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The Inverse Square Law and Surface Area

The Inverse Square Law and Surface Area. Calculating Distances to Stars. Measuring Distances. There are several techniques used to measure distances to stars. The distance to the very closest stars can be measured by trigonometric parallax. The diagram shows Earth Orbit around the Sun.

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The Inverse Square Law and Surface Area

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  1. The Inverse Square Law and Surface Area Calculating Distances to Stars

  2. Measuring Distances There are several techniques used to measure distances to stars. The distance to the very closest stars can be measured by trigonometric parallax

  3. The diagram shows Earth Orbit around the Sun. The position of a nearby star changes by a tiny amount over a six month period. This allows us to use trigonometry to find its distance. The angles are extremely small. This direct method is the most accurate way of determining distance

  4. Using The Inverse Square Law. Every instant a star radiates its energy into space The energy which was at the surface is distributed at the surface of an expanding sphere

  5. D The amount of electromagnetic energy at every instant emitted by the star is at the surface of an expanding sphere whose radius is the distance to the star star Earth This energy distribution obeys an inverse square law. This is because the surface area of this sphere is 4πD2

  6. The Inverse Square Law The power received from a star per metre squared at the Earth is called the intensity (I) of the star’s radiation This is related to the power of the star in this way Where D is the radius of the sphere i.e. the distance from Earth to the star

  7. The Intensity of the Sun • At Earth the measured power output of the Sun is 1.3 kWm-2. • The mean distance to the Sun is 150 million km. • Calculate the power output of the Sun.

  8. Stars of Known Power Output • There are several classes of stars with known power output. • Stars which have the same surface temperature ( and spectral characteristics) as the sun all have the same power output • We can readily calculate the power output of nearby stars and classify their power output and compare them with more distant stars The following very bright objects of known luminosity can be identified in distant galaxies • Cepheid Variable Stars • Supernovae

  9. The Sun has a power output of 3.91 x 1026W. Knowing this and knowing its surface temperature allows us to calculate its surface area using P=σ.AT4 This now allows us to calculate the radius of the Sun as A =4πr2. Calculate the solar radius.

  10. Comparing the Power Output of the Sun with other Stars (Using Stefan’s Law: P=σAT4)

  11. Sources Which Have The Same Power Output As The Sun Example: calculate the radius of a star which has a surface temperature of 3 700 degrees and the same power output as the Sun. What is the diameter of this star taking the solar diameter as 1 unit? (Ps = 3.91 x 1026W) (Ds = 1.4 million km) Apply Stefan’s Law P=σAT4 .

  12. What is the diameter of a star, the maximum intensity of which occurs at 625nm • Steps: • Use Wien’s law λmax T = 2.9 x 10-3mK to calculate the temperature of the star • Apply Stefan’s Law to calculate its radius.

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