Gravitation
Every mass attracts every other mass. From a falling apple to orbiting satellites and the dance of the planets — explore the universal force with live simulators, full derivations and real exam questions.
Newton's Law of Universal Gravitation
Two point masses $m_1$ and $m_2$ separated by a distance $r$ attract each other along the line joining them with a force:
$$F = \dfrac{G\,m_1 m_2}{r^2}$$where $G = 6.674\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$ is the universal gravitational constant.
- The force is always attractive and acts along the line joining the centres.
- It obeys Newton's third law — the two forces are equal and opposite.
- It is a central force and independent of the medium between the masses.
- It follows an inverse-square law: $F\propto\dfrac{1}{r^2}$.
Inverse-square law explorer
Change the masses and their separation and watch the gravitational force respond. Double the distance and the force drops to a quarter — the hallmark of the inverse-square law.
Gravitational Field Strength
The gravitational field strength (intensity) at a point is the force experienced by a unit mass placed there:
$$E = \dfrac{F}{m} = \dfrac{GM}{r^2}$$At the surface of the Earth (mass $M$, radius $R$) this field strength equals the acceleration due to gravity:
$$g = \dfrac{GM}{R^2} \quad\Rightarrow\quad GM = gR^2$$Variation of $g$
$g$ is maximum at the surface. It falls off as the inverse square with altitude, but decreases linearly with depth, reaching zero at the centre of the Earth.
Derivation — variation with height & depth
Let $g=\dfrac{GM}{R^2}$ be the surface value for an Earth of mass $M$, radius $R$ and (assumed uniform) density $\rho$.
Step 1 — At height $h$ above the surface
The distance from the centre is $R+h$, so:
$$g_h = \frac{GM}{(R+h)^2} = g\left(\frac{R}{R+h}\right)^2 = g\left(1+\frac{h}{R}\right)^{-2}$$For $h\ll R$, expanding to first order:
Step 2 — At depth $d$ below the surface
Only the sphere of radius $(R-d)$ beneath contributes. With uniform density, $M' = M\dfrac{(R-d)^3}{R^3}$, so:
$$g_d = \frac{GM'}{(R-d)^2} = \frac{GM(R-d)}{R^3} = g\,\frac{R-d}{R}$$At the centre $d=R$, giving $g=0$; the decrease with depth is linear.
How $g$ changes from the centre outward
Slide a point from the Earth's centre out into space. Inside the Earth $g$ rises linearly to its surface peak; outside it falls away as the inverse square.
Gravitational Potential & Energy
- Gravitational potential $V$ is the work done per unit mass in bringing a small mass from infinity to that point.
- Potential energy $U = mV$; both are negative because the field is attractive (bound system).
- Field and potential are linked by $E = -\dfrac{dV}{dr}$.
Derivation — potential & potential energy
Bring a unit mass from infinity to a point P a distance $r$ from a mass $M$, moving against the gravitational pull.
Step 1 — Work done by the field
At distance $x$ the field is $\dfrac{GM}{x^2}$. The potential is the work done by the field on a unit mass coming from infinity to $r$:
$$V = -\int_{\infty}^{r}\frac{GM}{x^2}\,dx = -GM\left[-\frac{1}{x}\right]_{\infty}^{r}$$Step 2 — Evaluate
$$V = -GM\left(\frac{1}{r}-0\right)$$The work needed to remove the mass $m$ entirely (to infinity) is therefore $+\dfrac{GMm}{r}$.
Kepler's Laws of Planetary Motion
- 1. Law of orbits: every planet moves in an ellipse with the Sun at one focus.
- 2. Law of areas: the line joining a planet to the Sun sweeps out equal areas in equal times — so a planet moves fastest at perihelion and slowest at aphelion. This is a statement of conservation of angular momentum.
- 3. Law of periods: the square of the period is proportional to the cube of the semi-major axis: $$T^2 \propto a^3$$
Derivation — Kepler's third law (circular orbit)
For a planet of mass $m$ in a circular orbit of radius $r$ about the Sun (mass $M$), gravity supplies the centripetal force.
Step 1 — Balance the forces
$$\frac{GMm}{r^2} = m\omega^2 r = m\left(\frac{2\pi}{T}\right)^2 r$$Step 2 — Solve for $T^2$
$$\frac{GM}{r^2} = \frac{4\pi^2 r}{T^2} \;\;\Rightarrow\;\; T^2 = \frac{4\pi^2}{GM}\,r^3$$Satellite Dynamics
- Close to the surface ($h\approx0$): $v_o=\sqrt{gR}\approx7.9\,$km/s, $T\approx84\,$min.
- Orbital velocity is independent of the satellite's mass.
- Total energy $E = -\dfrac{GMm}{2r}$ (negative → bound).
Derivation — orbital velocity & period
A satellite of mass $m$ moves in a circular orbit of radius $r=R+h$; gravity provides the centripetal force.
Step 1 — Centripetal balance
$$\frac{GMm}{r^2} = \frac{mv_o^2}{r} \;\;\Rightarrow\;\; v_o = \sqrt{\frac{GM}{r}}$$Step 2 — Using $GM=gR^2$
Step 3 — Period
$$T = \frac{2\pi r}{v_o} = 2\pi\sqrt{\frac{r^3}{GM}}$$Satellite orbit simulator
Raise the orbit and watch the satellite slow down and take longer to circle the Earth. Find the special altitude (~35,786 km) where the period hits 24 hours — a geostationary orbit.
Escape Velocity
- For Earth, $v_e\approx11.2\,$km/s — about $\sqrt2$ times the surface orbital speed.
- Escape velocity is independent of the mass and direction of projection of the body.
- It depends only on the planet's mass $M$ and radius $R$.
Derivation — escape velocity
A body of mass $m$ is thrown up from the surface so that it just reaches infinity with zero speed. Use conservation of energy.
Step 1 — Energy at the surface = energy at infinity
Total energy at launch must be zero (it just escapes, arriving at infinity with $v=0$, $U=0$):
$$\frac{1}{2}mv_e^2 + \left(-\frac{GMm}{R}\right) = 0$$Step 2 — Solve for $v_e$
$$\frac{1}{2}mv_e^2 = \frac{GMm}{R} \;\;\Rightarrow\;\; v_e = \sqrt{\frac{2GM}{R}}$$Step 3 — In terms of $g$ and $v_o$
Escape velocity across the Solar System
Pick a world and see how fast you must launch to break free. A tiny Moon needs just 2.4 km/s, while the mighty Sun demands over 600 km/s.
Practical Application — GPS
The Global Positioning System is a constellation of about 24–31 satellites in ~20,200 km orbits (period ≈ 12 h), each carrying an atomic clock and broadcasting its position and time.
- A receiver measures the travel time of each signal to find its distance from that satellite ($d = c\times t$).
- Distances from three satellites fix the position by trilateration; a fourth corrects the receiver's clock.
- The satellites' orbits obey exactly the gravitation laws of this chapter — and relativistic corrections (special + general) are applied to keep the clocks accurate to nanoseconds.