Force of gravitation is equal to the centrifugal force for a body in circular motion.

If the law of gravitation becomes an inverse cube law, then we can write, for a planet of mass $m$ revolving around the sun of mass $M_s$.
$$
F=\frac{G M_s m_p}{r^3}=\frac{m v^2}{r} \quad\left(\because v^2=\frac{G M_s}{r^2}\right)
$$
where $M_s=$ mass of sun, $r$ is radius of orbiting planet, $m_p=$ mass of planet
$v($ orbital speed $)=\frac{\sqrt{G M_s}}{r} \Rightarrow v \propto \frac{1}{r}$
Time period of revolution of a planet
$$
T=\frac{2 \pi r}{v}=\frac{2 \pi r}{\frac{\sqrt{G M_s}}{a}}=\frac{2 \pi r^2}{\sqrt{G M_s}}
$$
or $T \propto r^2$
or $T^2 \propto r^4$
For elliptical orbit the condition on is $T^2 \propto r^3$ from Kepler's law.
Hence, orbit of planetary motion will not be elliptical. As force
$$
F=\left(\frac{G M_p}{R^3}\right) m=g^{\prime} m
$$
(where $R=$ radius of earth, $m=$ mass of a body on earth)
New acceleration due to gravity,
$$
g^{\prime}=\frac{G M_p}{R^3}
$$
As $g^{\prime}$, acceleration due to gravity is constant for a planet or earth hence path followed by a projectile will be approximately parabolic. (as $T \propto r^2$ ).