In this hypothetical universe, the gravitational force is proportional to the inverse cube of the distance between two objects, i.e., $ F \propto \frac{1}{r^3} $. This is different from the familiar inverse square law, which leads to various differences in orbital mechanics.

---

### **Change in Gravitational Force**

Let’s denote the original distance of the planet from the star as $ r_1 $. The gravitational force at that distance is:

$$
F_1 = \frac{k}{r_1^3}
$$

After being pushed to a new distance of $ r_2 = 2r_1 $, the new gravitational force is:

$$
F_2 = \frac{k}{(2r_1)^3} = \frac{k}{8r_1^3}
$$

Thus, the new force is:

$$
F_2 = \frac{1}{8} F_1
$$

---

### **Effect on the Planet’s Orbit**

To determine the **likely outcome** of the new orbit, we must understand how the energy of the system changes when the planet is displaced to a new radius without an appropriate velocity adjustment.

In the original circular orbit, the total mechanical energy is **zero**, as is typical for gravitational systems with inverse square forces, but in this case, due to the inverse **cube** force:

- The gravitational potential energy is $ U = -\frac{k}{2r^2} $.
- The kinetic energy $ T = \frac{1}{2} m v^2 $, and for a circular orbit, the gravitational force provides the necessary centripetal acceleration: $ F = \frac{m v^2}{r} $.
- Substituting $ F = \frac{k}{r^3} $, we solve for $ v $:

$$
\frac{k}{r^3} = \frac{m v^2}{r} \Rightarrow v^2 = \frac{k}{m r^2} \Rightarrow T = \frac{1}{2} m v^2 = \frac{k}{2r^2}
$$

So total energy $ E = T + U = \frac{k}{2r^2} - \frac{k}{2r^2} = 0 $, as expected.

---

### **What Happens When the Planet is Pushed to Twice the Distance**

If the planet is **suddenly pushed into an orbit exactly twice as far** (i.e., $ r_2 = 2r_1 $) **without changing its velocity**, the energy of the system changes. The new kinetic energy is:

$$
T = \frac{k}{2r_1^2}
$$

And the gravitational potential is:

$$
U = -\frac{k}{2(2r_1)^2} = -\frac{k}{8r_1^2}
$$

So total energy becomes:

$$
E = \frac{k}{2r_1^2} - \frac{k}{8r_1^2} = \frac{3k}{8r_1^2} > 0
$$

This is a **positive** total energy, indicating that the system is **unbound** and the planet will **escape** the gravitational pull of the star—regardless of the shape of the trajectory (it may be a hyperbolic or parabolic escape orbit, depending on initial conditions).

---

### **Conclusion**

- The **gravitational force** is reduced to **1/8th** of its original value when the distance is doubled in a $ 1/r^3 $ regime.
- Due to the **positive energy** associated with the new configuration, the **planet will not remain in a bounded orbit** and will **escape** from the star.

---

### **Final Answer**

The new gravitational force is \boxed{\dfrac{1}{8}} of the original. The likely outcome is that the planet will escape the star's gravitational pull and not maintain a stable orbit.