Specific orbital energy ε = −GM - 2a: Worked examples

3.2.11 · D3 · Physics › Orbital Mechanics & Astrodynamics › Specific orbital energy ε = −GM - 2a

Yeh page parent topic ka drill hall hai. Hum ek master formula aur Vis-viva equation lekar har tarah ki situation hunt karte hain jo exam mein aa sakti hai — $ε$ ke har sign ke saath, degenerate circle, escape edge, unbound hyperbola, ek real mission problem, aur ek exam-style trap. Kuch bhi skip nahi kiya, taaki koi bhi case aisa na ho jo tune pehle practice na kiya ho.

Neeche jo bhi hai wo sirf do tools par tikaa hai jo tere paas pehle se hain:

The scenario matrix

Kuch bhi calculate karne se pehle, chalte hain har cell lay out karte hain jo yeh topic hit kar sakta hai. Baad ke har example mein uska cell(s) tag kiya gaya hai.

#	Case class	Kya special hai isme	Filled by
C1	$ε < 0$ , circle ( $r_{p} = r_{a}$ )	degenerate ellipse, $r = a$	Ex 1
C2	$ε < 0$ , general ellipse	$r_{p} \neq = r_{a}$ , speed differs by point	Ex 2, Ex 3
C3	$ε = 0$ , parabola	escape edge, $a \to \infty$	Ex 4
C4	$ε > 0$ , hyperbola	unbound, negative $a$	Ex 5
C5	Limiting behaviour	$a \to \infty$ , $r \to \infty$ , sign flips	Ex 4, Ex 5
C6	Real-world word problem	ek Hohmann transfer orbit burn	Ex 6
C7	Exam-style twist	given $v, r$ → orbit classify karo	Ex 7
C8	Inverse / consistency check	given $ε$ → $a$ recover karo, Kepler's Third Law se cross-check	Ex 8

$ε$ ke teen signs (rows C1–C4) hi backbone hain. Poora matrix ek picture mein — padhte waqt ise khula rakhna.

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

Example 1 — Circular LEO (cell C1)

Worked example 400 km altitude par circular orbit

Ek satellite Earth ke around $r = a = 6.78 \times 1 0^{6}$ m radius par circle karti hai. $ε$ aur $v$ nikalo.

Forecast: compute karne se pehle andaza lagao — kya $ε$ negative hoga? Kya $v$ kareeb $7.7$ km/s hogi? Apna guess likh lo.

Step 1. Circle recognize karo ⇒ $r_{p} = r_{a} = r$ , isliye $2 a = r_{p} + r_{a} = 2 r$ , jisse $a = r$ milta hai. Yeh step kyun? Circle ek degenerate ellipse hai jahan dono extreme distances ek hi hoti hain; sirf tabhi "radius = $a$ " hold karta hai.

Step 2. Energy: $ε = - \frac{GM}{2 a} = - \frac{3.986 \times 1 0 ^{14}}{2 ( 6.78 \times 1 0 ^{6} )} = - 2.94 \times 1 0^{7} J/kg$ . Yeh step kyun? Seedha substitution — formula ko sirf $a$ ki geometry chahiye.

Step 3. Vis-viva se speed, $r = a$ ke saath: $v^{2} = GM (\frac{2}{r} - \frac{1}{r}) = \frac{GM}{r}$ . Yeh step kyun? Dono terms partly cancel ho jaati hain, aur sirf circles ke liye famous $v = GM / r$ milta hai.

Step 4. $v = 3.986 \times 1 0^{14} /6.78 \times 1 0^{6} = 7.67 \times 1 0^{3}$ m/s.

Verify: units $(m^{3} / s^{2}) / m = m^{2} / s^{2} = m/s$ ✔. Aur $7.67$ km/s textbook LEO speed hai ✔. Yeh bhi: $ε = \frac{v ^{2}}{2} - \frac{GM}{r} = 2.94 \times 1 0^{7} - 5.88 \times 1 0^{7} = - 2.94 \times 1 0^{7}$ ✔ Step 2 se match karta hai.

Example 2 — General ellipse: extremes se energy (cell C2)

Worked example GEO transfer orbit energy

Ek orbit ka perigee $r_{p} = 6.78 \times 1 0^{6}$ m aur apogee $r_{a} = 4.22 \times 1 0^{7}$ m hai. $ε$ aur $a$ nikalo.

Forecast: kya yahan $ε$ , Ex 1 ke LEO circle se zyada negative hoga ya kam negative? (Bada loop ⇒ zero ke paas — change ka sign guess karo.)

Step 1. $ε = - \frac{GM}{r _{p} + r _{a}}$ use karo (parent ka Step-5 shortcut, kyunki $r_{p} + r_{a} = 2 a$ ). Yeh step kyun? Jab dono extremes mile hon, yeh pehle $a$ calculate karne ka step skip kar deta hai.

Step 2. $r_{p} + r_{a} = 4.898 \times 1 0^{7}$ m, isliye $ε = - \frac{3.986 \times 1 0 ^{14}}{4.898 \times 1 0 ^{7}} = - 8.14 \times 1 0^{6} J/kg$ .

Step 3. $a$ recover karo: $a = \frac{r _{p} + r _{a}}{2} = 2.449 \times 1 0^{7}$ m. Yeh step kyun? Confirm karta hai ki semi-major axis extremes ka average hai, na koi ek wali.

Verify: $- \frac{GM}{2 a} = - \frac{3.986 \times 1 0 ^{14}}{2 ( 2.449 \times 1 0 ^{7} )} = - 8.14 \times 1 0^{6}$ ✔ Step 2 jaisa hi. Aur $- 8.14 \times 1 0^{6}$ Ex 1 ke $- 2.94 \times 1 0^{7}$ se kam negative hai — bada loop, richer account, jaise forecast kiya tha ✔.

Example 3 — Us ellipse mein apogee vs perigee par speed (cell C2)

Worked example Satellite sabse fast kahan hoti hai?

Same orbit jaise Ex 2 ( $a = 2.449 \times 1 0^{7}$ m). Perigee par $v_{p}$ aur apogee par $v_{a}$ nikalo.

Forecast: $v_{p}$ aur $v_{a}$ mein se kaunsa bada hai? Energy bank ke hisaab se: paas = fast. Ratio bhi guess karo.

Step 1. Perigee: $v_{p}^{2} = GM (\frac{2}{r _{p}} - \frac{1}{a}) = 3.986 \times 1 0^{14} (\frac{2}{6.78 \times 1 0 ^{6}} - \frac{1}{2.449 \times 1 0 ^{7}})$ . Yeh step kyun? Vis-viva hi woh ek tool hai jo speed ko single point $r$ se jodta hai.

Step 2. $\frac{2}{r _{p}} = 2.950 \times 1 0^{- 7}$ , $\frac{1}{a} = 4.083 \times 1 0^{- 8}$ ; difference $= 2.542 \times 1 0^{- 7}$ . Isliye $v_{p}^{2} = 1.013 \times 1 0^{8}$ , $v_{p} = 1.007 \times 1 0^{4}$ m/s $\approx 10.1$ km/s.

Step 3. Apogee: $v_{a}^{2} = GM (\frac{2}{r _{a}} - \frac{1}{a}) = 3.986 \times 1 0^{14} (4.739 \times 1 0^{- 8} - 4.083 \times 1 0^{- 8}) = 2.614 \times 1 0^{6}$ , isliye $v_{a} = 1.617 \times 1 0^{3}$ m/s $\approx 1.6$ km/s. Yeh step kyun? Same formula, naya $r$ — apogee door hai, isliye $2/ r$ term chhoti ho jaati hai, speed gir jaati hai.

Verify (Conservation of angular momentum): dono ends par $v ⊥ r$ hai, isliye $r_{p} v_{p}$ ko $r_{a} v_{a}$ ke barabar hona chahiye. $r_{p} v_{p} = 6.78 \times 1 0^{6} \times 1.007 \times 1 0^{4} = 6.83 \times 1 0^{10}$ ; $r_{a} v_{a} = 4.22 \times 1 0^{7} \times 1.617 \times 1 0^{3} = 6.82 \times 1 0^{10}$ ✔ (rounding tak equal). Perigee ~6× fast hai, jaise forecast kiya tha.

Example 4 — Escape edge: parabola (cells C3, C5)

Worked example Earth ko barely chhod ke jaana

$r = 6.78 \times 1 0^{6}$ m se, kis speed par $ε = 0$ hoga (parabolic escape)? $a$ kya hoga?

Forecast: kya escape speed, Ex 1 ki circular speed se badi hai ya chhoti? Kitna fark hoga (ek factor guess karo)?

Step 1. $ε = \frac{v ^{2}}{2} - \frac{GM}{r}$ mein $ε = 0$ set karo: $v_{esc} = \frac{2 GM}{r}$ . Yeh step kyun? Zero total energy bound aur unbound ke beech ki boundary hai — yahi Escape velocity condition hai.

Step 2. $v_{esc} = \frac{2 ( 3.986 \times 1 0 ^{14} )}{6.78 \times 1 0 ^{6}} = 1.085 \times 1 0^{4}$ m/s $\approx 10.8$ km/s.

Step 3. Semi-major axis: $a = - \frac{GM}{2 ε} \to - \frac{GM}{0} = \infty$ . Yeh step kyun? $ε \to 0^{-}$ se $a \to + \infty$ ho jaata hai: ellipse itna stretch ho jaata hai ki kabhi band nahi hota — yeh parabola hai. Yahi limiting behaviour hai (cell C5).

Verify: $v_{esc} / v_{circ} = 1.085 \times 1 0^{4} /7.67 \times 1 0^{3} = 1.414 \approx 2$ ✔ — classic result: same radius par escape speed, circular speed ka $2$ guna hoti hai, jaise forecast kiya tha.

Example 5 — Unbound hyperbola: negative $a$ (cells C4, C5)

Worked example Ek probe jo bahut zyada energy ke saath bahar pheki gayi

$r = 6.78 \times 1 0^{6}$ m par ek probe $v = 1.20 \times 1 0^{4}$ m/s se chal rahi hai. Dikhao ki yeh escape karti hai, $ε$ aur $a$ nikalo, aur door jaane par speed ( $r \to \infty$ ) bhi nikalo.

Forecast: kyunki $v > v_{esc}$ (Ex 4 se), compute karne se pehle $ε$ ka sign aur $a$ ka sign predict karo.

Step 1. $ε = \frac{v ^{2}}{2} - \frac{GM}{r} = \frac{( 1.20 \times 1 0 ^{4} ) ^{2}}{2} - \frac{3.986 \times 1 0 ^{14}}{6.78 \times 1 0 ^{6}}$ . Yeh step kyun? Jab kisi point par speed diya ho, geometry nahi, tabhi $ε$ ka $v, r$ wala form kaam karta hai.

Step 2. $= 7.20 \times 1 0^{7} - 5.879 \times 1 0^{7} = 1.321 \times 1 0^{7} J/kg > 0$ . Yeh step kyun? Positive $ε$ ⇒ unbound (hyperbola) — probe hamesha ke liye chali jaayegi.

Step 3. $a = - \frac{GM}{2 ε} = - \frac{3.986 \times 1 0 ^{14}}{2 ( 1.321 \times 1 0 ^{7} )} = - 1.508 \times 1 0^{7}$ m — negative, hyperbola ki formal signature.

Step 4. Infinity par speed: jab $r \to \infty$ , $v^{2} = GM (\frac{2}{r} - \frac{1}{a}) \to - \frac{GM}{a} = 2 ε$ , isliye $v_{\infty} = 2 ε = 2 (1.321 \times 1 0^{7}) = 5.14 \times 1 0^{3}$ m/s. Yeh step kyun? Infinity par PE $\to 0$ ho jaata hai, isliye sirf leftover kinetic energy $ε$ bachti hai — yahi limiting behaviour hai (cell C5).

Verify: $\frac{1}{2} v_{\infty}^{2} = 0.5 (5.14 \times 1 0^{3})^{2} = 1.321 \times 1 0^{7} = ε$ ✔ (infinity par saari energy kinetic hai). Signs forecast se match karte hain: $ε > 0$ , $a < 0$ ✔.

Example 6 — Word problem: ek Hohmann transfer burn (cell C6)

Worked example LEO se GEO tak raise karna — perigee kick

Ek Hohmann transfer orbit ek circular LEO $r_{1} = 6.78 \times 1 0^{6}$ m se start hoti hai aur ek transfer ellipse par jaana chahiye jiska apogee GEO $r_{2} = 4.22 \times 1 0^{7}$ m par touch kare. Perigee burn par kitna speed change $Δ v$ chahiye?

Forecast: burn apogee raise karega, isliye $Δ v > 0$ hoga. Guess karo — yeh $1$ , $2$ , ya $3$ km/s ke paas hoga?

Step 1. Circular LEO speed: $v_{circ} = GM / r_{1} = 7.67 \times 1 0^{3}$ m/s (Ex 1 se). Yeh step kyun? Yeh burn se pehle ki speed hai.

Step 2. Transfer ellipse ka $a = (r_{1} + r_{2}) /2 = 2.449 \times 1 0^{7}$ m hai (same orbit jaise Ex 2). Iske perigee par speed $v_{p} = 1.007 \times 1 0^{4}$ m/s hai (Ex 3 se). Yeh step kyun? Burn same point $r_{1}$ par speed change karta hai, satellite ko circle se ellipse par le jaata hai — same $r$ , different $a$ .

Step 3. $Δ v = v_{p} - v_{circ} = 1.007 \times 1 0^{4} - 7.67 \times 1 0^{3} = 2.40 \times 1 0^{3}$ m/s $\approx 2.4$ km/s.

Verify: direction check — humne perigee par speed badhaayi, isliye energy $- 2.94 \times 1 0^{7}$ (Ex 1) se $- 8.14 \times 1 0^{6}$ (Ex 2) tak badhi, yaani kam negative = bada orbit ✔. $Δ v \approx 2.4$ km/s forecast band mein hai ✔.

Example 7 — Exam twist: raw $v, r$ se classify karo (cell C7)

Worked example "Kya yeh orbit bound hai?"

Ek body $r = 1.00 \times 1 0^{7}$ m par measure ki gayi hai, speed $v = 9.00 \times 1 0^{3}$ m/s. Orbit classify karo aur $a$ nikalo.

Forecast: pehle $v$ ko local escape speed $2 GM / r$ se compare karo — bound hai ya nahi?

Step 1. Local escape speed: $2 (3.986 \times 1 0^{14}) /1 0^{7} = 8.929 \times 1 0^{3}$ m/s. Yeh step kyun? Yahi yardstick hai: neeche ⇒ ellipse, barabar ⇒ parabola, upar ⇒ hyperbola.

Step 2. $v = 9.00 \times 1 0^{3} > 8.929 \times 1 0^{3}$ , isliye hyperbolic (unbound) — lekin barely. Yeh step kyun? Margin bahut chhota hai; $ε$ ka sign decide karta hai, isliye exactly compute karo.

Step 3. $ε = \frac{1}{2} (9.00 \times 1 0^{3})^{2} - \frac{3.986 \times 1 0 ^{14}}{1 0 ^{7}} = 4.050 \times 1 0^{7} - 3.986 \times 1 0^{7} = 6.40 \times 1 0^{5} J/kg > 0$ ✔ hyperbola.

Step 4. $a = - \frac{GM}{2 ε} = - \frac{3.986 \times 1 0 ^{14}}{2 ( 6.40 \times 1 0 ^{5} )} = - 3.11 \times 1 0^{8}$ m.

Verify: $ε > 0$ ↔ $a < 0$ ✔ (dono ko agree karna chahiye). Forecast (barely unbound) sahi nikla — $71$ m/s ki kami ise wapas bound ellipse bana deti.

Example 8 — Inverse + Kepler's Third Law se consistency (cell C8)

Worked example Energy se period tak

Ek orbit ka $ε = - 8.14 \times 1 0^{6}$ J/kg hai (Ex 2 wali ellipse). $a$ recover karo, phir orbital period $T$ nikalo aur Kepler's Third Law se cross-check karo.

Forecast: kyunki $a \approx 2.45 \times 1 0^{7}$ m hai (GEO se chhota hai), guess karo — $T$ ek din se zyada hoga ya kam?

Step 1. Energy formula invert karo: $a = - \frac{GM}{2 ε} = - \frac{3.986 \times 1 0 ^{14}}{2 ( - 8.14 \times 1 0 ^{6} )} = 2.449 \times 1 0^{7}$ m ✔ Ex 2 se match karta hai. Yeh step kyun? Confirm karta hai ki energy↔geometry link ek true two-way street hai.

Step 2. Kepler's Third Law: $T = 2 π \frac{a ^{3}}{GM}$ . Yeh step kyun? Period sirf $a$ par depend karta hai (isliye sirf $ε$ par) — wahi " $a$ decides everything" theme.

Step 3. $a^{3} = (2.449 \times 1 0^{7})^{3} = 1.469 \times 1 0^{22}$ ; $a^{3} / GM = 3.685 \times 1 0^{7}$ ; $= 6.070 \times 1 0^{3}$ ; $T = 2 π (6.070 \times 1 0^{3}) = 3.814 \times 1 0^{4}$ s.

Verify: $T \approx 3.81 \times 1 0^{4}$ s $= 10.6$ hours — ek din se kam, jaise forecast kiya tha (is GTO ka $a$ , GEO ke $4.22 \times 1 0^{7}$ m se chhota hai). Consistent chain: $ε \to a \to T$ sab agree karte hain ✔.

Recall Kaun sa cell kaun sa tha?

Ek nazar mein every-scenario coverage. Circle jahan $r = a$ — kaun sa example? ::: Ex 1 (cell C1) $r_{p} + r_{a}$ se general ellipse energy? ::: Ex 2 (cell C2) Ellipse ke around speed changes (angular-momentum check)? ::: Ex 3 (cell C2) $ε = 0$ escape / parabola case? ::: Ex 4 (cells C3, C5) Positive $ε$ , negative $a$ , hyperbola? ::: Ex 5 (cells C4, C5) Hohmann burn ke liye $Δ v$ ? ::: Ex 6 (cell C6) Raw $v, r$ se orbit classify karo? ::: Ex 7 (cell C7) Energy → $a$ → period cross-check? ::: Ex 8 (cell C8)