Specific orbital energy ε = −GM - 2a: Exercises

This page is a self-test ladder for the parent topic. Every problem hides its answer inside a collapsible solution — read the problem, try it, then open the solution.

Throughout, we use only tools built in the parent note:

$ε = \frac{v ^{2}}{2} - \frac{GM}{r}$ — the energy per kilogram of an orbit ( $v$ = speed, $r$ = distance to the central mass's centre).
$ε = - \frac{GM}{2 a}$ — that same number written using only $a$ , the semi-major axis (half the long width of the ellipse).
$v^{2} = GM (\frac{2}{r} - \frac{1}{a})$ — the Vis-viva equation, speed anywhere from $r$ and $a$ .
$2 a = r_{p} + r_{a}$ — the major axis spans perigee-to-apogee (see figure).

Figure — Specific orbital energy ε = −GM - 2a

The figure names every symbol you will use below: the two extreme points ( $r_{p}$ perigee, $r_{a}$ apogee), the centre $C$ , the focus $F$ (where Earth sits), and the fact that $r_{p} + r_{a} = 2 a$ .

L1 — Recognition

Recall Solution 1.1

WHAT the sign means. The sign of $ε$ is the entire "bound-or-free" verdict: $ε < 0 \Rightarrow ellipse (trapped), ε = 0 \Rightarrow parabola, ε > 0 \Rightarrow hyperbola (escaping) .$ Here $ε > 0$ , so the orbit is a hyperbola — the probe will coast away forever with leftover speed. WHY $a < 0$ . From $ε = - GM /2 a$ we get $a = - GM / (2 ε)$ . With $ε > 0$ the fraction is negative, so $a < 0$ . Negative $a$ is the formal signature of an unbound path, nothing physical to measure with a ruler.

Recall Solution 1.2

Neither — they are equal. The word "specific" means per unit mass: $ε = E_{total} / m$ . Dividing out $m$ makes the body's own mass cancel, exactly like all objects falling at the same rate. Both have $ε = - GM /2 a$ with the same $a$ , so the same $ε$ .

L2 — Application

Recall Solution 2.1

WHAT is the radius? Distance from Earth's centre, not the surface: $r = R_{\oplus} + h = 6.371 \times 1 0^{6} + 0.550 \times 1 0^{6} = 6.921 \times 1 0^{6} m .$ WHY $r = a$ here. In a circle perigee = apogee = radius, so $2 a = r_{p} + r_{a} = 2 r$ , giving $a = r$ . Apply vis-viva with $a = r$ : $v^{2} = GM (\frac{2}{r} - \frac{1}{r}) = \frac{GM}{r} = \frac{3.986 \times 1 0 ^{14}}{6.921 \times 1 0 ^{6}} = 5.759 \times 1 0^{7} .$ $v = 7.59 \times 1 0^{3} m/s \approx 7.59 km/s .$

Recall Solution 2.2

WHAT are the two distances (from centre)? $r_{p} = 6.371 \times 1 0^{6} + 0.300 \times 1 0^{6} = 6.671 \times 1 0^{6} m,$ $r_{a} = 6.371 \times 1 0^{6} + 3.000 \times 1 0^{6} = 9.371 \times 1 0^{6} m .$ WHY use $r_{p} + r_{a}$ directly. The parent's Step 4 gave $ε = - GM / (r_{p} + r_{a})$ before ever mentioning $a$ : $r_{p} + r_{a} = 1.6042 \times 1 0^{7} m, ε = - \frac{3.986 \times 1 0 ^{14}}{1.6042 \times 1 0 ^{7}} = - 2.485 \times 1 0^{7} J/kg .$ Then $a$ is half of that span: $a = (r_{p} + r_{a}) /2 = 8.021 \times 1 0^{6}$ m. Check: $- GM /2 a = - 2.485 \times 1 0^{7}$ ✓.

L3 — Analysis

Recall Solution 3.1

WHY vis-viva twice. It links speed to a single point, so evaluate it at $r_{p}$ then $r_{a}$ . $v_{p}^{2} = GM (\frac{2}{r _{p}} - \frac{1}{a}) = 3.986 \times 1 0^{14} (\frac{2}{6.671 \times 1 0 ^{6}} - \frac{1}{8.021 \times 1 0 ^{6}}) .$ $= 3.986 \times 1 0^{14} (2.9980 \times 1 0^{- 7} - 1.2467 \times 1 0^{- 7}) = 3.986 \times 1 0^{14} \times 1.7513 \times 1 0^{- 7} = 6.982 \times 1 0^{7} .$ $v_{p} = 8.356 \times 1 0^{3} m/s \approx 8.36 km/s .$ Now apogee: $v_{a}^{2} = 3.986 \times 1 0^{14} (\frac{2}{9.371 \times 1 0 ^{6}} - \frac{1}{8.021 \times 1 0 ^{6}}) = 3.986 \times 1 0^{14} (2.1342 \times 1 0^{- 7} - 1.2467 \times 1 0^{- 7})$ $= 3.986 \times 1 0^{14} \times 8.875 \times 1 0^{- 8} = 3.538 \times 1 0^{7}, v_{a} = 5.948 \times 1 0^{3} m/s \approx 5.95 km/s .$ WHAT IT LOOKS LIKE. Fast when close (perigee), slow when far (apogee) — energy hiding in height, exactly the parent's swoop picture. Check Conservation of angular momentum: $r_{p} v_{p} = 6.671 \times 1 0^{6} \times 8.356 \times 1 0^{3} = 5.574 \times 1 0^{10},$ $r_{a} v_{a} = 9.371 \times 1 0^{6} \times 5.948 \times 1 0^{3} = 5.574 \times 1 0^{10} .$ Equal ✓ — $h$ is conserved.

Recall Solution 3.2

Speed. For a circle $v = GM / r$ . Doubling $r$ : $\frac{v ( 2 r )}{v ( r )} = \frac{GM /2 r}{GM / r} = \frac{1}{2} = 0.707.$ The higher orbit is slower by a factor $1/ 2 \approx 0.707$ . Energy. $ε = - GM /2 a$ with $a = r$ : doubling $r$ halves $∣ ε ∣$ , i.e. $ε$ becomes less negative (closer to zero, more loosely bound). Higher orbit = more total energy, yet less speed — the extra energy went into potential energy.

L4 — Synthesis

Recall Solution 4.1

WHAT is the transfer's $a$ ? Its major axis spans both circles: $a_{t} = \frac{r _{1} + r _{2}}{2} = \frac{6.921 \times 1 0 ^{6} + 4.216 \times 1 0 ^{7}}{2} = 2.4541 \times 1 0^{7} m .$ Speeds on the circles (vis-viva with $a = r$ ): $v_{c 1} = GM / r_{1} = 3.986 \times 1 0^{14} /6.921 \times 1 0^{6} = 7.589 \times 1 0^{3} m/s,$ $v_{c 2} = GM / r_{2} = 3.986 \times 1 0^{14} /4.216 \times 1 0^{7} = 3.075 \times 1 0^{3} m/s .$ Speeds on the transfer ellipse at its two ends (vis-viva with $a = a_{t}$ ): $v_{t, p}^{2} = GM (\frac{2}{r _{1}} - \frac{1}{a _{t}}) = 3.986 \times 1 0^{14} (2.8898 \times 1 0^{- 7} - 4.0748 \times 1 0^{- 8}) = 9.897 \times 1 0^{7},$ $v_{t, p} = 9.948 \times 1 0^{3} m/s .$ $v_{t, a}^{2} = GM (\frac{2}{r _{2}} - \frac{1}{a _{t}}) = 3.986 \times 1 0^{14} (4.7438 \times 1 0^{- 8} - 4.0748 \times 1 0^{- 8}) = 2.667 \times 1 0^{6},$ $v_{t, a} = 1.633 \times 1 0^{3} m/s .$ The burns are the speed gaps at each end: $Δ v_{1} = v_{t, p} - v_{c 1} = 9948 - 7589 = 2359 m/s,$ $Δ v_{2} = v_{c 2} - v_{t, a} = 3075 - 1633 = 1442 m/s,$ $Δ v_{total} = 2359 + 1442 = 3801 m/s \approx 3.80 km/s .$ WHY two burns. The first speeds you up so the ellipse reaches $r_{2}$ ; the second speeds you up again (you arrive slow at apogee) to circularise. Energy language: each burn raises $ε$ to the next orbit's value.

L5 — Mastery

Recall Solution 5.1

(a) $ε$ from the one snapshot — $ε$ is constant, so one $(r, v)$ pair fixes it: $ε = \frac{v _{1}^{2}}{2} - \frac{GM}{r _{1}} = \frac{( 8.5 \times 1 0 ^{3} ) ^{2}}{2} - \frac{3.986 \times 1 0 ^{14}}{7.0 \times 1 0 ^{6}} = 3.6125 \times 1 0^{7} - 5.6943 \times 1 0^{7} = - 2.082 \times 1 0^{7} J/kg .$ (b) $a$ from $ε$ : $a = - \frac{GM}{2 ε} = - \frac{3.986 \times 1 0 ^{14}}{2 ( - 2.082 \times 1 0 ^{7} )} = 9.573 \times 1 0^{6} m .$ (c) Type: $ε < 0$ (and $a > 0$ ) ⇒ ellipse, a bound orbit. (d) Apogee from $2 a = r_{p} + r_{a}$ : $r_{a} = 2 a - r_{p} = 2 (9.573 \times 1 0^{6}) - 6.6 \times 1 0^{6} = 1.2546 \times 1 0^{7} m \approx 1.25 \times 1 0^{7} m .$

Recall Solution 5.2

(a) WHY $ε = 0$ for escape. Escape means "coast to infinity with speed $\to 0$ ," i.e. total energy exactly zero. Setting $ε = \frac{v ^{2}}{2} - \frac{GM}{r} = 0$ gives $v_{esc} = \frac{2 GM}{r _{1}} = \frac{2 \times 3.986 \times 1 0 ^{14}}{7.0 \times 1 0 ^{6}} = 1.1389 \times 1 0^{8} = 1.0672 \times 1 0^{4} m/s \approx 10.67 km/s .$ (b) Extra speed needed: $Δ v = v_{esc} - v_{1} = 10672 - 8500 = 2172 m/s \approx 2.17 km/s .$ Sanity: $v_{esc} = 2 v_{circ}$ . Circular speed here is $GM / r_{1} = 7546$ m/s and $2 \times 7546 = 10672$ ✓.

Recall Solution 5.3

WHY period depends only on $a$ . Kepler's Third Law says $T^{2} = \frac{4 π ^{2}}{GM} a^{3}$ — period is fixed by $a$ alone, and so is $ε = - GM /2 a$ . Same $a$ ⇒ same energy ⇒ same period, regardless of eccentricity. Compute: $T = 2 π \frac{a ^{3}}{GM} = 2 π \frac{( 9.573 \times 1 0 ^{6} ) ^{3}}{3.986 \times 1 0 ^{14}} = 2 π \frac{8.773 \times 1 0 ^{20}}{3.986 \times 1 0 ^{14}} = 2 π 2.2010 \times 1 0^{6} .$ $T = 2 π \times 1483.6 = 9322 s \approx 155 min \approx 2.6 h .$

Recall Quick self-check ledger (open only to grade yourself)

1.1 hyperbola, $a < 0$ ::: 1.2 equal ::: 2.1 $\approx 7.59$ km/s ::: 2.2 $ε = - 2.485 \times 1 0^{7}$ J/kg, $a = 8.02 \times 1 0^{6}$ m 3.1 $v_{p} \approx 8.36$ , $v_{a} \approx 5.95$ km/s ::: 3.2 speed $\times 0.707$ , $∣ ε ∣$ halved 4.1 $Δ v_{1} \approx 2359$ , $Δ v_{2} \approx 1442$ , total $\approx 3801$ m/s 5.1 $ε = - 2.08 \times 1 0^{7}$ , $a = 9.57 \times 1 0^{6}$ m, ellipse, $r_{a} = 1.25 \times 1 0^{7}$ m 5.2 $v_{esc} \approx 10.67$ km/s, $Δ v \approx 2172$ m/s ::: 5.3 $T \approx 9322$ s ≈ 155 min