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)).

Question 1

Step 1: Convert units.
$r = 5 \, cm = 0.05 \, \mathrm{m}$, $\rho = 1.67 \times 10^{3} \, \mathrm{kg/m^3}$.

Step 2: Acceleration due to gravity on surface of uniform sphere:
$g = \frac{4}{3} \pi G \rho r$
where $G = 6.67 \times 10^{-11} \, m^3 \, kg^{-1 \, s^{-2}}$.

Step 3: Compute $G \rho$:
$G \rho = (6.67 \times 10^{-11})(1.67 \times 10^{3}) = 1.114 \times 10^{-7} \, m^3 \, s^{-2 \, m^{-3}} = 1.114 \times 10^{-7} \, s^{-2}$

Step 4: Compute $G \rho r$:
$G \rho r = (1.114 \times 10^{-7})(0.05) = 5.57 \times 10^{-9} \, m/s^2$

Step 5: Compute $\frac{4}{3} \pi = 4.1888$.
$g = 4.1888 \times 5.57 \times 10^{-9} = 2.33 \times 10^{-8} \, m/s^2$

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

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

Step 1: Mass from surface gravity:
$g = \frac{G M}{r^2} \implies M = \frac{g r^2}{G}$

Step 2: Compute $r^2 = (0.05)^2 = 2.5 \times 10^{-3} \, \mathrm{m^2}$.
$g r^2 = (6.7 \times 10^{-7})(2.5 \times 10^{-3}) = 1.675 \times 10^{-9} \, m^3/s^2$

Step 3:
$M = \frac{1.675 \times 10^{-9}}{6.67 \times 10^{-11}} = 2.51 \times 10^{1} \, kg = 25.1 \, kg$

Step 4: Volume $V = \frac{4}{3} \pi r^3$.
$r^3 = (0.05)^3 = 1.25 \times 10^{-4} \, \mathrm{m^3}$,
$V = 4.1888 \times 1.25 \times 10^{-4} = 5.236 \times 10^{-4} \, m^3$

Step 5: Density:
$\rho = \frac{M}{V} = \frac{25.1}{5.236 \times 10^{-4}} = 4.80 \times 10^{4} \, kg/m^3$

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

Question 3(a)

Step 1: Gravitational field due to point mass $m$ at $\vec{r}'$:
$\vec{g}(\vec{r}) = -G m \frac{\vec{r} - \vec{r}'}{|\vec{r} - \vec{r}'|^3}$

Step 2: For continuous distribution:
$\vec{g}(\vec{r}) = -G \int \rho(\vec{r}') \frac{\vec{r} - \vec{r}'}{|\vec{r} - \vec{r}'|^3} \, dV'$

Step 3: Take divergence ($\vec{R} = \vec{r} - \vec{r}'$):
$\nabla \cdot \vec{g} = -G \int \rho(\vec{r}') \nabla \cdot \left( \frac{\vec{R}}{R^3} \right) dV'$
Known: $\nabla \cdot \left( \frac{\vec{R}}{R^3} \right) = 4\pi \delta^3(\vec{R})$.

Step 4:
$\nabla \cdot \vec{g} = -4\pi G \rho(\vec{r})$

Step 5: By divergence theorem, for volume $V$ with surface $S$:
$\oint_S \vec{g} \cdot d\vec{A} = \int_V (\nabla \cdot \vec{g}) \, dV = -4\pi G \int_V \rho \, dV = -4\pi G M_enc$

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

Question 3(b) Hollow sphere, inner radius $a$, outer $b$, uniform $\rho$. Spherical symmetry, $\vec{g} = g(r) \hat{r}$.

Step 1: For $r < a$ (cavity), Gaussian sphere radius $r$: $M_enc = 0$.
$g(r) (4\pi r^2) = -4\pi G (0) \implies g(r) = 0$

Step 2: For $a < r < b$, $M_enc = \rho \frac{4}{3} \pi (r^3 - a^3)$.
$g(r) (4\pi r^2) = -4\pi G \rho \frac{4}{3} \pi (r^3 - a^3)$
$g(r) = -\frac{4\pi G \rho}{3} \frac{r^3 - a^3}{r^2}$

Step 3: For $r > b$, $M_enc = \rho \frac{4}{3} \pi (b^3 - a^3)$.
$g(r) = -G \frac{M_enc}{r^2} = -\frac{4\pi G \rho}{3} \frac{b^3 - a^3}{r^2}$

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}}}

Question 4

Step 1: Orbital radius $r = R_e + h = 6.4 \times 10^6 + 3.6 \times 10^7 = 4.24 \times 10^7 \, \mathrm{m}$.
$T = 24 \times 3600 = 8.64 \times 10^4 \, \mathrm{s}$.

Step 2: For circular orbit:
$\frac{G M m}{r^2} = m \frac{v^2}{r}, \quad v = \frac{2\pi r}{T}$
$\frac{G M}{r^3} = \left( \frac{2\pi}{T} \right)^2 \implies G M = \frac{4\pi^2 r^3}{T^2}$

Step 3: $r^3 = (4.24 \times 10^7)^3 = 7.622 \times 10^{22} \, \mathrm{m^3}$.
$4\pi^2 = 39.48$.
$4\pi^2 r^3 = 39.48 \times 7.622 \times 10^{22} = 3.008 \times 10^{24} \, \mathrm{m^3}$.

Step 4: $T^2 = (8.64 \times 10^4)^2 = 7.465 \times 10^9 \, \mathrm{s^2}$.
$G M = \frac{3.008 \times 10^{24}}{7.465 \times 10^9} = 4.027 \times 10^{14} \, m^3/s^2$

Step 5:
$M = \frac{4.027 \times 10^{14}}{6.67 \times 10^{-11}} = 6.04 \times 10^{24} \, kg \approx 6 \times 10^{24} \, kg$

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

Question 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: $T^2 \propto a^3$, where $T$ is orbital period, $a$ is semi-major axis.

For artificial satellites orbiting Earth: $T^2 \propto a^3$ (same form, with central mass $M_\ Earth$ constant).

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