Exercises — Vis-viva equation v² = GM(2 - r − 1 - a) — derivation

• $r$ = the instantaneous distance from the focus (planet centre) to the satellite. This changes continuously around an ellipse.
• $a$ = the semi-major axis, half the longest diameter of the orbit. This is fixed for one orbit.
• $GM$ = the standard gravitational parameter, a fixed property of the central body. Only $r$ changes as the satellite orbits.

A circle is the special ellipse where the semi-major axis equals the radius, so $a=r$. Substitute: $v^2=GM\left(\frac{2}{r}-\frac{1}{r}\right)=GM\cdot \frac{1}{r}=\frac{GM}{r}\Rightarrow v=\sqrt{\frac{GM}{r}}.$

$r=6378+400=6778\text{km}$. $v=\sqrt{\frac{398600}{6778}}=\sqrt{58.81}=7.67\text{km/s}.$

The two apsides sit at opposite ends of the long axis, so they add to the full length $2a$: $a=\frac{r_p+r_a}{2}=\frac{7000+42000}{2}=24500\text{km}.$

$v_p^2=398600\left(\frac{2}{7000}-\frac{1}{24500}\right)=398600\left(2.8571\times 10^{-4}-4.0816\times 10^{-5}\right)$ $=398600\times 2.4490\times 10^{-4}=97.61\Rightarrow v_p=9.88\text{km/s}.$ Fast, as expected — perigee is the closest, fastest point.

$v_a^2=398600\left(\frac{2}{42000}-\frac{1}{24500}\right)=398600\left(4.7619\times 10^{-5}-4.0816\times 10^{-5}\right)$ $=398600\times 6.803\times 10^{-6}=2.712\Rightarrow v_a=1.647\text{km/s}.$ Angular-momentum check: $r_p{v}_p=7000\times 9.88=69160$; $r_a{v}_a=42000\times 1.647=69174$. Equal to rounding. ✔

Escape means "just barely reach infinity with zero leftover speed", i.e. the orbit is a parabola, whose size is infinite: $a\to \infty$, so $\frac{1}{a}\to 0$. $v_{esc}^2=GM\left(\frac{2}{r}-0\right)=\frac{2GM}{r}\Rightarrow v_{esc}=\sqrt{\frac{2GM}{r}}.$ See Escape Velocity. Note $v_{esc}=\sqrt{2}{v}_{circ}$ at the same $r$.

Rearrange vis-viva for $\frac{1}{a}$: $\frac{1}{a}=\frac{2}{r}-\frac{v^2}{GM}=\frac{2}{8000}-\frac{81}{398600}=2.5\times 10^{-4}-2.0321\times 10^{-4}=4.679\times 10^{-5}.$ $a=\frac{1}{4.679\times 10^{-5}}=21372\text{km}.$ $a>0$ and finite, so the orbit is a bound ellipse. (Its energy is $\epsilon =-GM/2a<0$.)

$\frac{1}{a}=\frac{2}{r}-\frac{v^2}{GM}=\frac{2}{10000}-\frac{144}{398600}=2.0\times 10^{-4}-3.6127\times 10^{-4}=-1.6127\times 10^{-4}.$ $a=\frac{1}{-1.6127\times 10^{-4}}=-6201\text{km}.$ $a<0$ ⇒ hyperbolic, unbound. The negative $a$ correctly forces $v^2>2GM/r$: here $v^2=144$ while $2GM/r=79.7$, so indeed $144>79.7$. Excess (hyperbolic) speed present.

This is the Hohmann Transfer Orbit logic — vis-viva used twice. Circular speed at $r_1$ ($a=r_1$): $v_{c1}=\sqrt{\frac{398600}{7000}}=\sqrt{56.943}=7.546\text{km/s}.$ Transfer orbit has $a_t=\frac{r_1+r_2}{2}=\frac{7000+42000}{2}=24500\text{km}$. Perigee speed on transfer ($r=r_1$): $v_p^2=398600\left(\frac{2}{7000}-\frac{1}{24500}\right)=97.61\Rightarrow v_p=9.880\text{km/s}.$ Departure burn: $\Delta v_1=v_p-v_{c1}=9.880-7.546=2.334\text{km/s}.$

Transfer apogee speed ($r=r_2$, $a_t=24500$): $v_a^2=398600\left(\frac{2}{42000}-\frac{1}{24500}\right)=2.712\Rightarrow v_a=1.647\text{km/s}.$ Target circular speed at $r_2$ ($a=r_2$): $v_{c2}=\sqrt{\frac{398600}{42000}}=\sqrt{9.490}=3.081\text{km/s}.$ Arrival burn (speed up to catch the faster circle): $\Delta v_2=v_{c2}-v_a=3.081-1.647=1.434\text{km/s}.$ Total: $\Delta v_{tot}=2.334+1.434=3.768\text{km/s}.$

Kepler's third law gives $T=2\pi \sqrt{a^3/GM}$: $T=2\pi \sqrt{\frac{24500^3}{398600}}=2\pi \sqrt{\frac{1.4706\times 10^{13}}{398600}}=2\pi \sqrt{3.690\times 10^7}$ $=2\pi \times 6074.5=38167\text{s}\approx 10.60\text{hr}.$ Transfer time (perigee → apogee) is half of that: $T/2\approx 19084\text{s}\approx 5.30\text{hr}.$

(a) $\epsilon =-\frac{GM}{2a}\Rightarrow a=-\frac{GM}{2\epsilon }=-\frac{398600}{2(-8.0)}=24913\text{km}.$ (b) Use $\epsilon =\frac{1}{2}{v}^2-\frac{GM}{r}$: $v^2=2\left(\epsilon +\frac{GM}{r}\right)=2\left(-8.0+\frac{398600}{9000}\right)=2(-8.0+44.289)=72.578\Rightarrow v=8.519\text{km/s}.$ (c) New speed at same $r$ from new energy: $v^{\prime 2}=2\left(-5.0+44.289\right)=78.578\Rightarrow v^{\prime }=8.864\text{km/s}.$ $\Delta v=v^{\prime }-v=8.864-8.519=0.345\text{km/s}\approx 345\text{m/s}.$

(a) Circular ($a=r=8000$): $v_c=\sqrt{\frac{398600}{8000}}=\sqrt{49.825}=7.059\text{km/s}.$ (b) Hyperbola, $a=-15000$, $r=8000$: $v_h^2=398600\left(\frac{2}{8000}-\frac{1}{-15000}\right)=398600\left(2.5\times 10^{-4}+6.667\times 10^{-5}\right)$ $=398600\times 3.1667\times 10^{-4}=126.22\Rightarrow v_h=11.235\text{km/s}.$ (c) $\Delta v=v_h-v_c=11.235-7.059=4.176\text{km/s}.$ (d) Escape speed here: $v_{esc}=\sqrt{2GM/r}=\sqrt{2\times 49.825}=\sqrt{99.65}=9.982\text{km/s}.$ Since $11.235>9.982$, the probe is unbound with hyperbolic excess speed $\sqrt{v_h^2-v_{esc}^2}=\sqrt{126.22-99.65}=\sqrt{26.57}=5.155\text{km/s}.$ ✔

Did the circular case reduce correctly? ::: $v=\sqrt{GM/r}$ when $a=r$. Did every bound orbit give a positive bracket $\frac{2}{r}-\frac{1}{a}$? ::: Yes — required for a real speed. Did the hyperbola give $a<0$ and $v>v_{esc}$? ::: Yes — the sign convention is doing its job. For each burn, was $r$ held fixed and only $v$ changed? ::: Yes — instantaneous burns are at a point.