156-158 - Luminosity, Intensity and Astronomical Distances

Total Power And Received Intensity

A star does not just have a vague "brightness". It transfers energy away every second as electromagnetic radiation. The total power radiated by the star is its luminosity, $L$, measured in watts.

A detector on Earth receives only a tiny share of that power. What matters at the detector is the power arriving per square metre. That received power per unit area is intensity, $I$, measured in $\mathrm{W\,m^{-2}}$.

The two quantities answer different questions:

Now imagine transparent spheres centred on a star. The same luminosity crosses every sphere, but a sphere twice as far away has four times the surface area. The same total power is spread more thinly.

For a source radiating uniformly in all directions, at distance $d$:

The physical assumption is doing work here: the radiation spreads uniformly and absorption is negligible unless the question says otherwise.

Misconception contrast: a star looking dimmer does not prove it is less luminous. It might be equally luminous but further away, or more luminous and much further away. Intensity is what reaches you; luminosity is what the source emits.

Using The Inverse-Square Equation

The equation

is an inverse-square relationship. If $L$ stays constant and $d$ doubles, $d^{2}$ becomes four times larger, so $I$ becomes one quarter as large.

Worked example: the Sun has luminosity about $3.83 \times 10^{26}\,\mathrm{W}$. The mean Earth-Sun distance is $1.50 \times 10^{11}\,\mathrm{m}$. Calculate the intensity of solar radiation at the top of Earth's atmosphere, assuming no absorption before it reaches Earth.

Decision point 1: use the distance from the source to the detector, not the radius of the Sun or Earth.

Decision point 2: the denominator is the area of an imaginary sphere centred on the Sun.

Work through the power of ten carefully:

That value is a power per square metre. It is not the Sun's total power; it is what each square metre facing the Sun would receive before atmospheric absorption.

For an Edexcel “calculate” question, the substitution and final unit matter. For a “determine” question where distance is unknown, rearrange before substituting:

Guided check: if two identical stars have the same luminosity and one is three times further away, the more distant star has $\left(\frac{1}{3}\right)^{2} = \frac{1}{9}$ of the received intensity.

Trigonometric Parallax

Parallax is a geometric method. Hold a finger in front of your face and look with one eye, then the other: the finger appears to shift against the distant background. Nearby stars do the same against much more distant stars as Earth moves around the Sun.

Astronomers observe a nearby star six months apart, when Earth is on opposite sides of its orbit. The star appears to shift against a background of very distant stars. The parallax angle, often called $p$, is half the total angular shift.

For a small parallax angle in radians, the geometry is approximately:

where $r$ is the Earth-Sun distance, about $1\,\mathrm{AU}$, and $d$ is the distance to the star. The angle must be in radians if this equation is used with SI distances.

Worked example: a nearby star has a measured parallax angle of $0.25\ \text{arcseconds}$. Determine its distance in parsecs.

Decision point 1: the quoted $0.25\ \text{arcseconds}$ is already the parallax angle, not the total six-month shift.

Decision point 2: the parsec is defined so that:

So:

If you used the SI geometry instead, you would first convert the angle to radians. The same physics is underneath both routes: a smaller parallax angle means a larger distance.

Practical-method reasoning matters here. The limiting issue is angular resolution: very distant stars have such tiny parallax shifts that measurement uncertainty can swamp the signal. Repeated observations across the year and comparison with distant background stars help separate true parallax from measurement scatter.