Exercises — Conservation of energy and angular momentum in gravitational field

Gravity central hai ($\vec{F}\parallel \vec{r}$) aur conservative hai. Central $\Rightarrow$ torque zero $\Rightarrow$ $L$ constant. Conservative $\Rightarrow$ $E=KE+U$ constant. (Yaad karo $U=U(r)=-GMm/r$, upar define ki gayi potential energy.)

• (a) speed — changes (Sun ke paas tez, door dheema).
• (b) kinetic energy — changes (andar girane par badhti hai).
• (c) total energy $E$ — constant ✔
• (d) angular momentum $L$ — constant ✔
• (e) potential energy $U=-GMm/r$ — changes (paas hone par zyada negative). Answer: sirf (c) aur (d) constant hain.

Sign chart $E=-\frac{GMm}{2a}$ use karo:

• $E<0\Rightarrow a>0$ finite $\Rightarrow$ bound ellipse. Toh $E_1$: ellipse.
• $E=0\Rightarrow a\to \infty \Rightarrow$ parabola (bahut mushkil se escape karti hai, infinity par zero speed ke saath pahunchti hai). Toh $E_2$: parabola.
• $E>0\Rightarrow a<0\Rightarrow$ hyperbola (speed bachake escape karti hai). Toh $E_3$: hyperbola.

Circle ke liye, orbit ka size hi radius hai: $a=r$. Vis-viva mein plug in karo: $v^2=GM\left(\frac{2}{r}-\frac{1}{a}\right)=GM\left(\frac{2}{r}-\frac{1}{r}\right)=\frac{GM}{r}.$ Dono terms partially cancel ho jaati hain, ek clean circular result milta hai. $v=\sqrt{\frac{3.99\times 10^{14}}{7.00\times 10^6}}=\sqrt{5.70\times 10^7}\approx 7.55\times 10^3\text{m/s}=7.55\text{km/s}.$ Yeh familiar low-Earth-orbit speed hai. ✔

"Barely escape" ka matlab hai parabolic case $E=0$, yani $a\to \infty$ isliye vis-viva mein $1/a$ term gayab ho jaata hai. Radius $r$ par total energy zero set karo: $0=\frac{1}{2}m{v}_{\text{esc}}^2-\frac{GMm}{r}\Rightarrow v_{\text{esc}}=\sqrt{\frac{2GM}{r}}.$ $v_{\text{esc}}=\sqrt{\frac{2(3.99\times 10^{14})}{1.00\times 10^7}}=\sqrt{7.98\times 10^7}\approx 8.93\times 10^3\text{m/s}=8.93\text{km/s}.$

Abhi nikali escape speed $v_{\text{esc}}=8.93\text{km/s}$ se compare karo. Kyunki $9.00>8.93$, hamare paas $v>v_{\text{esc}}$ hai, isliye $E>0$. Yeh hyperbolic path par escape karega. Energy per kilogram quick check: $\epsilon =\frac{1}{2}{v}^2-\frac{GM}{r}=\frac{1}{2}(9000{)}^2-\frac{3.99\times 10^{14}}{10^7}=4.05\times 10^7-3.99\times 10^7=+6.0\times 10^5\text{J/kg}>0.$ Positive $\Rightarrow$ hyperbola. ✔

(a) Long axis $2a=r_p+r_a$ hai (figure s01 mein gray double-arrow dekho): $a=\frac{r_p+r_a}{2}=\frac{6.60\times 10^6+1.32\times 10^7}{2}=9.90\times 10^6\text{m}.$ (b) $r_p=a(1-e)$ se: $e=1-\frac{r_p}{a}=1-\frac{6.60\times 10^6}{9.90\times 10^6}=\mathrm{0.3333.}$ (c) Perihelion par, vis-viva $r=r_p$ ke saath use karo: $v_p^2=GM\left(\frac{2}{r_p}-\frac{1}{a}\right)=3.99\times 10^{14}\left(\frac{2}{6.60\times 10^6}-\frac{1}{9.90\times 10^6}\right).$ Andar: $\frac{2}{6.60\times 10^6}=3.030\times 10^{-7}$, $\frac{1}{9.90\times 10^6}=1.010\times 10^{-7}$, difference $=2.020\times 10^{-7}$. $v_p^2=3.99\times 10^{14}\times 2.020\times 10^{-7}=8.06\times 10^7\Rightarrow v_p=8.98\times 10^3\text{m/s}.$ (d) Aphelion par, $r=r_a$: $v_a^2=3.99\times 10^{14}\left(\frac{2}{1.32\times 10^7}-\frac{1}{9.90\times 10^6}\right).$ Andar: $\frac{2}{1.32\times 10^7}=1.515\times 10^{-7}$, minus $1.010\times 10^{-7}$ $=5.05\times 10^{-8}$. $v_a^2=3.99\times 10^{14}\times 5.05\times 10^{-8}=2.015\times 10^7\Rightarrow v_a=4.49\times 10^3\text{m/s}.$ (e) $r_p{v}_p=6.60\times 10^6\times 8.98\times 10^3=5.93\times 10^{10}$; $r_a{v}_a=1.32\times 10^7\times 4.49\times 10^3=5.93\times 10^{10}$. Equal ✔ — yeh $L/m$ conserved hai, kyunki turning points par $L=mrv$.

Minimum wahan hota hai jahan slope zero ho. Differentiate karke zero set karo — yehi kaaran hai ki hum derivative use karte hain: yeh tug-of-war curve ka flat spot dhundh nikalta hai. $\frac{d{U}_{\text{eff}}}{dr}=-\frac{L^2}{m{r}^3}+\frac{GMm}{r^2}=0\Rightarrow \frac{GMm}{r^2}=\frac{L^2}{m{r}^3}.$ Dono sides ko $r^3$ se multiply karo: $GMmr=\frac{L^2}{m}$, toh $\boxed{r_0=\frac{L^2}{GM{m}^2}}.$ Is radius par object ko kisi bhi direction mein koi radial push nahi milta, isliye $\dot{r}$ hamesha zero rehta hai: yeh circular orbit hai. Kue ke bottom par baithne ka matlab hai orbit stable hai — $r$ ko thoda nudge karo aur yeh $r_0$ ke aas paas oscillate karega. Woh chhoti radial wobble ek thodi eccentric ellipse (low $e$) correspond karti hai; yeh large disturbance ki poori stretched ellipse se alag hai — sirf ek chhota nudge kue ke bottom ke paas rehta hai. Figure s03 dekho.

(a) Energy size fix karti hai: $E=-\frac{GMm}{2a}$, toh $a=-\frac{GMm}{2E}=-\frac{(3.99\times 10^{14})(1000)}{2(-2.00\times 10^9)}=\frac{3.99\times 10^{17}}{4.00\times 10^9}=9.98\times 10^7\text{m}.$ (b) Turning point par $\dot{r}=0$, isliye poori speed tangential hai: $v=\frac{L}{mr}$. Ise energy equation mein daalo: $E=\frac{1}{2}m{\left(\frac{L}{mr}\right)}^2-\frac{GMm}{r}=\frac{L^2}{2m{r}^2}-\frac{GMm}{r}.$ $r^2$ se multiply karo aur $r$ mein quadratic ke roop mein rearrange karo: $E{r}^2+GMmr-\frac{L^2}{2m}=0.$ Numbers: $E=-2.00\times 10^9$; $GMm=3.99\times 10^{17}$; $\frac{L^2}{2m}=\frac{(4.00\times 10^{13}{)}^2}{2000}=\frac{1.60\times 10^{27}}{2000}=8.00\times 10^{23}$. $-2.00\times 10^9{r}^2+3.99\times 10^{17}r-8.00\times 10^{23}=0.$ $-2.00\times 10^9$ se divide karo: $r^2-1.995\times 10^{8}r+4.00\times 10^{14}=0.$ $r=\frac{1.995\times 10^8\pm \sqrt{(1.995\times 10^8{)}^2-4(4.00\times 10^{14})}}{2}.$ Discriminant: $(1.995\times 10^8{)}^2=3.980\times 10^{16}$; minus $1.600\times 10^{15}$ deta hai $3.820\times 10^{16}$; square root $=1.955\times 10^8$. $r=\frac{1.995\times 10^8\pm 1.955\times 10^8}{2}.$ Do roots do turning points hain: $r_a=\frac{3.950\times 10^8}{2}=1.975\times 10^8\text{m}$ (aphelion) aur $r_p=\frac{4.00\times 10^6}{2}=2.00\times 10^6\text{m}$ (perihelion). Chhota root perihelion hai: $r_p\approx 2.00\times 10^6\text{m}$. (Sanity check: $r_p+r_a=2.00\times 10^6+1.975\times 10^8=1.995\times 10^8=2a$ ✔.) (c) Perihelion speed: $v_p=\frac{L}{m{r}_p}=\frac{4.00\times 10^{13}}{1000\times 2.00\times 10^6}=2.00\times 10^4\text{m/s}=20.0\text{km/s}.$

Step 1 — $r_1$ par circular speed (initial): $a=r_1$, $v_{\text{circ}}=\sqrt{\frac{GM}{r_1}}=\sqrt{\frac{3.99\times 10^{14}}{7.00\times 10^6}}=7.55\times 10^3\text{m/s}.$ Step 2 — transfer-orbit size: transfer ellipse ka $a_t=\frac{r_1+r_2}{2}=\frac{7.00\times 10^6+4.22\times 10^7}{2}=2.46\times 10^7\text{m}.$ Step 3 — transfer ke perihelion $r_1$ par chahiye speed (vis-viva with $a=a_t$): $v_{\text{trans}}^2=GM\left(\frac{2}{r_1}-\frac{1}{a_t}\right)=3.99\times 10^{14}\left(\frac{2}{7.00\times 10^6}-\frac{1}{2.46\times 10^7}\right).$ Andar: $2.857\times 10^{-7}-4.065\times 10^{-8}=2.451\times 10^{-7}$. Toh $v_{\text{trans}}^2=9.78\times 10^7$, $v_{\text{trans}}=9.89\times 10^3\text{m/s}.$ Step 4 — boost: dono velocities $r_1$ par tangential hain, isliye $\Delta v=v_{\text{trans}}-v_{\text{circ}}=9.89\times 10^3-7.55\times 10^3=2.34\times 10^3\text{m/s}\approx 2.34\text{km/s}.$ Yeh exactly Hohmann transfer ka pehla burn hai.

Do expressions ek hi constant $E$ describe karte hain — toh inhe equal set karo, jisse hum $E$ ko poori tarah eliminate kar sakein aur sirf geometry rakhe: $\frac{1}{2}m{v}^2-\frac{GMm}{r}=-\frac{GMm}{2a}.$ Step 1 — kinetic term left par isolate karne ke liye dono sides mein $\frac{GMm}{r}$ add karo: $\frac{1}{2}m{v}^2=\frac{GMm}{r}-\frac{GMm}{2a}.$ Step 2 — har term mein factor $m$ hai; ise divide out karo (trajectory orbiting mass ki parwah nahi karti): $\frac{1}{2}{v}^2=\frac{GM}{r}-\frac{GM}{2a}=GM\left(\frac{1}{r}-\frac{1}{2a}\right).$ Step 3 — $v^2$ free karne ke liye dono sides ko $2$ se multiply karo: $v^2=2GM\left(\frac{1}{r}-\frac{1}{2a}\right)=GM\left(\frac{2}{r}-\frac{1}{a}\right).■$ Ek saas mein logic: energy conservation pehla equation kisi bhi point par deta hai; only-$a$ energy law doosra deta hai; inhe equate karne se invisible constant $E$ ki jagah visible $a$ aa jaata hai.

Perihelion aur aphelion par $\dot{r}=0$, isliye poori velocity tangential hai aur $v=\frac{L}{mr}$. Har turning point par energy: $E=\frac{L^2}{2m{r}^2}-\frac{GMm}{r}.$ Yeh $r_p$ aur $r_a$ dono par same $E$ dena chahiye (energy conserved hai), isliye yeh dono is equation ke roots hain: $E{r}^2+GMmr-\frac{L^2}{2m}=0(\text{upar ki line ko }{r}^2\text{ se multiply karo}).$ Quadratic $E{r}^2+GMmr-\frac{L^2}{2m}=0$ ke liye, roots ka sum hai $r_p+r_a=-\frac{GMm}{E}$ ($a{r}^2+br+c$ ke $-b/a$ se). Lekin geometrically $r_p+r_a=2a$. Equal karo: $2a=-\frac{GMm}{E}\Rightarrow \boxed{E=-\frac{GMm}{2a}}.$ Dhyaan do: $L$ sirf roots ke product mein aaya tha, aur $e$ sum mein kabhi aaya hi nahi. Dono cancel ho gaye — $E$ sirf $a$ par aur kisi cheez par nahi depend karta. Jo do orbits same $a$ lekin bilkul alag $e$ rakhte hain woh same energy share karte hain.

(a) Circular speed: $v_{\text{circ}}=\sqrt{GM/r}=7.55\times 10^3\text{m/s}$ (L2·Q1 se). Same radius par escape speed: $v_{\text{esc}}=\sqrt{2GM/r}$. Calculate karo: $2GM/r=2\cdot 3.99\times 10^{14}/7.00\times 10^6=1.140\times 10^8$, toh $v_{\text{esc}}=\sqrt{1.140\times 10^8}=1.068\times 10^4\text{m/s}.$ $\Delta v=v_{\text{esc}}-v_{\text{circ}}=1.068\times 10^4-7.55\times 10^3=3.13\times 10^3\text{m/s}\approx 3.13\text{km/s}.$ (b) Algebraically: $v_{\text{esc}}=\sqrt{\frac{2GM}{r}}=\sqrt{2}\cdot \sqrt{\frac{GM}{r}}=\sqrt{2}{v}_{\text{circ}}.$ Isliye $\Delta v=v_{\text{esc}}-v_{\text{circ}}=(\sqrt{2}-1)v_{\text{circ}}\approx 0.414v_{\text{circ}}.$ Check: $0.414\times 7.55\times 10^3=3.13\times 10^3\text{m/s}$ ✔. Toh circular orbit se escape karne mein hamesha tumhari existing speed se 41.4% zyada speed lagti hai — ek beautiful mass- aur planet-independent constant.

(a) $L=0$ hone par tangential motion bilkul nahi hoti — object seedha neeche Earth ke centre se guzarne wali line par girta hai. Yeh eccentricity $e\to 1$ ki degenerate ellipse hai: ek needle-thin "orbit" jo radial line par flat ho gayi. Iska turning point (aphelion) drop radius $r_0$ hai, aur far focus aur centre line ke ends se coincide karte hain. Semi-major axis fall distance ka aadha hai centre se $r_0$ tak: kyunki plunge $r_0$ se $r=0$ ki taraf jaata hai, $a=r_0/2=1.00\times 10^7\text{m}$. (b) Bhale hi $L=0$ ho, energy conservation (isliye vis-viva) tab bhi lagti hai — vis-viva ne kabhi $L$ use nahi kiya, usne $a$ use kiya. $a=1.00\times 10^7$ ke saath $r=7.00\times 10^6$ par apply karo: $v^2=GM\left(\frac{2}{r}-\frac{1}{a}\right)=3.99\times 10^{14}\left(\frac{2}{7.00\times 10^6}-\frac{1}{1.00\times 10^7}\right).$ Andar: $2.857\times 10^{-7}-1.000\times 10^{-7}=1.857\times 10^{-7}$. Toh $v^2=3.99\times 10^{14}\times 1.857\times 10^{-7}=7.41\times 10^7$, deta hai $v=8.61\times 10^3\text{m/s}\approx 8.61\text{km/s}.$ (Cross-check raw energy drop se: $\frac{1}{2}{v}^2=GM(1/r-1/r_0)=3.99\times 10^{14}(1.429\times 10^{-7}-5.00\times 10^{-8})=3.71\times 10^7$, toh $v^2=7.41\times 10^7$ ✔ — dono methods agree karte hain.)

Bina dekhe, kya tum kar sakte ho: (1) naam bata sako ki kaunsi do quantities conserved hain aur kyun; (2) vis-viva se circular aur escape speed calculate kar sako; (3) $(E,L)$ ko $(a,r_p,v_p)$ mein convert kar sako; (4) vis-viva aur $E=-GMm/2a$ derive kar sako; (5) mass-free escape ratio $(\sqrt{2}-1)$ bata sako; (6) $L=0$ radial-plunge limit handle kar sako? Agar koi bhi shaky lage, us level par wapas jao.