Gravitational Fields

$\boxed{F = \frac{GMm}{r^2}}$

Where:

• $G = 6.67 \times 10^{-11}$ N m² kg⁻² (gravitational constant)
• $M$, $m$ = masses (kg)
• $r$ = distance between centres (m)

The force is always attractive and acts along the line joining the centres.

$\boxed{g = \frac{F}{m} = \frac{GM}{r^2}}$

Units: N/kg or m/s²

• At Earth's surface: $g = 9.81$ N/kg
• $g$ decreases with $r^2$ (inverse square law)
• Inside a uniform sphere: $g$ increases linearly from centre to surface

Field Lines

• Point towards the mass (direction of force on test mass)
• Closer lines = stronger field
• Radial field for point/spherical masses
• Uniform field near Earth's surface (parallel lines, constant $g$)

$\boxed{V = -\frac{GM}{r}}$

• Units: J/kg
• Always negative (zero at infinity)
• The work done per unit mass to bring a test mass from infinity to that point

Gravitational Potential Energy

$E_p = mV = -\frac{GMm}{r}$

$g = -\frac{dV}{dr}$

$g$ = negative gradient of the $V$ vs $r$ graph.

For a circular orbit, gravitational force provides centripetal force:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

$v = \sqrt{\frac{GM}{r}}$

Period of Orbit

$T = 2\pi r/v = 2\pi\sqrt{\frac{r^3}{GM}}$

This gives Kepler's Third Law: $T^2 \propto r^3$

Geostationary Orbit

• Period = 24 hours (synchronous with Earth)
• Orbits above the equator
• Used for communications satellites
• $r \approx 42{,}200$ km from Earth's centre

The minimum speed to escape a gravitational field (reach infinity with zero KE):

$\frac{1}{2}mv^2 = \frac{GMm}{r}$ $\boxed{v_{\text{esc}} = \sqrt{\frac{2GM}{r}}}$

• $F = GMm/r^2$; $g = GM/r^2$; $V = -GM/r$
• $g = -dV/dr$
• Orbital speed: $v = \sqrt{GM/r}$; Kepler: $T^2 \propto r^3$
• Escape velocity: $v = \sqrt{2GM/r}$
• All potentials are negative; zero at infinity