force is **1/8th** of the original force.

**Reasoning:** Because the force is proportional to the *cube* of the inverse distance, doubling the distance reduces the force by a factor of $2^3$, which is 8. (For comparison, in our universe, doubling the distance would reduce the force to $1/2^2$, or 1/4th of the original).

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### Part 2: The Likely Outcome for the Planet's New Orbit

This is the more interesting part. For a planet to maintain a stable circular orbit, the gravitational force must be exactly equal to the centripetal force required to keep it moving in a circle.

*   **Centripetal Force ($F_c$):** $F_c = mv^2 / r$ (where *m* is the planet's mass, *v* is its orbital velocity, and *r* is its orbital radius). This formula is based on kinematics and remains the same in any universe.
*   **Gravitational Force ($F_g$):** $F_g = G'Mm / r^3$ (where *M* is the star's mass, and *G'* is the gravitational constant for this universe).

For a stable circular orbit, $F_g = F_c$. Let's set them equal to find the required velocity for such an orbit:

$G'Mm / r^3 = mv^2 / r$

We can cancel *m* from both sides and simplify:

$G'M / r^2 = v^2$
$v = \sqrt{G'M / r^2}$
$v = \frac{\sqrt{G'M}}{r}$

This result is crucial. In this inverse-cube universe, the velocity required for a stable circular orbit is proportional to **1/r**.

Now, let's analyze the situation:

1.  **Original Orbit:** The planet was at distance $r_{old}$ with an orbital velocity of $v_{old} = \frac{\sqrt{G'M}}{r_{old}}$.
2.  **The Push:** The planet is "suddenly pushed" to a new distance of $r_{new} = 2r_{old}$. A "sudden push" in physics problems usually implies that the object's position is changed, but it initially retains its velocity from just before the push. So, its velocity is still $v_{old}$.
3.  **The Problem:** At this new distance of $2r_{old}$, what is the velocity *required* for a stable circular orbit? According to our formula:
    $v_{required} = \frac{\sqrt{G'M}}{r_{new}} = \frac{\sqrt{G'M}}{2r_{old}}$

    If we compare this to the original velocity, we see that $v_{required} = \frac{1}{2} v_{old}$.

**The Conclusion and Outcome:**

The planet is now at a distance where it needs to be moving at **half its original speed** to maintain a stable circular orbit. However, it is actually still moving at its **full original speed**.

Because the planet is moving twice as fast as it should be for a stable orbit at that distance, its kinetic energy vastly overwhelms the star's gravitational pull (which is now only 1/8th as strong). The planet has far too much momentum to be held in any closed orbit (circular or elliptical).

Therefore, the most likely outcome is that **the planet will fly away from the star on an unbound, hyperbolic trajectory, escaping the star system forever.**