Advanced University Physics

Proof: Variation of $g$ with Altitude

For altitudes $h \ll R_E$, a first-order approximation can be found using the binomial expansion $(1+x)^n \approx 1+nx$ for small $x$. We can write $g(h)$ as:

Where $g_0 = GM_E/R_E^2$ is the acceleration at the surface. Using the binomial approximation with $x=h/R_E$ and $n=-2$:

Proof: Variation of $g$ with Depth

Assuming the Earth has a uniform density $\rho$, the gravitational force on a mass $m$ at a distance $r$ from the center ($r < R_E$) is due only to the mass enclosed within the sphere of radius $r$, according to the Shell Theorem. The mass enclosed is $M_{enc} = M_E (r^3/R_E^3)$. The gravitational force is $F_g(r) = G M_{enc}m/r^2$.

Since $F_g = mg(r)$, the gravitational acceleration inside the Earth is:

This shows that $g$ decreases linearly as one approaches the center of the Earth, becoming zero at the center.

Proof: Total Energy of a Circular Orbit

For a satellite in a circular orbit, the gravitational force provides the centripetal force: $G Mm/r^2 = mv^2/r$. The kinetic energy is $K = \frac{1}{2} mv^2 = GMm/(2r)$. The potential energy is $U = - GMm/r$. The total energy is the sum, a result known as the Virial Theorem for this system:

The negative energy signifies a bound system. An object with $E \ge 0$ is unbound and will follow a parabolic or hyperbolic trajectory.

Proof: Escape Velocity

Escape velocity is the minimum speed an object must have to escape a gravitational field, i.e., to reach $r=\infty$ with zero final speed. We use conservation of energy. The minimum energy for escape is $E_{total}=0$. Initial energy at radius $R$ must equal the final energy at infinity:

Proof of Kepler's Second Law (Law of Areas)

This law is a direct consequence of the conservation of angular momentum. Since gravity is a central force, the torque $\vec{\tau} = \vec{r} \times \vec{F}_g$ is zero, which implies the angular momentum $\vec{L} = \vec{r} \times \vec{p}$ is constant. The infinitesimal area $dA$ swept by the position vector $\vec{r}$ in time $dt$ is $dA = \frac{1}{2} |\vec{r} \times d\vec{r}| = \frac{1}{2} |\vec{r} \times \vec{v}dt|$. The rate of change of area is:

Proof of Kepler's Third Law (Law of Periods)

The total area of an ellipse is $A = \pi ab$, where $a$ is the semi-major axis. The orbital period is $T = A / (dA/dt)$. Using $dA/dt = L/2m$ and expressions for an ellipse relating $L$ and $a$, we find:

Squaring both sides gives the celebrated result:

Proof of Shell Theorem via Gauss's Law for Gravity

Gauss's Law for Gravity states that the flux of the gravitational field through any closed surface is proportional to the enclosed mass: $\oint \vec{g} \cdot d\vec{A} = -4\pi G M_{enc}$.

Outside the shell ($r > R$): We choose a spherical Gaussian surface of radius $r$. By symmetry, $\vec{g}$ is radial. So, $g(4\pi r^2) = -4\pi G M_{enc}$. The enclosed mass is the entire shell, $M_{enc}=M$.

Inside the shell ($r < R$): The enclosed mass is zero, $M_{enc}=0$, so $g=0$. This elegantly proves both parts of the Shell Theorem.