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Show that the radius of gyration about an axis through the center of a hollow cylinder of external radius 'a' and internal radius 'b' is sqrt(2/5 * (a⁵ - b⁵) / (a³ - b³)).

Physics

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23 steps

Part 1· 18 steps

  1. Convert units.

    r=5 cm=0.05 mr = 5 \, \mathrm{cm} = 0.05 \, \mathrm{m}, ρ=1.67×103 kg/m3\rho = 1.67 \times 10^{3} \, \mathrm{kg/m^3}.

  2. Acceleration due to gravity on surface of uniform sphere

    g=43πGρrg = \frac{4}{3} \pi G \rho r

    where G=6.67×10−11 m3 kg−1 s−2G = 6.67 \times 10^{-11} \, \mathrm{m^3 \, kg^{-1} \, s^{-2}}.

  3. Compute GρG \rho

    Gρ=(6.67×10−11)(1.67×103)=1.114×10−7 m3 s−2 m−3=1.114×10−7 s−2G \rho = (6.67 \times 10^{-11})(1.67 \times 10^{3}) = 1.114 \times 10^{-7} \, \mathrm{m^3 \, s^{-2} \, m^{-3}} = 1.114 \times 10^{-7} \, \mathrm{s^{-2}}

  4. Compute GρrG \rho r

    Gρr=(1.114×10−7)(0.05)=5.57×10−9 m/s2G \rho r = (1.114 \times 10^{-7})(0.05) = 5.57 \times 10^{-9} \, \mathrm{m/s^2}

  5. Compute 43π=4.1888\frac{4}{3} \pi = 4.1888.

    g=4.1888×5.57×10−9=2.33×10−8 m/s2g = 4.1888 \times 5.57 \times 10^{-9} = 2.33 \times 10^{-8} \, \mathrm{m/s^2}

    2.33 \times 10^{-8} m/s^2

    Question 2 (assuming same r=0.05 mr = 0.05 \, \mathrm{m})

  6. Mass from surface gravity

    g=GMr2  ⟹  M=gr2Gg = \frac{G M}{r^2} \implies M = \frac{g r^2}{G}

  7. Compute r2=(0.05)2=2.5×10−3 m2r^2 = (0.05)^2 = 2.5 \times 10^{-3} \, \mathrm{m^2}.

    gr2=(6.7×10−7)(2.5×10−3)=1.675×10−9 m3/s2g r^2 = (6.7 \times 10^{-7})(2.5 \times 10^{-3}) = 1.675 \times 10^{-9} \, \mathrm{m^3/s^2}

  8. Step 8

    M=1.675×10−96.67×10−11=2.51×101 kg=25.1 kgM = \frac{1.675 \times 10^{-9}}{6.67 \times 10^{-11}} = 2.51 \times 10^{1} \, \mathrm{kg} = 25.1 \, \mathrm{kg}

  9. Volume V=43πr3V = \frac{4}{3} \pi r^3.

    r3=(0.05)3=1.25×10−4 m3r^3 = (0.05)^3 = 1.25 \times 10^{-4} \, \mathrm{m^3},

    V=4.1888×1.25×10−4=5.236×10−4 m3V = 4.1888 \times 1.25 \times 10^{-4} = 5.236 \times 10^{-4} \, \mathrm{m^3}

  10. Density

    ρ=MV=25.15.236×10−4=4.80×104 kg/m3\rho = \frac{M}{V} = \frac{25.1}{5.236 \times 10^{-4}} = 4.80 \times 10^{4} \, \mathrm{kg/m^3}

    M = 25.1 kg, \rho = 4.80 \times 10^{4} kg/m^3

    Question 3(a)

  11. Gravitational field due to point mass mm at r⃗′\vec{r}'

    g⃗(r⃗)=−Gmr⃗−r⃗′∣r⃗−r⃗′∣3\vec{g}(\vec{r}) = -G m \frac{\vec{r} - \vec{r}'}{|\vec{r} - \vec{r}'|^3}

  12. For continuous distribution

    g⃗(r⃗)=−G∫ρ(r⃗′)r⃗−r⃗′∣r⃗−r⃗′∣3 dV′\vec{g}(\vec{r}) = -G \int \rho(\vec{r}') \frac{\vec{r} - \vec{r}'}{|\vec{r} - \vec{r}'|^3} \, dV'

  13. Take divergence (R⃗=r⃗−r⃗′\vec{R} = \vec{r} - \vec{r}')

    ∇⋅g⃗=−G∫ρ(r⃗′)∇⋅(R⃗R3)dV′\nabla \cdot \vec{g} = -G \int \rho(\vec{r}') \nabla \cdot \left( \frac{\vec{R}}{R^3} \right) dV'

    Known: ∇⋅(R⃗R3)=4πδ3(R⃗)\nabla \cdot \left( \frac{\vec{R}}{R^3} \right) = 4\pi \delta^3(\vec{R}).

  14. Step 14

    ∇⋅g⃗=−4πGρ(r⃗)\nabla \cdot \vec{g} = -4\pi G \rho(\vec{r})

  15. By divergence theorem, for volume VV with surface SS

    ∮Sg⃗⋅dA⃗=∫V(∇⋅g⃗) dV=−4πG∫Vρ dV=−4πGMenc\oint_S \vec{g} \cdot d\vec{A} = \int_V (\nabla \cdot \vec{g}) \, dV = -4\pi G \int_V \rho \, dV = -4\pi G M_\mathrm{enc}

    \oint \vec{g} \cdot d\vec{A} = -4\pi G M_enc

    Question 3(b) Hollow sphere, inner radius aa, outer bb, uniform ρ\rho. Spherical symmetry, g⃗=g(r)r^\vec{g} = g(r) \hat{r}.

  16. For r<ar < a (cavity), Gaussian sphere radius rr

    Menc=0M_\mathrm{enc} = 0.

    g(r)(4πr2)=−4πG(0)  ⟹  g(r)=0g(r) (4\pi r^2) = -4\pi G (0) \implies g(r) = 0

  17. For a<r<ba < r < b, Menc=ρ43π(r3−a3)M_\mathrm{enc} = \rho \frac{4}{3} \pi (r^3 - a^3).

    g(r)(4πr2)=−4πGρ43π(r3−a3)g(r) (4\pi r^2) = -4\pi G \rho \frac{4}{3} \pi (r^3 - a^3) g(r)=−4πGρ3r3−a3r2g(r) = -\frac{4\pi G \rho}{3} \frac{r^3 - a^3}{r^2}

  18. For r>br > b, Menc=ρ43π(b3−a3)M_\mathrm{enc} = \rho \frac{4}{3} \pi (b^3 - a^3).

    g(r)=−GMencr2=−4πGρ3b3−a3r2g(r) = -G \frac{M_\mathrm{enc}}{r^2} = -\frac{4\pi G \rho}{3} \frac{b^3 - a^3}{r^2}

    r<a: g=0;
    a<r<b: g=-\frac{4\pi G\rho}{3}\frac{r^{3}-a^{3}}{r^{2}};
    r>b: g=-\frac{4\pi G\rho}{3}\frac{b^{3}-a^{3}}{r^{2}}

Part 4· 5 steps

  1. Orbital radius r=Re+h=6.4×106+3.6×107=4.24×107 mr = R_e + h = 6.4 \times 10^6 + 3.6 \times 10^7 = 4.24 \times 10^7 \, \mathrm{m}.

    T=24×3600=8.64×104 sT = 24 \times 3600 = 8.64 \times 10^4 \, \mathrm{s}.

  2. For circular orbit

    GMmr2=mv2r,v=2πrT\frac{G M m}{r^2} = m \frac{v^2}{r}, \quad v = \frac{2\pi r}{T} GMr3=(2πT)2  ⟹  GM=4π2r3T2\frac{G M}{r^3} = \left( \frac{2\pi}{T} \right)^2 \implies G M = \frac{4\pi^2 r^3}{T^2}

  3. r3=(4.24×107)3=7.622×1022 m3r^3 = (4.24 \times 10^7)^3 = 7.622 \times 10^{22} \, \mathrm{m^3}.

    4π2=39.484\pi^2 = 39.48.
    4π2r3=39.48×7.622×1022=3.008×1024 m34\pi^2 r^3 = 39.48 \times 7.622 \times 10^{22} = 3.008 \times 10^{24} \, \mathrm{m^3}.

  4. T2=(8.64×104)2=7.465×109 s2T^2 = (8.64 \times 10^4)^2 = 7.465 \times 10^9 \, \mathrm{s^2}.

    GM=3.008×10247.465×109=4.027×1014 m3/s2G M = \frac{3.008 \times 10^{24}}{7.465 \times 10^9} = 4.027 \times 10^{14} \, \mathrm{m^3/s^2}

  5. Step 5

    M=4.027×10146.67×10−11=6.04×1024 kg≈6×1024 kgM = \frac{4.027 \times 10^{14}}{6.67 \times 10^{-11}} = 6.04 \times 10^{24} \, \mathrm{kg} \approx 6 \times 10^{24} \, \mathrm{kg}

    M \approx 6 \times 10^{24} kg

Part 5

Kepler's second law: The radius vector from the Sun to a planet sweeps out equal areas in equal time intervals. This implies conservation of angular momentum.

Kepler's third law for planets orbiting the Sun: T2∝a3T^2 \propto a^3, where TT is orbital period, aa is semi-major axis.

For artificial satellites orbiting Earth: T2∝a3T^2 \propto a^3 (same form, with central mass M EarthM_\ Earth constant).

T^{2} \propto a^{3}}

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