### Short answer
- New gravitational force: one-eighth of the old (\(F \propto 1/r^3 \Rightarrow F_{\text{new}} = F_{\text{old}}/8\) when \(r \to 2r\)).
- Likely orbit: circular orbits in a \(1/r^3\) force are marginal/unstable. An outward kick almost always leads to runaway: the planet keeps moving outward and escapes unless the kick is exquisitely tuned to give exactly the circular speed \(v \propto 1/r\) with no radial velocity. Any mismatch sends it either outward to infinity (slightly too much energy) or inward (slightly too little).

Reasoning (brief):
- For a circular orbit under \(F = k/r^3\), you need \(v^2/r = k/r^3 \Rightarrow v = \sqrt{k}/r\) and angular momentum \(L = mrv = m\sqrt{k}\), independent of radius. That means circular orbits exist only for this single \(L\) and are neutrally stable; any radial perturbation doesn’t restore the radius, causing runaway behavior.

- Summary
  - Force at 2× radius: \(1/8\) of original.
  - New orbit: not stably circular; unless perfectly tuned, an outward push leads to escape (runaway outward).