{
  "index": "1946-B-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "3. In a solid sphere of radius \\( R \\) the density \\( \\rho \\) is a function of \\( r \\), the distance from the center of the sphere. If the magnitude of the gravitational force of attraction due to the sphere at any point inside the sphere is \\( k r^{2} \\), where \\( k \\) is a constant, find \\( \\rho \\) as a function of \\( r \\). Find also the magnitude of the force of attraction at a point outside the sphere at a distance \\( r \\) from the center. (Assume that the magnitude of the force of attraction at a point \\( P \\) due to a thin spherical shell is zero if \\( P \\) is inside the shell, and is \\( m / r^{2} \\) if \\( P \\) is outside the shell, \\( m \\) being the mass of the shell, and \\( r \\) the distance of \\( P \\) from the center.)",
  "solution": "Solution. Let \\( P \\) be a point at distance \\( r \\) from the center. We regard the solid sphere as the union of many thin concentric spherical shells. The force of attraction at \\( P \\) is the sum of the forces of attraction due to the shells. All of these forces act in the same direction so we can simply add their magnitudes.\n\nThe shell at distance \\( \\dot{s} \\) from the center and thickness \\( \\Delta s \\) has (approximately) volume \\( 4 \\pi s^{2} \\Delta s \\) and mass \\( 4 \\pi \\rho(s) s^{2} \\Delta s \\), so the force of attraction at \\( \\boldsymbol{P} \\) due to this shell is approximately\n\\[\n\\frac{4 \\pi \\rho(s) s^{2} \\Delta s}{r^{2}}\n\\]\nif \\( r>s \\). It is zero if \\( r<s \\). The total force of attraction at \\( P \\) is therefore (exactly)\n\\[\nF=\\left\\{\\begin{array}{ll}\n\\int_{0}^{r} \\frac{4 \\pi \\rho(s) s^{2} d s}{r^{2}} & \\text { if } r \\leq R \\\\\n\\int_{0}^{R} \\frac{4 \\pi \\rho(s) s^{2} d s}{r^{2}} & \\text { if } r>R\n\\end{array}\\right.\n\\]\n\nSince we are given that \\( F=k r^{2} \\) for \\( r \\leq R \\), we have\n\\[\nk r^{2}=\\int_{0}^{r} \\frac{4 \\pi \\rho(s) s^{2} d s}{r^{2}} \\quad \\text { for } r \\leq R,\n\\]\nand we must solve for \\( \\rho \\). Multiplying by \\( r^{2} \\), and then differentiating with respect to \\( r \\), we obtain\n\\[\n4 k r^{3}=4 \\pi \\rho(r) r^{2} .\n\\]\n\nTherefore\n\\[\n\\rho(r)=\\frac{k}{\\pi} r\n\\]\nis the required formula for \\( \\rho \\).\nSubstituting this in (1), we have\n\\[\nF=\\frac{4 \\pi}{r^{2}} \\int_{0}^{R} \\frac{k s}{\\pi} s^{2} d s=\\frac{k R^{4}}{r^{2}} \\text { for } r>R\n\\]\n\nRemark. We could obtain the last formula directly because we were given that \\( F \\) is \\( k R^{2} \\) at the surface and we know it falls off inversely with \\( r^{2} \\) outside the sphere.",
  "vars": [
    "r",
    "s",
    "F",
    "\\\\rho",
    "P"
  ],
  "params": [
    "R",
    "k",
    "m"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "r": "radialdist",
        "s": "shellrad",
        "F": "totalforce",
        "\\rho": "densityfunc",
        "P": "pointloc",
        "R": "sphererad",
        "k": "forceconst",
        "m": "shellmass"
      },
      "question": "3. In a solid sphere of radius \\( sphererad \\) the density \\( densityfunc \\) is a function of \\( radialdist \\), the distance from the center of the sphere. If the magnitude of the gravitational force of attraction due to the sphere at any point inside the sphere is \\( forceconst radialdist^{2} \\), where \\( forceconst \\) is a constant, find \\( densityfunc \\) as a function of \\( radialdist \\). Find also the magnitude of the force of attraction at a point outside the sphere at a distance \\( radialdist \\) from the center. (Assume that the magnitude of the force of attraction at a point \\( pointloc \\) due to a thin spherical shell is zero if \\( pointloc \\) is inside the shell, and is \\( shellmass / radialdist^{2} \\) if \\( pointloc \\) is outside the shell, \\( shellmass \\) being the mass of the shell, and \\( radialdist \\) the distance of \\( pointloc \\) from the center.)",
      "solution": "Solution. Let \\( pointloc \\) be a point at distance \\( radialdist \\) from the center. We regard the solid sphere as the union of many thin concentric spherical shells. The force of attraction at \\( pointloc \\) is the sum of the forces of attraction due to the shells. All of these forces act in the same direction so we can simply add their magnitudes.\n\nThe shell at distance \\( \\dot{shellrad} \\) from the center and thickness \\( \\Delta shellrad \\) has (approximately) volume \\( 4 \\pi shellrad^{2} \\Delta shellrad \\) and mass \\( 4 \\pi densityfunc(shellrad) shellrad^{2} \\Delta shellrad \\), so the force of attraction at \\( \\boldsymbol{pointloc} \\) due to this shell is approximately\n\\[\n\\frac{4 \\pi densityfunc(shellrad) shellrad^{2} \\Delta shellrad}{radialdist^{2}}\n\\]\nif \\( radialdist>shellrad \\). It is zero if \\( radialdist<shellrad \\). The total force of attraction at \\( pointloc \\) is therefore (exactly)\n\\[\n totalforce=\\left\\{\\begin{array}{ll}\n \\int_{0}^{radialdist} \\frac{4 \\pi densityfunc(shellrad) shellrad^{2} d shellrad}{radialdist^{2}} & \\text { if } radialdist \\leq sphererad \\\\\n \\int_{0}^{sphererad} \\frac{4 \\pi densityfunc(shellrad) shellrad^{2} d shellrad}{radialdist^{2}} & \\text { if } radialdist>sphererad\n \\end{array}\\right.\n\\]\n\nSince we are given that \\( totalforce=forceconst radialdist^{2} \\) for \\( radialdist \\leq sphererad \\), we have\n\\[\n forceconst radialdist^{2}=\\int_{0}^{radialdist} \\frac{4 \\pi densityfunc(shellrad) shellrad^{2} d shellrad}{radialdist^{2}} \\quad \\text { for } radialdist \\leq sphererad,\n\\]\nand we must solve for \\( densityfunc \\). Multiplying by \\( radialdist^{2} \\), and then differentiating with respect to \\( radialdist \\), we obtain\n\\[\n 4 forceconst radialdist^{3}=4 \\pi densityfunc(radialdist) radialdist^{2} .\n\\]\n\nTherefore\n\\[\n densityfunc(radialdist)=\\frac{forceconst}{\\pi} radialdist\n\\]\nis the required formula for \\( densityfunc \\).\nSubstituting this in (1), we have\n\\[\n totalforce=\\frac{4 \\pi}{radialdist^{2}} \\int_{0}^{sphererad} \\frac{forceconst shellrad}{\\pi} shellrad^{2} d shellrad=\\frac{forceconst sphererad^{4}}{radialdist^{2}} \\text { for } radialdist>sphererad\n\\]\n\nRemark. We could obtain the last formula directly because we were given that \\( totalforce \\) is \\( forceconst sphererad^{2} \\) at the surface and we know it falls off inversely with \\( radialdist^{2} \\) outside the sphere."
    },
    "descriptive_long_confusing": {
      "map": {
        "r": "cactusleaf",
        "s": "pebblestone",
        "F": "lanternship",
        "\\\\rho": "wandering",
        "P": "quilldragon",
        "R": "moonlitsea",
        "k": "rivercloud",
        "m": "stardustox"
      },
      "question": "3. In a solid sphere of radius \\( moonlitsea \\) the density \\( wandering \\) is a function of \\( cactusleaf \\), the distance from the center of the sphere. If the magnitude of the gravitational force of attraction due to the sphere at any point inside the sphere is \\( rivercloud cactusleaf^{2} \\), where \\( rivercloud \\) is a constant, find \\( wandering \\) as a function of \\( cactusleaf \\). Find also the magnitude of the force of attraction at a point outside the sphere at a distance \\( cactusleaf \\) from the center. (Assume that the magnitude of the force of attraction at a point \\( quilldragon \\) due to a thin spherical shell is zero if \\( quilldragon \\) is inside the shell, and is \\( stardustox / cactusleaf^{2} \\) if \\( quilldragon \\) is outside the shell, \\( stardustox \\) being the mass of the shell, and \\( cactusleaf \\) the distance of \\( quilldragon \\) from the center.)",
      "solution": "Solution. Let \\( quilldragon \\) be a point at distance \\( cactusleaf \\) from the center. We regard the solid sphere as the union of many thin concentric spherical shells. The force of attraction at \\( quilldragon \\) is the sum of the forces of attraction due to the shells. All of these forces act in the same direction so we can simply add their magnitudes.\n\nThe shell at distance \\( \\dot{pebblestone} \\) from the center and thickness \\( \\Delta pebblestone \\) has (approximately) volume \\( 4 \\pi pebblestone^{2} \\Delta pebblestone \\) and mass \\( 4 \\pi wandering(pebblestone) pebblestone^{2} \\Delta pebblestone \\), so the force of attraction at \\( \\boldsymbol{quilldragon} \\) due to this shell is approximately\n\\[\n\\frac{4 \\pi wandering(pebblestone) pebblestone^{2} \\Delta pebblestone}{cactusleaf^{2}}\n\\]\nif \\( cactusleaf>pebblestone \\). It is zero if \\( cactusleaf<pebblestone \\). The total force of attraction at \\( quilldragon \\) is therefore (exactly)\n\\[\nlanternship=\\left\\{\\begin{array}{ll}\n\\int_{0}^{cactusleaf} \\frac{4 \\pi wandering(pebblestone) pebblestone^{2} d pebblestone}{cactusleaf^{2}} & \\text { if } cactusleaf \\leq moonlitsea \\\\\n\\int_{0}^{moonlitsea} \\frac{4 \\pi wandering(pebblestone) pebblestone^{2} d pebblestone}{cactusleaf^{2}} & \\text { if } cactusleaf>moonlitsea\n\\end{array}\\right.\n\\]\n\nSince we are given that \\( lanternship=rivercloud cactusleaf^{2} \\) for \\( cactusleaf \\leq moonlitsea \\), we have\n\\[\nrivercloud cactusleaf^{2}=\\int_{0}^{cactusleaf} \\frac{4 \\pi wandering(pebblestone) pebblestone^{2} d pebblestone}{cactusleaf^{2}} \\quad \\text { for } cactusleaf \\leq moonlitsea,\n\\]\nand we must solve for \\( wandering \\). Multiplying by \\( cactusleaf^{2} \\), and then differentiating with respect to \\( cactusleaf \\), we obtain\n\\[\n4 rivercloud cactusleaf^{3}=4 \\pi wandering(cactusleaf) cactusleaf^{2} .\n\\]\n\nTherefore\n\\[\nwandering(cactusleaf)=\\frac{rivercloud}{\\pi} cactusleaf\n\\]\nis the required formula for \\( wandering \\).\nSubstituting this in (1), we have\n\\[\nlanternship=\\frac{4 \\pi}{cactusleaf^{2}} \\int_{0}^{moonlitsea} \\frac{rivercloud pebblestone}{\\pi} pebblestone^{2} d pebblestone=\\frac{rivercloud moonlitsea^{4}}{cactusleaf^{2}} \\text { for } cactusleaf>moonlitsea\n\\]\n\nRemark. We could obtain the last formula directly because we were given that \\( lanternship \\) is \\( rivercloud moonlitsea^{2} \\) at the surface and we know it falls off inversely with \\( cactusleaf^{2} \\) outside the sphere."
    },
    "descriptive_long_misleading": {
      "map": {
        "r": "nearness",
        "s": "coredepth",
        "F": "weakness",
        "\\rho": "emptiness",
        "P": "extension",
        "R": "corepoint",
        "k": "variable",
        "m": "lightness"
      },
      "question": "3. In a solid sphere of radius \\( corepoint \\) the density \\( \\emptiness \\) is a function of \\( nearness \\), the distance from the center of the sphere. If the magnitude of the gravitational force of attraction due to the sphere at any point inside the sphere is \\( variable nearness^{2} \\), where \\( variable \\) is a constant, find \\( \\emptiness \\) as a function of \\( nearness \\). Find also the magnitude of the force of attraction at a point outside the sphere at a distance \\( nearness \\) from the center. (Assume that the magnitude of the force of attraction at a point \\( extension \\) due to a thin spherical shell is zero if \\( extension \\) is inside the shell, and is \\( lightness / nearness^{2} \\) if \\( extension \\) is outside the shell, \\( lightness \\) being the mass of the shell, and \\( nearness \\) the distance of \\( extension \\) from the center.)",
      "solution": "Solution. Let \\( extension \\) be a point at distance \\( nearness \\) from the center. We regard the solid sphere as the union of many thin concentric spherical shells. The force of attraction at \\( extension \\) is the sum of the forces of attraction due to the shells. All of these forces act in the same direction so we can simply add their magnitudes.\n\nThe shell at distance \\( \\dot{coredepth} \\) from the center and thickness \\( \\Delta coredepth \\) has (approximately) volume \\( 4 \\pi coredepth^{2} \\Delta coredepth \\) and mass \\( 4 \\pi \\emptiness(coredepth) coredepth^{2} \\Delta coredepth \\), so the force of attraction at \\( \\boldsymbol{extension} \\) due to this shell is approximately\n\\[\n\\frac{4 \\pi \\emptiness(coredepth) coredepth^{2} \\Delta coredepth}{nearness^{2}}\n\\]\nif \\( nearness>coredepth \\). It is zero if \\( nearness<coredepth \\). The total force of attraction at \\( extension \\) is therefore (exactly)\n\\[\nweakness=\\left\\{\\begin{array}{ll}\n\\int_{0}^{nearness} \\frac{4 \\pi \\emptiness(coredepth) coredepth^{2} d coredepth}{nearness^{2}} & \\text { if } nearness \\leq corepoint \\\\\n\\int_{0}^{corepoint} \\frac{4 \\pi \\emptiness(coredepth) coredepth^{2} d coredepth}{nearness^{2}} & \\text { if } nearness>corepoint\n\\end{array}\\right.\n\\]\n\nSince we are given that \\( weakness=variable nearness^{2} \\) for \\( nearness \\leq corepoint \\), we have\n\\[\nvariable nearness^{2}=\\int_{0}^{nearness} \\frac{4 \\pi \\emptiness(coredepth) coredepth^{2} d coredepth}{nearness^{2}} \\quad \\text { for } nearness \\leq corepoint,\n\\]\nand we must solve for \\( \\emptiness \\). Multiplying by \\( nearness^{2} \\), and then differentiating with respect to \\( nearness \\), we obtain\n\\[\n4 variable nearness^{3}=4 \\pi \\emptiness(nearness) nearness^{2} .\n\\]\n\nTherefore\n\\[\n\\emptiness(nearness)=\\frac{variable}{\\pi} nearness\n\\]\nis the required formula for \\( \\emptiness \\).\nSubstituting this in (1), we have\n\\[\nweakness=\\frac{4 \\pi}{nearness^{2}} \\int_{0}^{corepoint} \\frac{variable coredepth}{\\pi} coredepth^{2} d coredepth=\\frac{variable corepoint^{4}}{nearness^{2}} \\text { for } nearness>corepoint\n\\]\n\nRemark. We could obtain the last formula directly because we were given that \\( weakness \\) is \\( variable corepoint^{2} \\) at the surface and we know it falls off inversely with \\( nearness^{2} \\) outside the sphere."
    },
    "garbled_string": {
      "map": {
        "r": "qzxwvtnp",
        "s": "hjgrksla",
        "F": "mplqzndf",
        "\\\\rho": "zvfkldmq",
        "P": "xjvrnsop",
        "R": "tkwmsnyg",
        "k": "lpshvdra",
        "m": "cgtrwqnv"
      },
      "question": "3. In a solid sphere of radius \\( tkwmsnyg \\) the density \\( zvfkldmq \\) is a function of \\( qzxwvtnp \\), the distance from the center of the sphere. If the magnitude of the gravitational force of attraction due to the sphere at any point inside the sphere is \\( lpshvdra \\, qzxwvtnp^{2} \\), where \\( lpshvdra \\) is a constant, find \\( zvfkldmq \\) as a function of \\( qzxwvtnp \\). Find also the magnitude of the force of attraction at a point outside the sphere at a distance \\( qzxwvtnp \\) from the center. (Assume that the magnitude of the force of attraction at a point \\( xjvrnsop \\) due to a thin spherical shell is zero if \\( xjvrnsop \\) is inside the shell, and is \\( cgtrwqnv / qzxwvtnp^{2} \\) if \\( xjvrnsop \\) is outside the shell, \\( cgtrwqnv \\) being the mass of the shell, and \\( qzxwvtnp \\) the distance of \\( xjvrnsop \\) from the center.)",
      "solution": "Solution. Let \\( xjvrnsop \\) be a point at distance \\( qzxwvtnp \\) from the center. We regard the solid sphere as the union of many thin concentric spherical shells. The force of attraction at \\( xjvrnsop \\) is the sum of the forces of attraction due to the shells. All of these forces act in the same direction so we can simply add their magnitudes.\n\nThe shell at distance \\( \\dot{hjgrksla} \\) from the center and thickness \\( \\Delta hjgrksla \\) has (approximately) volume \\( 4 \\pi hjgrksla^{2} \\Delta hjgrksla \\) and mass \\( 4 \\pi zvfkldmq(hjgrksla) hjgrksla^{2} \\Delta hjgrksla \\), so the force of attraction at \\( \\boldsymbol{xjvrnsop} \\) due to this shell is approximately\n\\[\n\\frac{4 \\pi zvfkldmq(hjgrksla) hjgrksla^{2} \\Delta hjgrksla}{qzxwvtnp^{2}}\n\\]\nif \\( qzxwvtnp>hjgrksla \\). It is zero if \\( qzxwvtnp<hjgrksla \\). The total force of attraction at \\( xjvrnsop \\) is therefore (exactly)\n\\[\nmplqzndf=\\left\\{\\begin{array}{ll}\n\\int_{0}^{qzxwvtnp} \\frac{4 \\pi zvfkldmq(hjgrksla) hjgrksla^{2} d hjgrksla}{qzxwvtnp^{2}} & \\text { if } qzxwvtnp \\leq tkwmsnyg \\\\\n\\int_{0}^{tkwmsnyg} \\frac{4 \\pi zvfkldmq(hjgrksla) hjgrksla^{2} d hjgrksla}{qzxwvtnp^{2}} & \\text { if } qzxwvtnp>tkwmsnyg\n\\end{array}\\right.\n\\]\n\nSince we are given that \\( mplqzndf=lpshvdra \\, qzxwvtnp^{2} \\) for \\( qzxwvtnp \\leq tkwmsnyg \\), we have\n\\[\nlpshvdra \\, qzxwvtnp^{2}=\\int_{0}^{qzxwvtnp} \\frac{4 \\pi zvfkldmq(hjgrksla) hjgrksla^{2} d hjgrksla}{qzxwvtnp^{2}} \\quad \\text { for } qzxwvtnp \\leq tkwmsnyg,\n\\]\nand we must solve for \\( zvfkldmq \\). Multiplying by \\( qzxwvtnp^{2} \\), and then differentiating with respect to \\( qzxwvtnp \\), we obtain\n\\[\n4 lpshvdra \\, qzxwvtnp^{3}=4 \\pi zvfkldmq(qzxwvtnp) qzxwvtnp^{2} .\n\\]\n\nTherefore\n\\[\nzvfkldmq(qzxwvtnp)=\\frac{lpshvdra}{\\pi} qzxwvtnp\n\\]\nis the required formula for \\( zvfkldmq \\).\nSubstituting this in (1), we have\n\\[\nmplqzndf=\\frac{4 \\pi}{qzxwvtnp^{2}} \\int_{0}^{tkwmsnyg} \\frac{lpshvdra \\, hjgrksla}{\\pi} hjgrksla^{2} d hjgrksla=\\frac{lpshvdra \\, tkwmsnyg^{4}}{qzxwvtnp^{2}} \\text { for } qzxwvtnp>tkwmsnyg\n\\]\n\nRemark. We could obtain the last formula directly because we were given that \\( mplqzndf \\) is \\( lpshvdra \\, tkwmsnyg^{2} \\) at the surface and we know it falls off inversely with \\( qzxwvtnp^{2} \\) outside the sphere."
    },
    "kernel_variant": {
      "question": "In $d$–dimensional Euclidean space with $d\\ge 3$ (so that the Newtonian force produced by a thin $(d-1)$–sphere of mass $m$ is $0$ for an interior point and $m/r^{\\,d-1}$ for an exterior point whose distance is $r$), consider a solid hyper-ball $B$ of radius $a>0$.  \nThe mass density is purely radial, $\\rho=\\rho(r)$ with $r=\\lvert x\\rvert$.  \nExperiments performed with a unit test mass placed at any point whose distance from the centre satisfies $0\\le r\\le a$ show that the inward radial gravitational force has magnitude  \n\n\\[\nF_{\\text{in}}(r)=c\\,r^{\\,d-1}\\,e^{\\beta r^{2}}\\bigl(1+\\gamma r^{2}\\bigr)^{p},\\qquad(0\\le r\\le a)\n\\tag{$\\star$}\n\\]\n\nwhere $c>0$, $\\beta\\ge 0$, $\\gamma\\ge 0$ and $p\\in\\mathbb N\\cup\\{0\\}$.\n\n\\textbf{Tasks}  \n(a)\\; Determine $\\rho(r)$ in closed form in terms of $d,c,\\beta,\\gamma$ and $p$.  \n\n(b)\\; The total mass of $B$ is prescribed to be $M_{0}$.  Derive the explicit relation that fixes $c$ in terms of $M_{0},a,\\beta,\\gamma,p$ and $d$. (Any equivalent representation is acceptable.)  \n\n(c)\\; Find the magnitude of the gravitational force $F_{\\text{out}}(r)$ felt by a unit test mass situated at any point with $r\\ge a$.  \n\n(d)\\; Specialise your answers to the concrete case  \n\n\\[\nd=4,\\qquad\\beta=0,\\qquad p=2,\\qquad\\gamma>0 .\n\\]\n\nGive $\\rho(r)$, the constant $c$ in terms of $M_{0}$, and the exterior force for this particular choice of parameters.",
      "solution": "\\textbf{Preliminaries}  \nDenote by  \n\\[\n\\Omega_{d}=\\frac{2\\pi^{d/2}}{\\Gamma\\!\\bigl(\\tfrac{d}{2}\\bigr)}\n\\]  \nthe surface area of the unit $(d-1)$–sphere in $\\mathbb R^{d}$.  \nFor a spherically symmetric mass distribution in $d$ dimensions the generalised shell theorem states\n\n\\begin{itemize}\n\\item[(i)] the mass $M(r)$ enclosed by the hypersphere of radius $r$ is  \n\\[\nM(r)=\\int_{0}^{r}\\Omega_{d}\\,s^{\\,d-1}\\rho(s)\\,ds,\n\\]\n\\item[(ii)] a unit test mass situated at distance $r$ from the centre is attracted with force magnitude  \n\\[\nF(r)=\\frac{M(r)}{r^{\\,d-1}} .\n\\tag{1}\n\\]\n\\end{itemize}\n\nHence  \n\\[\nM(r)=r^{\\,d-1}F(r),\\qquad \n\\frac{dM}{dr}=\\Omega_{d}\\,r^{\\,d-1}\\rho(r).\n\\tag{2}\n\\]\n\n\\bigskip\n\\textbf{Step 1.  Density $\\rho(r)$ (task (a))}  \n\nInsert the empirical force law \\eqref{star} into \\eqref{2}:\n\n\\[\nM(r)=c\\,r^{\\,d-1}\\,r^{\\,d-1}\\,e^{\\beta r^{2}}\\bigl(1+\\gamma r^{2}\\bigr)^{p}\n     =c\\,r^{\\,2d-2}\\,e^{\\beta r^{2}}\\bigl(1+\\gamma r^{2}\\bigr)^{p}.\n\\tag{3}\n\\]\n\nDifferentiate:\n\n\\[\n\\begin{aligned}\n\\frac{dM}{dr}\n&=c\\,e^{\\beta r^{2}}\\bigl(1+\\gamma r^{2}\\bigr)^{p}\\,\\frac{d}{dr}\\bigl[r^{\\,2d-2}\\bigr] \\\\\n&\\quad +c\\,r^{\\,2d-2}\\bigl(1+\\gamma r^{2}\\bigr)^{p}\\,\\frac{d}{dr}\\!\\bigl[e^{\\beta r^{2}}\\bigr] \\\\\n&\\quad +c\\,r^{\\,2d-2}e^{\\beta r^{2}}\\,\\frac{d}{dr}\\!\\bigl[(1+\\gamma r^{2})^{p}\\bigr].\n\\end{aligned}\n\\]\n\nThe elementary derivatives are  \n\n\\[\n\\frac{d}{dr}\\!\\bigl[r^{\\,2d-2}\\bigr]=(2d-2)r^{\\,2d-3},\\qquad\n\\frac{d}{dr}\\!\\bigl[e^{\\beta r^{2}}\\bigr]=2\\beta r\\,e^{\\beta r^{2}},\\qquad\n\\frac{d}{dr}\\!\\bigl[(1+\\gamma r^{2})^{p}\\bigr]=2p\\gamma r\\,(1+\\gamma r^{2})^{p-1}.\n\\]\n\nFactorising $e^{\\beta r^{2}}(1+\\gamma r^{2})^{p-1}$ gives  \n\n\\[\n\\frac{dM}{dr}\n=2c\\,r^{\\,2d-3}\\,e^{\\beta r^{2}}\\,(1+\\gamma r^{2})^{p-1}\n\\Bigl[(d-1)(1+\\gamma r^{2})+\\beta r^{2}(1+\\gamma r^{2})+p\\gamma r^{2}\\Bigr].\n\\tag{4}\n\\]\n\nUsing the second relation in \\eqref{2} we obtain\n\n\\[\n\\boxed{\\;\n\\rho(r)=\\frac{2c}{\\Omega_{d}}\\,r^{\\,d-2}\\,e^{\\beta r^{2}}\\,(1+\\gamma r^{2})^{p-1}\n\\Bigl[(d-1)(1+\\gamma r^{2})+\\beta r^{2}(1+\\gamma r^{2})+p\\gamma r^{2}\\Bigr]\n\\;}.\n\\tag{5}\n\\]\n\n\\bigskip\n\\textbf{Step 2.  Fixing $c$ from the prescribed total mass $M_{0}$ (task (b))}  \n\nBecause $M(r)$ is explicit, simply set $r=a$ in \\eqref{3}:\n\n\\[\nM_{0}=M(a)=c\\,a^{\\,2d-2}\\,e^{\\beta a^{2}}\\bigl(1+\\gamma a^{2}\\bigr)^{p},\n\\]\n\nwhence  \n\n\\[\n\\boxed{\\; \nc=\\dfrac{M_{0}}{a^{\\,2d-2}\\,e^{\\beta a^{2}}\\bigl(1+\\gamma a^{2}\\bigr)^{p}}\n\\;}.\n\\tag{6}\n\\]\n\n\\emph{Consistency check.}  \nInsert \\eqref{5} into the definition of $M(r)$, switch to the variable $t=r^{2}$, and integrate:\n\n\\[\n\\begin{aligned}\nM(a)&=\\Omega_{d}\\int_{0}^{a}r^{\\,d-1}\\rho(r)\\,dr\\\\\n    &=c\\int_{0}^{a^{2}}\\frac{d}{dt}\\Bigl[t^{\\,d-1}(1+\\gamma t)^{p}e^{\\beta t}\\Bigr]\\,dt\\\\\n    &=c\\,a^{\\,2d-2}\\,e^{\\beta a^{2}}\\bigl(1+\\gamma a^{2}\\bigr)^{p}=M_{0},\n\\end{aligned}\n\\]\n\nwhich indeed reproduces \\eqref{6}.\n\n\\bigskip\n\\textbf{Step 3.  Exterior field (task (c))}  \n\nFor $r\\ge a$ the entire mass $M_{0}$ acts as though concentrated at the centre, hence by \\eqref{1}\n\n\\[\n\\boxed{\\;\nF_{\\text{out}}(r)=\\dfrac{M_{0}}{r^{\\,d-1}},\\qquad r\\ge a\n\\;}.\n\\tag{7}\n\\]\n\n\\bigskip\n\\textbf{Step 4.  Special case $d=4,\\ \\beta=0,\\ p=2$ (task (d))}  \n\nThe surface area of the unit $3$–sphere is $\\Omega_{4}=2\\pi^{2}$.  \nInsert $d=4,\\ \\beta=0,\\ p=2$ into \\eqref{5}:\n\n\\[\n\\begin{aligned}\n\\rho_{4}(r)\n&=\\frac{2c}{2\\pi^{2}}\\,r^{2}\\,(1+\\gamma r^{2})^{1}\n\\Bigl[3\\bigl(1+\\gamma r^{2}\\bigr)+2\\gamma r^{2}\\Bigr]\\\\[2mm]\n&=\\frac{c}{\\pi^{2}}\\,r^{2}\\bigl(1+\\gamma r^{2}\\bigr)\\bigl(3+5\\gamma r^{2}\\bigr).\n\\end{aligned}\n\\tag{8}\n\\]\n\nFrom \\eqref{6} one finds\n\n\\[\n\\boxed{\\;\nc=\\dfrac{M_{0}}{a^{6}\\bigl(1+\\gamma a^{2}\\bigr)^{2}}\n\\;}.\n\\tag{9}\n\\]\n\nFinally, with $d=4$, equation \\eqref{7} gives the exterior force\n\n\\[\n\\boxed{\\;\nF_{\\text{out}}(r)=\\dfrac{M_{0}}{r^{3}},\\qquad r\\ge a\n\\;}.\n\\tag{10}\n\\]\n\nAll requested formulae have now been obtained.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.407110",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensions:  The problem is set in an arbitrary d-dimensional Euclidean space, obliging the solver to generalise the classical three–dimensional shell theorem, work with the surface measure Ω_d and keep the dimensional parameter symbolic throughout.\n\n2. More variables & parameters:  Five independent parameters (c, β, γ, p, d) appear multiplicatively and in exponents, greatly complicating differentiation and integration compared with the single-parameter force laws of the original statements.\n\n3. Interaction of multiple concepts:  The solution requires  \n •  understanding the n-dimensional Newtonian field,  \n •  translating a measured force law into an integral equation for the enclosed mass,  \n •  differentiating products of several radial functions, and  \n •  relating the result back to mass density with appropriate geometric factors.\n\n4. Special functions:  While part (a) can be written in elementary form, part (b) forces the solver to manipulate non-elementary integrals and introduces the incomplete Gamma function (or, equivalently, hypergeometric functions) to express c explicitly.\n\n5. Multi-step reasoning:  The path “force ⇒ enclosed mass ⇒ derivative ⇒ density ⇒ re-integration ⇒ normalisation” is longer and more intricate than in the original problem, and each step is algebraically heavier because all quantities depend on d and the extra parameters β, γ, p.\n\n6. Layered specialisation:  The final sub-problem (d) shows how the general machinery collapses to concrete formulas, testing the solver’s ability to simplify higher-level expressions in a specific setting.\n\nThese added layers of abstraction and computation make the enhanced kernel variant markedly more challenging than both the original textbook exercise and the existing kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "In d–dimensional Euclidean space with d ≥ 3 (so that the Newtonian force produced by a thin (d–1)-sphere of mass m is 0 for an interior point and m / r^{d-1} for an exterior point whose distance is r), consider a solid hyper-ball B of radius a.  \nThe mass density ρ(r) is radial, i.e. ρ(r)=ρ(|x|).  Experiments performed with a unit test mass placed at any point whose distance from the centre satisfies 0 ≤ r ≤ a show that the inward radial gravitational force has magnitude  \n\n  F_in(r)=c r^{d-1} e^{β r²}(1+γ r²)^{p},                            (★)  \n\nwhere c>0, β≥0, γ≥0 and p is a non-negative integer.\n\nTasks  \n(a)  Determine ρ(r) in closed form in terms of d, c, β, γ and p.  \n(b)  The total mass of B is prescribed to be M₀.  Derive the explicit relation that fixes c in terms of M₀, a, β, γ, p and d. (Any equivalent representation is acceptable.)  \n(c)  Find the magnitude of the gravitational force F_out(r) felt by a unit test mass situated at any point with r ≥ a.  \n(d)  Specialise your answers to the concrete case  \n\n  d = 4, β = 0, p = 2, γ>0.\n\nGive ρ(r), the constant c in terms of M₀ and the exterior force for this particular choice of parameters.",
      "solution": "Preliminaries  \nLet Ω_d = 2π^{d/2}/Γ(d/2) be the surface area of the unit (d–1)-sphere.  \nFor a spherically–symmetric mass distribution in d dimensions the generalised “shell theorem’’ states that  \n\n(i)  the mass M(r) enclosed by the hyper-sphere of radius r is  \n\n  M(r)=∫₀^{r} Ω_d s^{d-1}ρ(s) ds,\n\n(ii)  a unit test mass situated at distance r is attracted with force magnitude  \n\n  F(r)=M(r)/r^{d-1}.                                                (1)\n\nHence  \n\n  M(r)=r^{d-1}F(r),   dM/dr = Ω_d r^{d-1}ρ(r).                   (2)\n\n\n\nStep 1. Density ρ(r) for general d (task (a))  \nInsert the empirical force law (★) into (2):\n\nM(r)=c r^{d-1}·r^{d-1}e^{β r²}(1+γ r²)^p\n    =c r^{2d-2} e^{β r²}(1+γ r²)^p.                               (3)\n\nDifferentiate:\n\ndM/dr\n  = c e^{β r²}(1+γ r²)^p d/dr[r^{2d-2}]\n  + c r^{2d-2}(1+γ r²)^p d/dr[e^{β r²}]\n  + c r^{2d-2}e^{β r²} d/dr[(1+γ r²)^p].\n\nCompute the three derivatives\n\nd/dr[r^{2d-2}]  = (2d-2) r^{2d-3},\nd/dr[e^{β r²}]   = 2β r e^{β r²},\nd/dr[(1+γ r²)^p] = 2pγ r(1+γ r²)^{p-1},\n\nand factor out e^{β r²}(1+γ r²)^{p-1}:\n\ndM/dr = 2c r^{2d-3} e^{β r²}(1+γ r²)^{p-1}\n       ×[(d-1)(1+γ r²)+β r²(1+γ r²)+pγ r²].                     (4)\n\nUsing the second relation in (2) we finally get\n\nρ(r)= 2c e^{β r²}(1+γ r²)^{p-1} r^{d-2}/Ω_d\n     ×[(d-1)(1+γ r²)+β r²(1+γ r²)+pγ r²].                      (5)\n\n\n\nStep 2. Fixing c from the prescribed total mass M₀ (task (b))  \nBecause M(r) has already been obtained explicitly, we can set r=a in (3):\n\nM₀ = M(a) = c a^{2d-2} e^{β a²}(1+γ a²)^p  ⇒ \n\n  c = M₀ /[a^{2d-2} e^{β a²}(1+γ a²)^p].                        (6)\n\nRemark (consistency check by direct integration).  \nUsing (5) in the definition of M(r) one finds\n\nΩ_d ∫₀^{a} r^{d-1}ρ(r) dr\n  = c ∫₀^{a²} d/dt[ t^{d-1}(1+γ t)^p e^{β t} ] dt    (t=r²)     \n  = c a^{2d-2}e^{β a²}(1+γ a²)^p = M₀,\n\nwhich reproduces (6).  Hence (6) is the unique correct relation; any more elaborate formula must reduce to it.\n\n\n\nStep 3. Exterior field (task (c))  \nFor r ≥ a the whole mass M₀ acts as though concentrated at the centre, so from (1)\n\n  F_out(r)= M₀ / r^{d-1},   r ≥ a.                            (7)\n\n\n\nStep 4. Special case d = 4, β = 0, p = 2 (task (d))  \n\nSurface area Ω_4 = 2π².  Substitute d=4, β=0, p=2 in (5):\n\nρ_4(r) = 2c r^{2}(1+γ r²) / (2π²) · [3(1+γ r²)+2γ r²]  \n       = (c/π²) r²(1+γ r²)(3+5γ r²).                            (8)\n\nFrom (6) the amplitude becomes\n\n  c = M₀ /[a^{6}(1+γ a²)²].                                     (9)\n\nFinally, with d=4, (7) yields the exterior force\n\n  F_out(r)= M₀ / r³,   r ≥ a.                                 (10)",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.348264",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimensions:  The problem is set in an arbitrary d-dimensional Euclidean space, obliging the solver to generalise the classical three–dimensional shell theorem, work with the surface measure Ω_d and keep the dimensional parameter symbolic throughout.\n\n2. More variables & parameters:  Five independent parameters (c, β, γ, p, d) appear multiplicatively and in exponents, greatly complicating differentiation and integration compared with the single-parameter force laws of the original statements.\n\n3. Interaction of multiple concepts:  The solution requires  \n •  understanding the n-dimensional Newtonian field,  \n •  translating a measured force law into an integral equation for the enclosed mass,  \n •  differentiating products of several radial functions, and  \n •  relating the result back to mass density with appropriate geometric factors.\n\n4. Special functions:  While part (a) can be written in elementary form, part (b) forces the solver to manipulate non-elementary integrals and introduces the incomplete Gamma function (or, equivalently, hypergeometric functions) to express c explicitly.\n\n5. Multi-step reasoning:  The path “force ⇒ enclosed mass ⇒ derivative ⇒ density ⇒ re-integration ⇒ normalisation” is longer and more intricate than in the original problem, and each step is algebraically heavier because all quantities depend on d and the extra parameters β, γ, p.\n\n6. Layered specialisation:  The final sub-problem (d) shows how the general machinery collapses to concrete formulas, testing the solver’s ability to simplify higher-level expressions in a specific setting.\n\nThese added layers of abstraction and computation make the enhanced kernel variant markedly more challenging than both the original textbook exercise and the existing kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}