In 2014, ISRO's Mars Orbiter Mission, Mangalyaan, went into orbit round Mars. It ran on sunlight caught by its solar panel. Mars is about \(1.52\) times as far from the Sun as the Earth is. At the Earth, each square metre that faces the Sun receives \(1361\,\text{W}\). How much does a square metre receive at Mars? Nobody had to go there to know.
An angle in three dimensions
A plane angle is measured by an arc. Draw a circle of radius \(r\) round the corner of the angle. The angle in radians is the arc it cuts off divided by the radius. A larger circle gives a longer arc in the same ratio, so the size of the circle does not matter. A full turn is \(2\pi\) radians.
A cone opening from a point needs the same idea one dimension up. Draw a sphere of radius \(r\) round the tip of the cone. The cone cuts a patch of area \(A\) out of the sphere. Its solid angle is that area divided by the square of the radius, \(\Omega = A/r^2\). On a sphere twice as large, the same cone cuts four times the area, so the ratio stays fixed.
A cone cuts an area \(A\) from a sphere of radius \(r\) centred on its tip. Its solid angle is \(\Omega = A/r^2\). Its unit is the steradian, sr, a pure number. The whole sphere round a point is \(4\pi r^2/r^2 = 4\pi\,\text{sr}\), and a hemisphere is \(2\pi\,\text{sr}\).
A small flat patch that faces the point squarely is almost a piece of a sphere, so it holds \(A/r^2\). A patch of \(1\,\text{m}^2\) seen squarely from \(10\,\text{m}\) holds \(0.01\,\text{sr}\). A patch tilted away holds less, \(A\cos\alpha/r^2\), where \(\alpha\) is the angle between its normal and the line to the point. It looks smaller from the point, exactly as its shadow does.
A detector window of area \(1\,\text{cm}^2\) faces a small radioactive source squarely from \(20\,\text{cm}\). The source sends its particles equally in all directions. What fraction of them enters the window, in units of \(10^{-4}\)?
One cone, spheres of every size
Now let a small source send out light evenly in all directions. Think of one narrow cone of solid angle \(\Omega\) opening from it. On a sphere of radius \(r\), the cone cuts a patch of area \(\Omega r^2\). Twice as far, the patch has four times the area. Three times as far, nine times.
The light inside the cone stays inside it. No light is made or lost on the way, so the same power crosses every one of those patches. The same power on \(r^2\) times the area means the power on each square metre falls as \(1/r^2\). This is the inverse-square law, and it is geometry, not a special property of light.
Whatever a point source sends out evenly in all directions, and nothing absorbs on the way, spreads over the sphere \(4\pi r^2\) round it. So the amount on each unit of area falls as the inverse square of the distance, \(I = P/4\pi r^2\).
Mangalyaan has its answer. At the Earth, \(1361\,\text{W}\) falls on each square metre. Mars is \(1.52\) times as far, so the same sunlight is spread over \(1.52^2 = 2.31\) times the area. Each square metre at Mars receives \(1361/2.31 = 589\,\text{W}\), less than half. The orbiter's panel had to be sized for that.
Jupiter is \(5.2\) times as far from the Sun as the Earth. At the Earth, sunlight delivers \(1361\,\text{W}\) to each square metre facing the Sun. How much does a square metre facing the Sun receive at Jupiter, in watts?
The solid angle of a cone
A cone with its tip at the centre of a sphere cuts a round cap from the sphere. Let the cone have the half-angle \(\theta\), the angle between its axis and its edge. Archimedes found the area of such a cap: it is \(2\pi Rh\), where \(h\) is the height of the cap along the axis. That is the curved area of a cylinder of radius \(R\) and height \(h\), which is a surprising and exact result.
\[ h = R - R\cos\theta = R\,(1 - \cos\theta) \]
Why. the rim of the cap lies in a plane a distance \(R\cos\theta\) from the centre, measured along the axis. The cap reaches out to \(R\).
\[ A = 2\pi Rh = 2\pi R^2(1 - \cos\theta) \]
Why. Archimedes' area of a cap, with the height from line 1.
\[ \Omega = \dfrac{A}{R^2} = 2\pi(1 - \cos\theta) \]
Read it. divide by \(R^2\), the definition of the solid angle. The radius cancels, as it must.
Check the formula at the two ends. At \(\theta = 90°\), \(\cos\theta = 0\) and \(\Omega = 2\pi\), the hemisphere. At \(\theta = 180°\), \(\cos\theta = -1\) and \(\Omega = 4\pi\), the whole sphere. For a narrow cone, \(1 - \cos\theta \approx \theta^2/2\) in radians, so \(\Omega \approx \pi\theta^2\). That is the area of a small disc of radius \(r\theta\) divided by \(r^2\).
A small lamp shines equally in all directions. A lampshade lets its light out only through a cone of half-angle \(60°\). What share of the lamp's light leaves through the cone?
Back to physics: a charge shares its flux
The field of a point charge falls as \(1/r^2\), and its lines leave it evenly in all directions. So Electric Flux and Gauss's Law finds that the charge sends the same flux \(q/\varepsilon_0\) through every sphere round it. Each cone from the charge takes the share \(\Omega/4\pi\). Any surface that fills the cone takes that share, flat or curved, because every line in the cone crosses it once.
A point charge with \(q/\varepsilon_0 = 6000\,\text{N m}^2/\text{C}\) sits on the axis of a disc of radius \(3\,\text{cm}\), \(4\,\text{cm}\) from its centre. Find the flux through the disc.
The disc fills a cone with its tip at the charge. The half-angle \(\theta\) runs from the axis to a line to the rim.
\(\cos\theta = 4/5 = 0.8\), so \(\Omega = 2\pi(1 - \cos\theta)\).
\(\Omega = 2\pi \times 0.2 = 0.4\pi = 1.26\,\text{sr}\). The share is \(0.4\pi/4\pi = 0.1\), so the flux is \(0.1 \times 6000 = 600\,\text{N m}^2/\text{C}\).
A tenth of the charge's flux passes through the disc, \(600\,\text{N m}^2/\text{C}\), exactly the answer of the physics chapter. No integral and no value of \(\varepsilon_0\) was needed, only the cone.
The same sharing gives the faces of a cube. A charge at the centre of a cube sees six faces that look exactly alike. So each face holds \(4\pi/6 = 2.09\,\text{sr}\) and takes a sixth of \(q/\varepsilon_0\). Sound from a loudspeaker, photons from a lamp and the Sun's heat share out in the same way.
The same charge, with \(q/\varepsilon_0 = 6000\,\text{N m}^2/\text{C}\), sits on the axis of a disc of radius \(5\,\text{cm}\), \(12\,\text{cm}\) from its centre. What is the flux through the disc, in N m²/C?
What solid angle does one face of a cube hold, seen from the centre of the cube, in steradians?
The inverse square needs a point source, one small compared with the distance, that sends evenly in all directions through a medium that absorbs nothing. Near a long straight source, such as a tube light or a busy highway, the spread is over a cylinder. There the intensity falls only as \(1/r\). Near a large flat source it hardly falls at all. Fog and walls that absorb or reflect also break the law.
In 1604 Kepler published a book on the optics of astronomy. In it he argued that light from a point spreads out over spheres. A sphere twice as large has four times the surface, so the same light is spread four times as thin. He stated that the strength of light falls as the square of the distance, the first clear statement of the inverse-square law. Newton later found the same law for gravity.
Electric Flux and Gauss's Law defines the solid angle. It finds the flux of a point charge through spheres, cones, discs and cube faces, and derives Gauss's law from it. Energy, Power, Intensity and Loudness finds how the sound of a small source falls with distance. Radiation, Convection and Cooling finds the solar constant from the Sun's power spread over a sphere. Photons and the Photoelectric Effect counts the photons of a lamp that land on each square metre. Gravitational Force and Field proves the shell theorems with the same cones.
- Sharing flux or light in proportion to the half-angle of a cone.Share by solid angle, \(\Omega = 2\pi(1 - \cos\theta)\), out of \(4\pi\).
- Dividing by the distance instead of its square.The area of a patch in a fixed cone grows as \(r^2\), so the intensity falls as \(1/r^2\).
- Using the inverse square next to a long or wide source.A line source spreads over a cylinder and falls as \(1/r\). The inverse square needs a source small compared with the distance.