**1.  How the gravitational pull changes**

In this “toy universe’’ the law of gravitation is  

\[
F(r)=\frac{GMm}{r^{3}}\; ,
\]

instead of the familiar \(GMm/r^{2}\).

If the planet is moved from its original radius \(r_0\) to twice that distance, \(r_1 = 2r_0\), the force from the star becomes  

\[
F(r_1)=\frac{GMm}{(2r_0)^{3}}
       =\frac{GMm}{8r_0^{3}}
       =\frac{1}{8}\,F(r_0) .
\]

So the gravity it feels drops by a factor of eight.

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**2.  What that means for the new orbit**

A *circular* orbit requires that the inward gravitational force supply the exact centripetal force needed to keep the planet moving round the star:

\[
\frac{GMm}{r^{3}}=\frac{m v^{2}}{r}\;\;\Longrightarrow\;\;
v^{2}=\frac{GM}{r^{2}} .
\]

Note that for an \(1/r^{3}\) force the speed which makes a *stable* circular orbit is

\[
v_c(r)=\sqrt{\frac{GM}{r^{2}}}\;\propto\;\frac{1}{r},
\]

and the angular momentum for such an orbit is

\[
L=m v_c r
  =m\sqrt{GM}\; ,
\]

**independent of the orbital radius**.

When the planet is abruptly “pushed’’ outward to \(2r_0\) its speed does *not* change instantaneously; it still has whatever velocity it had on the old circular orbit:

\[
v_{\text{after}}=v_c(r_0)=\sqrt{\frac{GM}{r_0^{2}}}\; .
\]

But the speed that would be required for a new circular orbit at the larger radius is only

\[
v_c(2r_0)=\sqrt{\frac{GM}{(2r_0)^{2}}}
         =\frac{1}{2}\,v_c(r_0)=\tfrac{1}{2}\,v_{\text{after}} .
\]

Thus the planet is *moving too fast* for a circular orbit at the new distance. The excess kinetic energy means the planet cannot hold onto a bounded, circular path; instead it will keep climbing outwards. In a potential that falls like \(-1/r^{2}\) (the integral of the \(1/r^{3}\) force), the effective radial “stiffness’’ is zero – the effective potential has a flat minimum, so any slight disturbance leads to runaway motion.

Hence the planet will not settle into a new circular orbit. Instead it will spiral outward and eventually escape to infinity (or, if the push were in the opposite direction, run inward until it crashes into the star). The orbit would be highly unstable and essentially impossible to maintain once perturbed.

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**Summary**

* The gravitational attraction at twice the distance is only \(1/8\) of its original value.
* Because the required centripetal speed for a circular orbit drops by a factor of two at the new radius, a planet that keeps its old speed (the same as before the push) is moving too fast.
* With this extra kinetic energy the planet will not remain on a circular orbit; the orbit is unstable, and the planet will continue to accelerate away from the star (or fall inward) rather than settling into a new, stable orbit.