The Inverse Square Law

Part II – The Small-Angle Formula
The small-angle formula uses the measure of an angle to determine the diameter of an object at a distance. In the following figure, θ (theta) is the angular size in radians, s is the actual linear size of an object, and d is the distance to the object.
For small angles, , therefore is the small-angle formula.

For example, suppose the angular diameter an observer measures is 0.01 radians and the linear size of the object being observed is known to be 3,480 kilometers (km). The distance can be measured using the small-angle formula as follows.

Therefore, the distance to the object being observed is 348,000 km, or 3.48 x 105 km.
Directions: Use the small angle formula to calculate the missing values in the chart below.
Angular Diameter (radians) Actual diameter (kilometers) Distance (kilometers)
0.01 3,480 348,000 or 3.48 x 105
0.009 1.5 x 108
3,474 3.84 x 105

Need help? Check out this tutorial video on this calculation.
In astronomy, scientists modify the small-angle formula to account for the arcsecond measurements used to account for Earth’s rotation to help track objects in the celestial sphere. In this modified formula, the angular diameter is measured in arcseconds and accounts for the difference by converting radians to arcseconds (1 radian = 206,265 arcseconds).

Use the modified small-angle formula to complete the table of values.
Object Cat’s Eye Nebula Horsehead Nebula Pleiades Cluster Globular Cluster M71
Angular diameter (arcseconds) 20 300 6,000 430
Actual diameter
Distance (light years) 3,300 1,630 440 13,000

What is the unit of measure used for the actual diameter in the table above?

Part III – Apparent Magnitudes
The apparent brightness of stars is reported using its apparent magnitude, a logarithmic unit where every five magnitudes corresponds to a change in brightness of 100. For example, a magnitude 1 star is 100 times brighter than a magnitude 6 star. Note the smaller the magnitude, the brighter the star and the larger the magnitude, the star appears dimmer.
As every five magnitudes corresponds to a factor of 100 in brightness, each magnitude step can be calculated as a factor of 1001/5, or 2.512 in brightness. Generally speaking, the brightness ratio is given by

where m2 – m1 is the difference in magnitude values. For example, a magnitude 3 star and a magnitude 5 star have a magnitude difference of 2, hence the magnitude 3 star appears to be (2.512)2 times brighter than the magnitude 5 star. (HINT: Check out this tutorial video on this calculation.)
How many times brighter is a magnitude 5 star than a magnitude 6 star?
How many times brighter is a magnitude 7 star than a magnitude 10 star?
How many times brighter is a magnitude 0 star than a magnitude 5 star?
Complete the following charts.
Difference in Magnitudes Brightness Ratio Difference in Magnitudes Brightness Ratio
0 1 7
1 2.5 8
2 6.3 9
3 10
4 15
5 20
6 25

Part IV – The Inverse Square Law
Light traveling from an object obeys the inverse square law, which states that light at any distance from a source (ie: star) is inversely proportional to the square of the distance from the source. In general, the apparent brightness of an object, b, with luminosity (in watts), L, may be determined at any distance (in meters), d, using the formula,

For example, the apparent brightness of the Sun as measured on Earth is determined using the above formula,

Hence, we consider the brightness of our Sun to have an apparent brightness of 1370 W/m2. However, measuring the Sun’s apparent brightness from Pluto would yield,

Thus, apparent brightness of an object is inversely proportional to its distance. In other words, as distance increases, the apparent brightness decreases.
Therefore, if two stars with identical luminosities are observed, then the ratio of their apparent brightness is given by,

For example, suppose star 1 and star 2 are two stars of the same type (and thus have the same luminosity), but star 1 has an apparent brightness in the night sky 230 times greater than that of star 2 (b1/b2 = 230). We could determine the relative distance of the stars from Earth using the formula above.

So, star 2 is approximately 15 times further away from Earth than star 1.
Check out the tutorial video on this calculation.
Complete the following table using the inverse square law comparing the brightness of two stars with the same luminosity. (Hint: The Magnitude Difference column should contain positive values for the magnitude difference between each set of stars.)
Spectral Type Star Name and (apparent brightness) Mag. Diff. Ratio of brightness Ratio of distances Conclusion
O9.5 V Mu Columbae (5.17)
Zeta Ophiuchi (2.56) 2.61 Mu Columbae is 3.3x farther than Zeta Ophiuchi
B3 V Alkaid (1.85)
Regulus (1.35)
A1 V Sirius (-1.46)
Merak (2.37) 3.83 Merak is 5.8x farther than Sirius
F0 V Porrima (3.65)
Alkalurops (4.31) 0.66 Alkalurops is 1.35x farther than Porrima
G2 V Rigel Kentaurus
(-0.02)
Sun (-26.72)
K5 V 61 Cygni A (5.2)
Kaffaljidhma (3.47)
M6 V Wolf 359 (13.53)
Ross 248 (12.29)