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

Sign ka matlab kya hai. $\epsilon$ ka sign poora "bound-or-free" faisla hai: $\epsilon <0\Rightarrow \text{ellipse (trapped)},\epsilon =0\Rightarrow \text{parabola},\epsilon >0\Rightarrow \text{hyperbola (escaping)}.$ Yahan $\epsilon >0$ hai, isliye orbit ek hyperbola hai — probe hamesha ke liye door chalata rahega, thodi speed bachegi. $a<0$ KYUN. $\epsilon =-GM/2a$ se $a=-GM/(2\epsilon )$ milta hai. $\epsilon >0$ hone par fraction negative hoga, isliye $a<0$. Negative $a$ ek unbound path ki formal pehchaan hai, ruler se measure karne layak kuch physical nahi.

Kisi ka nahi — dono equal hain. "Specific" ka matlab hai per unit mass: $\epsilon =E_{\text{total}}/m$. $m$ se divide karne par body ka apna mass cancel ho jaata hai, bilkul waisi tarah jaise saari cheezein same rate se girti hain. Dono ka $\epsilon =-GM/2a$ hai same $a$ ke saath, isliye same $\epsilon$.

Radius kya hai? Distance Earth ke centre se, surface se nahi: $r=R_{\oplus }+h=6.371\times 10^6+0.550\times 10^6=6.921\times 10^6\text{m}.$ $r=a$ KYUN yahan. Circle mein perigee = apogee = radius, isliye $2a=r_p+r_a=2r$, jisse $a=r$ milta hai. Vis-viva apply karo $a=r$ ke saath: $v^2=GM\left(\frac{2}{r}-\frac{1}{r}\right)=\frac{GM}{r}=\frac{3.986\times 10^{14}}{6.921\times 10^6}=5.759\times 10^7.$ $v=7.59\times 10^3\text{m/s}\approx 7.59km/s.$

Do distances (centre se) kya hain? $r_p=6.371\times 10^6+0.300\times 10^6=6.671\times 10^6\text{m},$ $r_a=6.371\times 10^6+3.000\times 10^6=9.371\times 10^6\text{m}.$ $r_p+r_a$ seedha use karne ki wajah. Parent ke Step 4 ne $\epsilon =-GM/(r_p+r_a)$ diya tha $a$ ka zikr karne se pehle: $r_p+r_a=1.6042\times 10^7\text{m},\epsilon =-\frac{3.986\times 10^{14}}{1.6042\times 10^7}=-2.485\times 10^7\text{J/kg}.$ Phir $a$ us span ka aadha hai: $a=(r_p+r_a)/2=8.021\times 10^6$ m. Check: $-GM/2a=-2.485\times 10^7$ ✓.

Vis-viva do baar KYUN. Yeh speed ko ek single point se link karta hai, isliye pehle $r_p$ par evaluate karo phir $r_a$ par. $v_p^2=GM\left(\frac{2}{r_p}-\frac{1}{a}\right)=3.986\times 10^{14}\left(\frac{2}{6.671\times 10^6}-\frac{1}{8.021\times 10^6}\right).$ $=3.986\times 10^{14}(2.9980\times 10^{-7}-1.2467\times 10^{-7})=3.986\times 10^{14}\times 1.7513\times 10^{-7}=6.982\times 10^7.$ $v_p=8.356\times 10^3\text{m/s}\approx 8.36km/s.$ Ab apogee: $v_a^2=3.986\times 10^{14}\left(\frac{2}{9.371\times 10^6}-\frac{1}{8.021\times 10^6}\right)=3.986\times 10^{14}(2.1342\times 10^{-7}-1.2467\times 10^{-7})$ $=3.986\times 10^{14}\times 8.875\times 10^{-8}=3.538\times 10^7,v_a=5.948\times 10^3\text{m/s}\approx 5.95km/s.$ Yeh kaisa dikhta hai. Paas hone par fast (perigee), door hone par slow (apogee) — energy height mein chhupi hui, bilkul parent ki swoop picture ki tarah. Conservation of angular momentum check karo: $r_p{v}_p=6.671\times 10^6\times 8.356\times 10^3=5.574\times 10^{10},$ $r_a{v}_a=9.371\times 10^6\times 5.948\times 10^3=5.574\times 10^{10}.$ Equal ✓ — $h$ conserved hai.

Speed. Circle ke liye $v=\sqrt{GM/r}$. $r$ double karne par: $\frac{v(2r)}{v(r)}=\sqrt{\frac{GM/2r}{GM/r}}=\sqrt{\frac{1}{2}}=\mathrm{0.707.}$ Upar wali orbit $1/\sqrt{2}\approx 0.707$ factor se slow hai. Energy. $\epsilon =-GM/2a$ jahan $a=r$ hai: $r$ double karne par $|\epsilon |$ aadha ho jaata hai, yaani $\epsilon$ kam negative ho jaata hai (zero ke paas, zyada loosely bound). Upar ki orbit = zyada total energy, phir bhi kam speed — extra energy potential energy mein gayi.

Transfer ka $a$ kya hai? Uska major axis dono circles ko span karta hai: $a_t=\frac{r_1+r_2}{2}=\frac{6.921\times 10^6+4.216\times 10^7}{2}=2.4541\times 10^7\text{m}.$ Circles par speeds (vis-viva jahan $a=r$): $v_{c1}=\sqrt{GM/r_1}=\sqrt{3.986\times 10^{14}/6.921\times 10^6}=7.589\times 10^3\text{m/s},$ $v_{c2}=\sqrt{GM/r_2}=\sqrt{3.986\times 10^{14}/4.216\times 10^7}=3.075\times 10^3\text{m/s}.$ Transfer ellipse par speeds uske dono ends par (vis-viva jahan $a=a_t$): $v_{t,p}^2=GM\left(\frac{2}{r_1}-\frac{1}{a_t}\right)=3.986\times 10^{14}(2.8898\times 10^{-7}-4.0748\times 10^{-8})=9.897\times 10^7,$ $v_{t,p}=9.948\times 10^3\text{m/s}.$ $v_{t,a}^2=GM\left(\frac{2}{r_2}-\frac{1}{a_t}\right)=3.986\times 10^{14}(4.7438\times 10^{-8}-4.0748\times 10^{-8})=2.667\times 10^6,$ $v_{t,a}=1.633\times 10^3\text{m/s}.$ Burns har end par speed ka gap hain: $\Delta v_1=v_{t,p}-v_{c1}=9948-7589=2359m/s,$ $\Delta v_2=v_{c2}-v_{t,a}=3075-1633=1442m/s,$ $\Delta v_{\text{total}}=2359+1442=3801m/s\approx 3.80\text{km/s}.$ Do burns KYUN. Pehla burn speed badhata hai taaki ellipse $r_2$ tak pahunche; doosra burn phir speed badhata hai (tum apogee par slow pahunchte ho) taaki circularise ho sake. Energy ki bhasha mein: har burn $\epsilon$ ko agle orbit ki value tak utha deta hai.

(a) Ek snapshot se $\epsilon$ — $\epsilon$ constant hai, isliye ek $(r,v)$ pair use fix kar deta hai: $\epsilon =\frac{v_1^2}{2}-\frac{GM}{r_1}=\frac{(8.5\times 10^3{)}^2}{2}-\frac{3.986\times 10^{14}}{7.0\times 10^6}=3.6125\times 10^7-5.6943\times 10^7=-2.082\times 10^7\text{J/kg}.$ (b) $\epsilon$ se $a$: $a=-\frac{GM}{2\epsilon }=-\frac{3.986\times 10^{14}}{2(-2.082\times 10^7)}=9.573\times 10^6\text{m}.$ (c) Type: $\epsilon <0$ (aur $a>0$) ⇒ ellipse, ek bound orbit. (d) Apogee $2a=r_p+r_a$ se: $r_a=2a-r_p=2(9.573\times 10^6)-6.6\times 10^6=1.2546\times 10^7\text{m}\approx 1.25\times 10^7\mathbf{m}.$

(a) Escape ke liye $\epsilon =0$ KYUN. Escape ka matlab hai "infinity tak coast karo jahan speed $\to 0$," yaani total energy exactly zero. $\epsilon =\frac{v^2}{2}-\frac{GM}{r}=0$ set karne par $v_{\text{esc}}=\sqrt{\frac{2GM}{r_1}}=\sqrt{\frac{2\times 3.986\times 10^{14}}{7.0\times 10^6}}=\sqrt{1.1389\times 10^8}=1.0672\times 10^4\text{m/s}\approx 10.67km/s.$ (b) Extra speed chahiye: $\Delta v=v_{\text{esc}}-v_1=10672-8500=2172m/s\approx 2.17\text{km/s}.$ Sanity check: $v_{\text{esc}}=\sqrt{2}{v}_{\text{circ}}$. Yahan circular speed $\sqrt{GM/r_1}=7546$ m/s hai aur $\sqrt{2}\times 7546=10672$ ✓.

Period sirf $a$ par KYUN depend karta hai. Kepler's Third Law kehta hai $T^2=\frac{4{\pi }^2}{GM}{a}^3$ — period sirf $a$ se fix hota hai, aur waise hi $\epsilon =-GM/2a$. Same $a$ ⇒ same energy ⇒ same period, eccentricity chahe kuch bhi ho. Compute karo: $T=2\pi \sqrt{\frac{a^3}{GM}}=2\pi \sqrt{\frac{(9.573\times 10^6{)}^3}{3.986\times 10^{14}}}=2\pi \sqrt{\frac{8.773\times 10^{20}}{3.986\times 10^{14}}}=2\pi \sqrt{2.2010\times 10^6}.$ $T=2\pi \times 1483.6=9322\text{s}\approx 155\mathbf{min}\approx 2.6\text{h}.$

1.1 hyperbola, $a<0$ ::: 1.2 equal ::: 2.1 $\approx 7.59$ km/s ::: 2.2 $\epsilon =-2.485\times 10^7$ J/kg, $a=8.02\times 10^6$ m 3.1 $v_p\approx 8.36$, $v_a\approx 5.95$ km/s ::: 3.2 speed $\times 0.707$, $|\epsilon |$ halved 4.1 $\Delta v_1\approx 2359$, $\Delta v_2\approx 1442$, total $\approx 3801$ m/s 5.1 $\epsilon =-2.08\times 10^7$, $a=9.57\times 10^6$ m, ellipse, $r_a=1.25\times 10^7$ m 5.2 $v_{\text{esc}}\approx 10.67$ km/s, $\Delta v\approx 2172$ m/s ::: 5.3 $T\approx 9322$ s ≈ 155 min