= Solution
In a fixed point-mass <Newtonian gravitational field>, a test <mass> $m$ experiences
$$
\mathbf F=-\frac{GMm}{r^2}\widehat{\mathbf r}.
$$
Because this is a <central force>, its <torque> about the attracting centre vanishes and <angular momentum> is conserved. The orbit lies in a fixed plane and sweeps out equal areas in equal times. Since the <force> is also conservative, the specific orbital <energy> is constant:
$$
\varepsilon=\frac12\dot r^2+\frac{j^2}{2r^2}-\frac{GM}{r},\qquad j=r^2\dot\varphi.
$$
The last two terms form an <effective potential>. They show how the angular-momentum barrier competes with attraction and determines the turning radii. For nonradial motion the orbit is a <conic section>:
$$
\boxed{r=\frac{p}{1+e\cos(\varphi-\varphi_0)},\qquad p=\frac{j^2}{GM},\qquad e^2=1+\frac{2\varepsilon j^2}{G^2M^2}.}
$$
Negative <energy> gives an <ellipse>, with $e<1$; zero <energy> gives a <parabola>; positive <energy> gives a <hyperbola>. The circular limit has $e=0$, $v_c=\sqrt{GM/r}$ and $\varepsilon=-GM/(2r)$. For a bound <ellipse> of semimajor axis $a$, $\varepsilon=-GM/(2a)$ and the <orbital period> is $T=2\pi\sqrt{a^3/(GM)}$. The local <escape speed> is $\sqrt{2GM/r}$. These are the point-mass, test-particle limits; extended mass distributions or noncentral perturbations change the orbit laws.
Back to article page
