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Published byElyse Sparling Modified over 2 years ago

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Announcements Today will be Project 1 presentation first then new material Homework Set 5: Chapter 5 # 44, 46, 47, 50, 52, 53 & 54 Exam 2 is in two weeks. Will cover Light and Astronomical Instruments (material covered since the 1 st Exam).

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The Blackbody Spectrum Stefan’s Law relates the temperature to the flux (total energy emitted per square meter) Wein’s Law relates the temperature to the wavelength of maximum intensity

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Blackbody Example Determine the wavelength of maximum intensity of the Earth and the total energy emitted by it. Assume it is a perfect blackbody with a temperature of 288 K and a radius of 6.378x10 6 m Note that this image is not the blackbody emission of the Earth. What you see here is reflected sunlight. As we will see, the blackbody emission of the Earth is mostly in the infrared range.

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Example Solution For wavelength of maximum intensity, use Wein’s law Next, find total energy emitted: flux times surface area This wavelength is in the far infrared range.

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The Doppler Effect + If the source is moving away from the observer - If the source is moving towards the observer The light is redshifted if the source is moving away from the observer, blueshifted if it is moving towards

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Example of Doppler Effect The New Horizons spacecraft, currently bound for Pluto, is receding away from the Sun at 15.98 kilometers per second. If the radio transmitter on the spacecraft transmits at 9.302 GHz, what is the frequency received on Earth?

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Example Solution The Doppler formula is in wavelengths and the problem gives frequencies so do some algebra on the Doppler formula to put it in frequencies. Now plug in numbers and solve. Use “+” since it is receding.

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Brightness-Distance Relationship Light from a point source spreads out over the surface of a sphere so the brightness is related to the inverse of the surface area of the sphere

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Examples of Distance- Luminosity relationship How much dimmer does the Sun appear as seen from Jupiter than Earth? Jupiter’s orbital radius is 5.2 AU. How much brighter does the Sun appear on Mercury than the Earth? The orbital radius of Mercury is 0.387 AU

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Example Solution Since you are asked to compare the brightness to that at Earth, take the ratio of the brightness at Jupiter (at 5.2AU from the Sun) to that at Earth (at 1.0 AU). The Sun is 0.3698 times as bright (or 27.04 times dimmer) at Jupiter than at Earth.

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Example Solution 2 Compare brightness at Mercury (at 0.387 AU from the Sun) to the brightness at Earth (at 1.0 AU) The Sun appears 6.68 times brighter in the skies over Mercury than it does at Earth

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The albedo of a planet is the fraction of sunlight it reflects back out into space

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The temperature of a planet is an energy balance The fraction of sunlight that isn’t reflected away is absorbed by the planet. This heats the surface causing blackbody radiation. When the amount absorbed balances the amount emitted, a stable temperature is reached.

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The energy balance equation The energy absorbed is the absorbing cross-section area of the planet times the amount of sunlight hitting the planet times the fraction absorbed (1 – a) where a is the albedo of the planet The energy emitted by blackbody radiation is the surface area of the planet times the blackbody flux

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Equating the emitted and absorbed energies and solving for T

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Planetary temperature example Determine the equilibrium surface temperature of Earth. Assume an average albedo of 0.30

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Example Solution This is approximately -18 ° C, well below the freezing point of water. Gasses in our atmosphere absorb some of the infrared blackbody radiation the Earth emits, shifting the balance to a higher temperature of around 15 ° C

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