Part 2 · Chapter 13

Gravitation

The one force that reaches across the whole universe — binding apples to Earth, moons to planets, and galaxies to galaxies

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 60 min
i What you'll learn
  • Newton's law of gravitation \(F = Gm_{1}m_{2}/r^{2}\), and the shell theorem that lets a sphere act as a point at its center.
  • How to add forces by superposition and find the gravitational acceleration \(a_{g}\) near and inside Earth.
  • Gravitational potential energy \(U = -GMm/r\) and the escape speed \(v = \sqrt{2GM/R}\).
  • Kepler's three laws of orbits, areas, and periods — and why the law of areas is just conservation of angular momentum.
  • The orbits and energy of satellites (\(E = -GMm/2r\)), and Einstein's reinterpretation of gravity as curved spacetime.
Section 13-1

What Is Physics?

One long-standing goal of physics is to understand the gravitational force — the force that holds you to Earth, the Moon to Earth, and Earth to the Sun. Its reach does not stop there. It binds the hundreds of billions of stars in our Milky Way, holds together the Local Group of galaxies (including Andromeda, \(2.3 \times 10^{6}\) light-years away), draws the Local Supercluster toward the Great Attractor, and tries to hold together the entire expanding universe. The same force, taken to an extreme, creates black holes: the collapsed remnants of massive stars whose surface gravity is so strong that not even light escapes. Our understanding of all of it begins with one elegant law written by Isaac Newton.

Section 13-2

Newton's Law of Gravitation

In 1665, the 23-year-old Newton made one of the great unifications in physics: the force that holds the Moon in orbit is the same force that makes an apple fall. He went further — every body in the universe attracts every other body. The familiar tug of Earth simply overwhelms the feeble attraction between everyday objects (you and a nearby apple attract each other with less than the weight of a dust speck).

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Newton's law of gravitation
F = G m1m2 / r²

Every particle attracts every other with a force proportional to the product of their masses and inversely proportional to the square of their separation. The gravitational constant is \(G = 6.67 \times 10^{-11}\,\mathrm{N\cdot m^{2}/kg^{2}}\). In vector form, \(\vec{F} = G\,m_{1}m_{2}/r^{2}\,\hat{r}\), directed along the line joining the particles.

The two forces of an attracting pair are equal and opposite — a third-law pair — so we speak of the gravitational force between two particles. No object placed between them can shield one from the other. And although the law is written for particles, Newton rescued the apple–Earth case with a theorem of remarkable economy:

The shell theorem (mass outside)
\[ \text{A uniform spherical shell attracts an external particle as if all its mass were at its center.} \]
Earth is a nest of such shells, so from the apple's point of view Earth behaves like a point mass at its center. This is why Eq. 13-1, written for particles, works for planets and apples alike.
Equal forces, unequal accelerations. Earth pulls down on an apple with ≈0.80 N; the apple pulls up on Earth with exactly 0.80 N. Yet the apple accelerates at \(9.8\,\mathrm{m/s^{2}}\) while Earth's acceleration (in the apple–Earth center-of-mass frame) is only about \(1 \times 10^{-25}\,\mathrm{m/s^{2}}\). Same force, wildly different masses.
Section 13-3

Gravitation and the Principle of Superposition

To find the net force on one particle from many, we use the principle of superposition: compute each pairwise force separately, then add them as vectors. There is no shielding and no interference — each pull acts as if the others were not there.

Superposition — sum of forces, then the continuous limit
\[ \vec{F}_{1,\text{net}} = \sum_{i=2}^{n} \vec{F}_{1i} \qquad \vec{F}_{1} = \int d\vec{F} \]
For a real extended body, divide it into differential masses dm, treat each as a particle, and integrate. If the body is a uniform sphere or shell, skip the integral — put its mass at its center and use Eq. 13-1.
Section 13-4

Gravitation Near Earth's Surface

Treating Earth as a uniform sphere of mass \(M\), the force on a particle of mass \(m\) at distance \(r\) from the center is \(F = GMm/r^{2}\). Setting \(F = ma_{g}\) gives the gravitational acceleration:

Gravitational acceleration at distance r
\[ a_{g} = \frac{GM}{r^{2}} \]
It falls off as the inverse square of distance from Earth's center — yet remains substantial even in orbit (≈8.70 m/s² at 400 km, the altitude of the old shuttle orbit).

The measured free-fall acceleration \(g\) differs slightly from \(a_{g}\) for three reasons: Earth's mass is not uniformly distributed, Earth is not a perfect sphere (it bulges at the equator), and Earth rotates. The rotation effect is the cleanest to quantify — a crate on a scale at the equator needs a small inward net force for its circular motion, so the scale reads slightly less than the true gravitational pull:

Rotation lowers the measured weight
\[ g = a_{g} - \omega^{2}R \]
The correction ω²R is largest at the equator and is only about 0.034 m/s² — small against 9.8, so neglecting the difference between g and a_g is usually justified.
Tidal stretching. Because \(a_{g}\) changes with \(r\), an extended body feels a slightly stronger pull on its near side than its far side. Differentiating \(a_{g} = GM/r^{2}\) gives \(da_{g} = -2GM\,dr/r^{3}\). For an astronaut in low orbit this head-to-feet difference is a negligible microgravity tickle; near a black hole the same formula yields a brutal, body-stretching "spaghettification."
Section 13-5

Gravitation Inside Earth

The shell theorem has a striking companion for the interior: a uniform shell exerts no net gravitational force on a particle inside it. The pulls from all parts of the shell cancel exactly. So for a particle at radius \(r\) inside a uniform Earth, only the mass within radius \(r\) matters — the outer shells contribute nothing.

Writing the enclosed mass as \(M_{\text{ins}} = \rho\,\tfrac{4}{3}\pi r^{3}\) and substituting into \(F = GmM_{\text{ins}}/r^{2}\), the \(r^{2}\) in the denominator is overpowered, and the force grows linearly with depth toward the center:

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Force inside a uniform Earth — Hooke's-law form
F = (4πGmρ / 3) r  ⟹  F⃗ = −K r⃗

The force is proportional to distance from the center and points inward — exactly Hooke's law. A capsule dropped through a pole-to-pole tunnel would oscillate like a mass on a spring, reaching zero force (not maximum) at the center. For the real non-uniform Earth, the force actually rises a little on the way down before falling, because the dense core dominates.

Section 13-6

Gravitational Potential Energy

Near Earth's surface we used \(U = mgy\) with constant \(g\). For two particles separated by an arbitrary distance \(r\), we choose the zero of potential energy at infinite separation. Because gravity does positive work as the particles approach, \(U\) is negative for every finite separation:

Gravitational potential energy of a pair
\[ U = -\frac{GMm}{r} \]
A property of the system, not of either particle alone. It approaches zero as r → ∞ and becomes more negative as the particles draw together. For many particles, sum Eq. 13-21 over every pair.

Gravity is conservative, so the work it does — and hence \(\Delta U\) — is independent of the path taken. Differentiating \(U(r)\) recovers the force, \(F = -dU/dr = -GMm/r^{2}\), neatly closing the loop back to Newton's law. One of the most useful consequences is the escape speed: set the total energy at the surface to zero (kinetic plus potential), since a projectile that just barely escapes arrives at infinity with neither.

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Escape speed from a body of mass M and radius R
v = √(2GM / R)

Independent of the projectile's mass and of its launch direction. Firing eastward helps only because the launch site already moves eastward with the planet's rotation (≈1500 km/h at Cape Canaveral).

Table 13-2 · Some escape speeds
BodyMass (kg)Radius (m)Escape speed (km/s)
Ceres (asteroid)1.17 × 10²¹3.8 × 10⁵0.64
Earth's Moon7.36 × 10²²1.74 × 10⁶2.38
Earth5.98 × 10²⁴6.37 × 10⁶11.2
Jupiter1.90 × 10²⁷7.15 × 10⁷59.5
Sun1.99 × 10³⁰6.96 × 10⁸618
Neutron star2 × 10³⁰1 × 10⁴2 × 10⁵
Section 13-7

Planets and Satellites: Kepler's Laws

Working from Tycho Brahe's painstaking naked-eye observations, Johannes Kepler distilled three empirical laws of planetary motion. Newton later showed they all follow from his law of gravitation. They hold equally for planets around the Sun and for satellites around any massive central body.

1 The law of orbits

All planets move in elliptical orbits with the Sun at one focus. A circle is the special case of zero eccentricity; Earth's orbit is nearly circular (\(e = 0.0167\)).

2 The law of areas

The line from Sun to planet sweeps out equal areas in equal times (\(dA/dt = L/2m\) = const). The planet moves fastest at perihelion, slowest at aphelion.

3 The law of periods

\(T^{2} \propto a^{3}\): the square of the period scales with the cube of the semimajor axis.

The second law is angular-momentum conservation in disguise. The area swept per unit time is \(dA/dt = \tfrac{1}{2}r^{2}\omega = L/2m\). Since gravity is a central force (it exerts no torque about the Sun), \(L\) is constant — and so is \(dA/dt\). Kepler's equal-areas rule and the conservation of \(L\) are the very same statement.
Kepler's law of periods (circular orbit radius r, or semimajor axis a)
\[ T^{2} = \left( \frac{4\pi^{2}}{GM} \right) r^{3} \]
Apply Newton's second law to a circular orbit (gravity supplies the centripetal force) and the result drops out. The bracket depends only on the central mass M, so T²/a³ is the same for every body orbiting it.
Table 13-3 · Kepler's law of periods for the solar system
PlanetSemimajor axis a (10¹⁰ m)Period T (y)T²/a³ (10⁻³⁴ y²/m³)
Mercury5.790.2412.99
Venus10.80.6153.00
Earth15.01.002.96
Mars22.81.882.98
Jupiter77.811.93.01
Saturn14329.52.98
Uranus28784.02.98
Neptune4501652.99
Pluto5902482.99

The near-constant final column is Kepler's third law made visible — one number, the same across the solar system, set entirely by the Sun's mass.

Section 13-8

Satellites: Orbits and Energy

As a satellite orbits, its kinetic and potential energies trade back and forth, but its mechanical energy stays constant. For a circular orbit, applying Newton's second law (\(GMm/r^{2} = mv^{2}/r\)) gives the kinetic energy, which combines with the potential energy into a clean result:

Energies of a satellite in a circular orbit
\[ K = \frac{GMm}{2r} \qquad U = -\frac{GMm}{r} \qquad E = K + U = -\frac{GMm}{2r} \]
Note E = −K: the total energy is the negative of the kinetic energy. For an elliptical orbit, replace r with the semimajor axis a, giving E = −GMm/2a.
Energy depends only on the size of the orbit, not its shape. All ellipses with the same semimajor axis \(a\) share the same total energy \(E = -GMm/2a\), regardless of eccentricity. A counter-intuitive corollary: firing a forward thruster (removing energy) drops the satellite into a lower, faster orbit — friction and thrust against the direction of motion speed an orbiting body up over the long run.
Section 13-9

Einstein and Gravitation

Einstein's route to general relativity began with a single thought: a person in free fall does not feel their own weight. This is the principle of equivalence — gravitation and acceleration are locally indistinguishable. A physicist sealed in a box cannot tell whether the box rests on Earth or accelerates through deep space at \(9.8\,\mathrm{m/s^{2}}\); the scale reads the same and a dropped object falls the same way.

From there Einstein recast gravity not as a force but as the curvature of spacetime produced by mass. Two ships sailing due south from the equator draw together not because a force pulls them, but because Earth's surface is curved; likewise two apples dropped side by side converge because mass curves the space around Earth. Light feels this curvature too — passing a galaxy or black hole, its path bends (gravitational lensing), sometimes wrapping a distant quasar's image into a luminous Einstein ring. Whether gravity is best described as curved spacetime, a force, or the exchange of hypothetical gravitons remains, at the cosmological and quantum extremes, not fully understood.

Worked Examples

Putting It to Work

1 Net force on a particle from two others (2D)

Problem. Particle 1 (\(m_{1} = 6.0\,\mathrm{kg}\)) sits at a corner. Particle 2 (\(m_{2} = 4.0\,\mathrm{kg}\)) is a distance \(a = 2.0\,\mathrm{cm}\) away along \(+y\); particle 3 (\(m_{3} = 4.0\,\mathrm{kg}\)) is a distance \(2a\) away along \(-x\). Find the net gravitational force on particle 1.

Solution. Each force points toward the responsible particle. Compute magnitudes with Eq. 13-1, then add as perpendicular components.

Magnitudes, then vector sum
\[\begin{gathered} F_{12} = \frac{Gm_{1}m_{2}}{a^{2}} \approx 4.0 \times 10^{-6}\,\mathrm{N} \;(+y), \qquad F_{13} = \frac{Gm_{1}m_{3}}{(2a)^{2}} \approx 1.0 \times 10^{-6}\,\mathrm{N} \;(-x) \\ F_{1,\text{net}} = \sqrt{F_{12}^{2} + F_{13}^{2}} \approx 4.1 \times 10^{-6}\,\mathrm{N}, \qquad \theta \approx 104^{\circ} \end{gathered}\]

The doubled distance to particle 3 quartered its pull. The net direction (104° from \(+x\)) lies between the two forces — a good sanity check, and a reminder that a calculator's \(\tan^{-1}\) needs the quadrant checked by hand.

2 Escape speed from Earth

Problem. Find the minimum speed a projectile needs at Earth's surface to escape to infinity. Use \(M = 5.98 \times 10^{24}\,\mathrm{kg}\), \(R = 6.37 \times 10^{6}\,\mathrm{m}\).

Solution. Set total energy at the surface to zero, since a just-escaping projectile arrives at infinity at rest with \(U = 0\).

Energy conservation: K + U = 0
\[ \tfrac{1}{2}mv^{2} - \frac{GMm}{R} = 0 \;\Rightarrow\; v = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2(6.67\times10^{-11})(5.98\times10^{24})}{6.37\times10^{6}}} \approx 11.2\,\mathrm{km/s} \]

The mass \(m\) cancelled — a marble and a meteor need the same escape speed.

3 Asteroid falling toward Earth — energy conservation

Problem. An asteroid heads straight at Earth with speed \(v_{i} = 12\,\mathrm{km/s}\) when it is \(10R_{E}\) from Earth's center. Neglecting the atmosphere, find its impact speed \(v_{f}\) at the surface.

Solution. Mechanical energy is conserved; Earth's recoil is negligible, so the kinetic energy is the asteroid's alone.

K_f + U_f = K_i + U_i
\[\begin{gathered} \tfrac{1}{2}v_{f}^{2} - \frac{GM}{R_{E}} = \tfrac{1}{2}v_{i}^{2} - \frac{GM}{10R_{E}} \\ v_{f}^{2} = v_{i}^{2} + \frac{2GM}{R_{E}}\left(1 - \tfrac{1}{10}\right) \;\Rightarrow\; v_{f} \approx 16\,\mathrm{km/s} \end{gathered}\]

Even a 5 m asteroid arriving at this speed could release Hiroshima-scale energy — a sobering reminder that escape speed and impact speed are the same physics run backward.

4 Kepler's law of periods — Comet Halley

Problem. Comet Halley orbits the Sun with period \(T = 76\,\mathrm{y}\) and perihelion distance \(R_{p} = 8.9 \times 10^{10}\,\mathrm{m}\). Find (a) its semimajor axis and aphelion distance, and (b) its eccentricity. Use \(M_{\odot} = 1.99 \times 10^{30}\,\mathrm{kg}\).

Solution. Solve the law of periods for \(a\) (with \(T = 2.4 \times 10^{9}\,\mathrm{s}\)), then use the ellipse geometry \(R_{a} = 2a - R_{p}\) and \(ea = a - R_{p}\).

a from the period, then R_a and e
\[\begin{gathered} a = \left( \frac{GMT^{2}}{4\pi^{2}} \right)^{1/3} \approx 2.7 \times 10^{12}\,\mathrm{m} \\ R_{a} = 2a - R_{p} \approx 5.3 \times 10^{12}\,\mathrm{m}, \qquad e = 1 - \frac{R_{p}}{a} \approx 0.97 \end{gathered}\]

An eccentricity near unity means a long, thin ellipse — Halley swings from just inside Venus's orbit out to near Pluto's.

5 Mechanical energy of an orbiting satellite

Problem. A 7.20 kg ball is in a circular orbit at altitude \(h = 350\,\mathrm{km}\). Find its mechanical energy. (\(R_{E} = 6370\,\mathrm{km}\).)

Solution. The orbital radius is \(r = R_{E} + h = 6.72 \times 10^{6}\,\mathrm{m}\) (not just the altitude). Then use \(E = -GMm/2r\).

Total orbital energy
\[ E = -\frac{GMm}{2r} = -\frac{(6.67\times10^{-11})(5.98\times10^{24})(7.20)}{2(6.72\times10^{6})} \approx -214\,\mathrm{MJ} \]

On the launchpad the same ball has \(E_{0} = U_{0} \approx -451\,\mathrm{MJ}\), so reaching orbit raises its mechanical energy by ≈237 MJ — a few dollars' worth of electricity. The real expense of spaceflight is clearly not the orbital energy itself.

Review

Chapter Summary

Law of gravitation

\(F = Gm_{1}m_{2}/r^{2}\) with \(G = 6.67\times10^{-11}\). A uniform sphere or shell acts on an external particle as a point mass at its center.

Superposition

Net force is the vector sum of pairwise forces, \(\vec{F}_{1,\text{net}} = \sum \vec{F}_{1i}\); for extended bodies, integrate \(\vec{F}_{1} = \int d\vec{F}\).

Acceleration near Earth

\(a_{g} = GM/r^{2}\). Measured \(g\) differs slightly due to non-uniform mass, oblateness, and rotation (\(g = a_{g} - \omega^{2}R\)).

Inside Earth

A shell exerts no force on an interior particle; only the enclosed mass \(M_{\text{ins}} = \rho\tfrac{4}{3}\pi r^{3}\) acts, giving a Hooke's-law force \(F \propto r\).

Energy & escape

\(U = -GMm/r\) (zero at infinity); escape speed \(v = \sqrt{2GM/R}\), independent of mass and direction.

Kepler's laws

Ellipses with the Sun at a focus; equal areas in equal times (= conserved \(L\)); \(T^{2} = (4\pi^{2}/GM)a^{3}\).

Orbital energy

Circular: \(K = GMm/2r\), \(E = -GMm/2r = -K\). Elliptical: \(E = -GMm/2a\), depending only on \(a\).

Einstein's view

Gravitation and acceleration are equivalent; gravity is the curvature of spacetime produced by mass, bending both matter and light.

Practice

Problems

Force problems reward careful vector bookkeeping and the inverse-square law; energy problems reward conservation of \(K + U\) with \(U = -GMm/r\); orbital problems hinge on Kepler's third law and \(E = -GMm/2a\). Useful constants: \(G = 6.67\times10^{-11}\,\mathrm{N\cdot m^{2}/kg^{2}}\), \(M_{E} = 5.98\times10^{24}\,\mathrm{kg}\), \(R_{E} = 6.37\times10^{6}\,\mathrm{m}\).

  1. A mass \(M\) is split into two pieces, \(m\) and \(M - m\), held a fixed distance apart. What ratio \(m/M\) maximizes the gravitational force between them?
  2. What separation between a 5.2 kg particle and a 2.4 kg particle gives a mutual gravitational attraction of \(2.3 \times 10^{-12}\,\mathrm{N}\)?
  3. Three 5.00 kg spheres lie in an L-shape: B at the origin, A a distance \(d_{1} = 0.300\,\mathrm{m}\) up the \(y\) axis, C a distance \(d_{2} = 0.400\,\mathrm{m}\) along the \(x\) axis. Find the magnitude and direction of the net gravitational force on B.
  4. (a) What will an object weigh on the Moon's surface if it weighs 100 N on Earth? (b) How many Earth radii from Earth's center must the same object be to weigh that Moon value?
  5. At what altitude above Earth's surface is the gravitational acceleration equal to \(4.9\,\mathrm{m/s^{2}}\)?
  6. A solid uniform sphere has mass \(1.0 \times 10^{4}\,\mathrm{kg}\) and radius 1.0 m. Find the gravitational force it exerts on a 1 kg particle at (a) 1.5 m and (b) 0.50 m from its center. (c) Write a general expression for \(r \le 1.0\,\mathrm{m}\).
  7. What is the escape speed from a spherical asteroid of radius 500 km whose surface gravitational acceleration is \(3.0\,\mathrm{m/s^{2}}\)?
  8. (a) What linear speed must an Earth satellite have to stay in a circular orbit 160 km above the surface? (b) What is its period of revolution?
  9. An orbiting satellite is to stay fixed over one spot on Earth's equator (a geosynchronous orbit). What is the altitude of that orbit?
  10. Comet Halley has period 76 y and perihelion distance \(8.9 \times 10^{10}\,\mathrm{m}\). Find (a) its semimajor axis, (b) its aphelion distance, and (c) its eccentricity.
  11. A 20 kg satellite is in a circular orbit of period 2.4 h and radius \(8.0 \times 10^{6}\,\mathrm{m}\) around a planet. If the surface gravitational acceleration is \(8.0\,\mathrm{m/s^{2}}\), what is the planet's radius?
  12. A projectile is fired vertically from Earth's surface at 10 km/s. Neglecting air drag, how far above the surface does it rise?
Tip: before reaching for an equation, decide which regime you are in. Outside a sphere, treat it as a point mass at its center; inside a uniform one, keep only the enclosed mass. For "how fast / how far" questions, conservation of \(K + U\) almost always beats integrating forces — and remember that \(U\) is negative, so a bound orbit has negative total energy. For orbital periods and sizes, Kepler's third law is usually one line away from the answer.