Exercises — Gravitational field intensity g = GM - r²

(a) $g$ wo gravitational force per unit mass hai jo ek choti test mass ko us point par feel hogi — yeh location ki property hai, visitor ki nahi. (b) Test mass par force $F=GMm/r^2$ hai, jo $m$ ke proportional hai. Field hai $g=F/m$, toh $m$ se divide karne par woh cancel ho jaata hai: $g=GM/r^2$. Jo bachta hai woh sirf source $M$ aur distance $r$ par depend karta hai.

$r=R_{\text{moon}}=1.74\times 10^6\text{m}$ use karo (Moon ke centre se distance): $g=\frac{G{M}_{\text{moon}}}{R_{\text{moon}}^2}=\frac{(6.67\times 10^{-11})(7.35\times 10^{22})}{(1.74\times 10^6{)}^2}$ Numerator: $6.67\times 7.35=49.0$, toh $49.0\times 10^{11}=4.90\times 10^{12}$. Denominator: $(1.74{)}^2=3.03$, toh $3.03\times 10^{12}$. $g=\frac{4.90\times 10^{12}}{3.03\times 10^{12}}\approx 1.62{\text{N kg}}^{-1}$ Yeh Earth ke $9.8$ ka roughly $\frac{1}{6}$ hai — isliye astronauts bounce karte hain.

Definition $\vec{g}=\vec{F}/m$ se, rearrange karo: $F=mg$.

• $2\text{kg}$: $F=2\times 9.8=19.6\text{N}$.
• $50\text{kg}$: $F=50\times 9.8=490\text{N}$. Field $g$ dono ke liye same $9.8$ hai — force sirf visitor ki mass ke saath scale karta hai.

Altitude $h=2R_{\oplus }$ matlab centre se distance $r=R_{\oplus }+h=3R_{\oplus }$. $g_0$ ko Earth ki surface gravity maano — $r=R_{\oplus }$ par field, jisko hum jaante hain $g_0=9.8{\text{N kg}}^{-1}$ hai. Hum ise reference ki tarah use karte hain taaki $G$ aur $M$ dobara plug in karne ki jagah ratio se scale kar sakein. Field $1/r^2$ ke hisaab se scale karta hai, toh us surface value $g_0$ ke comparison mein: $g=g_0{\left(\frac{R_{\oplus }}{3R_{\oplus }}\right)}^2=g_0\cdot \frac{1}{9}=\frac{9.8}{9}\approx 1.09{\text{N kg}}^{-1}$ $3R$ kyun, $2R$ kyun nahi? Altitude surface se measure hoti hai; formula mein $r$ centre se hota hai, toh $R_{\oplus }$ add karo.

$g=GM/r^2=1.0$ set karo aur $r$ ke liye solve karo: $r=\sqrt{\frac{G{M}_{\oplus }}{g}}=\sqrt{\frac{(6.67\times 10^{-11})(5.97\times 10^{24})}{1.0}}$ Numerator: $6.67\times 5.97=39.8$, toh $GM=3.98\times 10^{14}$. $r=\sqrt{3.98\times 10^{14}}=\sqrt{39.8\times 10^{13}}\approx 1.996\times 10^7\text{m}$ Yeh roughly $2.0\times 10^7\text{m}$ hai, centre se roughly $3.13R_{\oplus }$.

$\frac{g_{\text{planet}}}{g_{\oplus }}=\frac{G{M}_p/R_p^2}{G{M}_{\oplus }/R_{\oplus }^2}=\frac{M_p}{M_{\oplus }}\cdot {\left(\frac{R_{\oplus }}{R_p}\right)}^2=(2)\cdot {\left(\frac{1}{2}\right)}^2=2\cdot \frac{1}{4}=\frac{1}{2}$ Toh $g_{\text{planet}}=\frac{1}{2}{g}_{\oplus }\approx 4.9{\text{N kg}}^{-1}$. Mass field ko double karta hai lekin radius (squared, denominator mein) use quarter kar deta hai — radius jeet jaata hai.

Figure mein, do amber dots equal masses $M$ hain jo $2a$ apart hain; beech mein chota cyan dot hamara test point hai, har mass se $a$ distance par (neeche ke do white "$a$" brackets yeh confirm karte hain). Akele har mass ek field produce karta hai jiska magnitude hai $g_{\text{each}}=\frac{GM}{a^2}$ jo apni mass ki taraf point karta hai. Figure mein do cyan arrows yeh contributions dikhate hain: left wala left taraf point karta hai (left mass ki taraf) aur right wala right taraf (right mass ki taraf) — bilkul opposite directions. $g_{\text{net}}=g_{\text{each}}-g_{\text{each}}=0$ Midpoint ek null point hai: fields symmetry se cancel ho jaati hain. Wahan exactly rakhi test mass par koi net gravitational pull nahi hoti (halaanki yeh ek unstable balance hai — thoda dhakelo toh woh nearer mass ki taraf gir jaati hai).

Figure mein, left par bada amber dot $4M$ hai aur right par chota dot $M$ hai; cyan dot null point mark karta hai, jahan do cyan arrows $4M$ ki field (left taraf khiinchti) aur $M$ ki field (right taraf khiinchti) ko oppose karte dikhaate hain. White brackets distances $x$ ($4M$ se) aur $d-x$ ($M$ se) label karte hain. Null point unke beech hi hona chahiye (dono fields apni apni mass ki taraf inward point karti hain, toh sirf beech mein hi woh oppose aur cancel ho sakti hain). Maano yeh $4M$ se $x$ distance par hai, toh $M$ se $d-x$ distance par. Magnitudes equal set karo: $\frac{G(4M)}{x^2}=\frac{GM}{(d-x{)}^2}$ $GM$ cancel karo: $\frac{4}{x^2}=\frac{1}{(d-x{)}^2}\Rightarrow 4(d-x{)}^2=x^2$ Dono sides ka positive square root lo ($x$ aur $d-x$ dono positive lengths hain): $2(d-x)=x\Rightarrow 2d-2x=x\Rightarrow 2d=3x\Rightarrow x=\frac{2d}{3}$ Null point heavy mass se $x=\frac{2}{3}d$ par hai — lighter mass ke zyada paas (jaisa figure mein dikhta hai, cyan dot $M$ ke nearer hai), jo sense banata hai: $M$ ke itna paas khada rehna padta hai taaki uski weak field, $4M$ ki strong field ke saath compete kar sake. Doosra root kyun reject karein? Negative root lene par $-2(d-x)=x\Rightarrow x=2d$ milta hai, jo segment ke bahar hai jahan dono fields same direction mein point karti hain — wahan cancellation nahi hogi.

Figure mein, horizontal cyan arrow $g_E$ (East) hai aur vertical cyan arrow $g_M$ (North) hai; amber arrow resultant $g_{\text{net}}$ hai, dashed rectangle ka diagonal. Kyunki $g_E$ aur $g_M$ right angle par milte hain, woh ek right triangle ki do legs form karte hain jiska hypotenuse $g_{\text{net}}$ hai — isliye hum unhe Pythagoras se combine karte hain, numbers add karke nahi: $g_{\text{net}}=\sqrt{g_E^2+g_M^2}=\sqrt{(3.0{)}^2+(4.0{)}^2}\times 10^{-3}=\sqrt{9+16}\times 10^{-3}=\sqrt{25}\times 10^{-3}$ $g_{\text{net}}=5.0\times 10^{-3}{\text{N kg}}^{-1}$ Direction, North of East measure kiya gaya (figure mein amber arc): $\theta =\arctan \left(\frac{g_M}{g_E}\right)=\arctan \left(\frac{4.0}{3.0}\right)=\arctan (1.333)\approx 53.1^{\circ }$ $\arctan$ kyun? Angle ka tangent opposite over adjacent $=g_M/g_E$ hota hai; $\arctan$ answer deta hai "kaunse angle ka yeh tangent hai?" Dono components positive hain (East aur North), toh answer first quadrant mein hai — koi sign correction needed nahi.

Newton's Second Law (F=ma) se, acceleration $a=F/m=g$ (mass cancel ho jaata hai — yahi cancellation field definition mein bhi hai). Toh acceleration simply hai $a=g=1.62{\text{m s}}^{-2}$ jo $10\text{kg}$ se independent hai. Rest se start karke, $v=at$: $v=1.62\times 3.0=4.86{\text{m s}}^{-1}$ $a$ numerically $g$ ke equal kyun hai? Kyunki $g=F/m$ (field) aur $a=F/m$ (Newton's 2nd law) ek hi $F/m$ hain. Isliye units ${\text{N kg}}^{-1}$ aur ${\text{m s}}^{-2}$ identical hain.

Yeh Gravitational Potential and Potential Energy se connect hota hai. $V=-GM{r}^{-1}$ ko $r$ ke respect mein differentiate karo: $\frac{dV}{dr}=-GM\cdot (-1)r^{-2}=\frac{GM}{r^2}$ Phir $g=-\frac{dV}{dr}=-\frac{GM}{r^2}$ Magnitude $GM/r^2$ hai (parent formula se match karta hai ✔). Minus sign kehta hai $g$ decreasing $r$ ki direction mein point karta hai — yaani inward, $M$ ki taraf. Field, potential landscape ki slope hai, aur cheezein source ki taraf downhill girti hain.

Electric Field Intensity E=kQ/r² dekho. Dono laws algebraically identical inverse-square fields hain: $g=\frac{GM}{r^2},E=\frac{kQ}{r^2}$

Match ::: role $G\leftrightarrow k$ ::: interaction ka coupling constant $M\leftrightarrow Q$ ::: "source" (mass vs charge) $g=F/m\leftrightarrow E=F/q$ ::: force per unit test property (mass vs charge)

Key physical difference: mass hamesha positive hoti hai, toh $g$ sirf attract karta hai (hamesha inward). Charge $+$ ya $-$ ho sakta hai, toh $E$ inward ya outward point kar sakta hai. Dono $1/r^2$ ke hisaab se dilute hote hain — geometric reason same hai — field lines $4\pi r^2$ area wale sphere par spread hoti hain, jise Gauss's Law for Gravity formalize karta hai.

(a) Maano null point lighter mass $M$ se $x$ distance par hai, toh $9M$ se $d-x$ par, jahan $d=1.0\times 10^8$ hai. Field magnitudes equal set karo: $\frac{GM}{x^2}=\frac{G(9M)}{(d-x{)}^2}\Rightarrow \frac{1}{x^2}=\frac{9}{(d-x{)}^2}$ $(d-x{)}^2=9x^2\Rightarrow d-x=3x\Rightarrow d=4x\Rightarrow x=\frac{d}{4}=2.5\times 10^7\text{m}$ Toh null point $M$ se $2.5\times 10^7\text{m}$ par hai (aur $9M$ se $7.5\times 10^7\text{m}$ par) — smaller mass ke paas, jaisi expected thi. (b) Ise $9M$ ki taraf nudge karo: ab woh $9M$ ke zyada paas hai, toh $9M$ ki field us par badhti hai jabki $M$ ki field ghatati hai — net pull $9M$ ki taraf hai, ise null point se aur door khiinchta hai. Woh bhaag jaata hai (return nahi karta). Null point ek unstable equilibrium hai — potential landscape mein ek saddle.

Centre par: sphere ka har chota chunk directly uski opposite side par ek identical chunk rakhta hai, same distance par. Unke do field contributions exactly opposite directions mein point karte hain aur pairs mein cancel ho jaate hain. Saare aisi pairs ka sum karne par centre par $g=0$ milta hai — pure symmetry se. Inside ($r<R$): shell theorem se (Gauss's Law for Gravity ka consequence), sirf radius $r$ ke andar wali mass contribute karti hai — ise enclosed mass $M_{\text{enc}}$ kaho, sphere ka woh hissa jo hamare point se centre ke zyada paas hai. Ek uniform sphere (constant density $\rho$) ke liye, yeh enclosed mass $M_{\text{enc}}=\rho \cdot \frac{4}{3}\pi r^3$ hai, toh yeh $r^3$ ki tarah badhti hai (jabki total mass $M=\rho \cdot \frac{4}{3}\pi R^3$ $r=R$ par full-sphere value hai). Phir $g_{\text{inside}}=\frac{G{M}_{\text{enc}}}{r^2}\propto \frac{r^3}{r^2}=r$ — $g$ centre se $0$ se linearly badhta hai. Surface par ($r=R$): ab $M_{\text{enc}}=M$ (poori mass), aur $g$ apne maximum par pahunchta hai, $g_{\max }=GM/R^2$. Outside ($r>R$): poori mass $M$ aise act karti hai jaise centre par concentrated ho, toh $g=GM/r^2$, jo $1/r^2$ se decrease karta hai — ordinary inverse-square falloff. Short mein: $g$ centre par $0$ se linearly climb karta hai, surface par $GM/R^2$ tak pahunchta hai, phir uske baad $1/r^2$ se decay karta hai.

Nahi. Potential aur field alag cheezein hain:

• Field $\vec{g}$ ek vector hai; opposing fields cancel ho kar zero ho sakti hain.
• Potential $V=-\frac{GM}{r}$ ek scalar hai aur mass ke liye hamesha negative hoti hai; do negatives cancel nahi ho sakte — woh ek zyada negative number mein add hote hain. Null point par, $x=2.5\times 10^7$ aur $d-x=7.5\times 10^7$ use karke: $V=-\frac{GM}{x}-\frac{G(9M)}{d-x}=-GM\left(\frac{1}{2.5\times 10^7}+\frac{9}{7.5\times 10^7}\right)$ $=-GM\left(4.0\times 10^{-8}+1.2\times 10^{-7}\right)=-GM(1.6\times 10^{-7}){\text{J kg}}^{-1}$ jo definitely zero nahi hai (yeh negative hai). Field ka zero hona matlab hai "yahan koi net pull nahi"; potential ka negative hona matlab hai "yahan se escape karne ke liye tumhe abhi bhi kaam karna padega." Zero force $\ne$ zero energy.