Satellites: Orbits and Energy — Question 10

Question 10

An effective-potential graph separates radial turning points from zero speed

A satellite has specific angular momentum $h^2=\mu r_0$ and energy $\epsilon=-\mu/(4r_0)$. Use $\dot r^2/2+V_{eff}(r)=\epsilon$, where $V_{eff}=h^2/(2r^2)-\mu/r$.

1. Find the two radial turning points and the allowed radial interval.

2. At a turning point, is the satellite at rest? Locate the stable circular radius for this fixed $h$.

Question 10 - Solution

An effective-potential graph separates radial turning points from zero speed

(a) Derivation and calculation

At a radial turning point, $\dot r=0$. Set $x=r/r_0$: $$\frac{r_0V_{\mathrm{eff}}}\mu=\frac1{2x^2}-\frac1x=-\frac14.$$ Multiplication by $4x^2$ gives $x^2-4x+2=0$, hence $$\boxed{r_\pm=(2\pm\sqrt2)r_0}.$$ The radial kinetic energy is nonnegative only for $$\boxed{r_-\le r\le r_+}.$$

(b) Result and interpretation

At either turning point, $\dot r=0$ but $v_t=h/r\ne0$. For a circular orbit at fixed $h$, $$V'_{\mathrm{eff}}=-\frac{h^2}{r^3}+\frac\mu{r^2}=0 \quad\Longrightarrow\quad\boxed{r_c=h^2/\mu=r_0}.$$ $$V_{\mathrm{min}}=-\frac\mu{2r_0},\qquad V''_{\mathrm{eff}}(r_0)=\frac\mu{r_0^3}>0.$$ The minimum is radially stable, and the given higher energy produces two distinct turning points.

Physical check

The energy lies between the negative minimum and zero, consistent with a bound noncircular orbit.