In this hypothetical universe where the force of gravity \( F \) is proportional to the inverse cube of the distance \( r \) between two objects, the gravitational force can be expressed as:

\[ F \propto \frac{1}{r^3} \]

Let's denote the initial distance between the planet and the star as \( r_1 \) and the new distance as \( r_2 = 2r_1 \).

### Gravitational Force Comparison:
Initially, the gravitational force \( F_1 \) is given by:

\[ F_1 \propto \frac{1}{r_1^3} \]

After the planet is pushed to a distance \( r_2 = 2r_1 \), the new gravitational force \( F_2 \) becomes:

\[ F_2 \propto \frac{1}{(2r_1)^3} = \frac{1}{8r_1^3} \]

Therefore, the ratio of the new force to the old force is:

\[ \frac{F_2}{F_1} = \frac{\frac{1}{8r_1^3}}{\frac{1}{r_1^3}} = \frac{1}{8} \]

So, the new gravitational force is \( \frac{1}{8} \) of the original gravitational force.

### Orbital Dynamics:
For a planet in a circular orbit, the gravitational force provides the necessary centripetal force to keep the planet in orbit. The centripetal force required for a circular orbit is given by:

\[ F_{\text{centripetal}} = \frac{mv^2}{r} \]

where \( m \) is the mass of the planet and \( v \) is its orbital velocity. In our universe with the inverse-cube law, the gravitational force is:

\[ F_{\text{grav}} \propto \frac{1}{r^3} \]

For a stable circular orbit, we set the gravitational force equal to the centripetal force:

\[ \frac{1}{r^3} \propto \frac{mv^2}{r} \]

This simplifies to:

\[ v^2 \propto \frac{1}{r^2} \]

or

\[ v \propto \frac{1}{r^{1/2}} \]

If the planet is moved to a new orbit at \( r_2 = 2r_1 \), the new orbital velocity \( v_2 \) will be:

\[ v_2 \propto \frac{1}{(2r_1)^{1/2}} = \frac{1}{\sqrt{2} r_1^{1/2}} = \frac{v_1}{\sqrt{2}} \]

where \( v_1 \) is the original orbital velocity.

The period \( T \) of the orbit is related to the velocity and the radius by:

\[ T = \frac{2\pi r}{v} \]

Substituting \( v \propto \frac{1}{r^{1/2}} \):

\[ T \propto r^{3/2} \]

Thus, if the distance doubles (\( r_2 = 2r_1 \)), the new period \( T_2 \) will be:

\[ T_2 \propto (2r_1)^{3/2} = 2^{3/2} r_1^{3/2} = 2\sqrt{2} r_1^{3/2} \]

Since \( T_1 \propto r_1^{3/2} \), the ratio of the new period to the old period is:

\[ \frac{T_2}{T_1} = 2\sqrt{2} \approx 2.83 \]

### Conclusion:
- The new gravitational force between the planet and the star is \( \frac{1}{8} \) of the original force.
- The planet's orbital velocity decreases by a factor of \( \sqrt{2} \).
- The period of the new orbit increases by a factor of approximately \( 2.83 \).

Given these changes, the planet will enter a new elliptical orbit that is more elongated than the original circular orbit, but it will not necessarily escape the star's gravitational influence unless additional factors (like other perturbations) come into play. The increased distance and reduced gravitational force suggest a slower, more elliptical orbit, but without further perturbations, the planet should remain bound to the star.