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Okay,

What is Aerosol Forcing?

Ragoat66 wrote:Okay,

What is Aerosol Forcing?

I didn't exactly find Aerosol Forcing's definition, but I did find Radiative forcing.

Radiative forcing is the difference between the incoming sunlight and the outgoing sunlight. A positive forcing number is warming the system, a negative forcing number is cooling it.

I don't know if that's the definition you want.

If I get it right, here's my question:

(This is actually from an invitational)

The Hubble Telescope has a diameter of 2.4 meters and orbits at a height above earth of 559 km. If it were used to observe the Earth, instead of space, what would be its theoretical limit of resolution in the visible spectrum, using Raleigh's criterion? (Disregard limits due to atmospheric effects).

A) About 15 mm
B) About 15 cm
C) About 1.5 m
D) About 15 m
E) About 150 m

Can someone help me out with this? How would you get the answer of (B)? I tried to look up Rayleigh criterion's formula but I still don't get how B comes out.

hearthstone224 wrote:If I get it right, here's my question:

(This is actually from an invitational)

The Hubble Telescope has a diameter of 2.4 meters and orbits at a height above earth of 559 km. If it were used to observe the Earth, instead of space, what would be its theoretical limit of resolution in the visible spectrum, using Raleigh's criterion? (Disregard limits due to atmospheric effects).

A) About 15 mm
B) About 15 cm
C) About 1.5 m
D) About 15 m
E) About 150 m

Can someone help me out with this? How would you get the answer of (B)? I tried to look up Rayleigh criterion's formula but I still don't get how B comes out.

Just interjecting to point out, generally posts about specific questions that you don't know the answer to belong in the event thread (viewtopic.php?f=227&p=308499#p308499), not the question marathon.

hearthstone224 wrote:If I get it right, here's my question:

(This is actually from an invitational)

The Hubble Telescope has a diameter of 2.4 meters and orbits at a height above earth of 559 km. If it were used to observe the Earth, instead of space, what would be its theoretical limit of resolution in the visible spectrum, using Raleigh's criterion? (Disregard limits due to atmospheric effects).

A) About 15 mm
B) About 15 cm
C) About 1.5 m
D) About 15 m
E) About 150 m

Can someone help me out with this? How would you get the answer of (B)? I tried to look up Rayleigh criterion's formula but I still don't get how B comes out.

Rayleigh's criterion says that the angular resolution of a circular aperture in radians is equal to 1.22 times the wavelength of light divided by the diameter of the aperture. When we plug in the values for this problem (assume wavelength of light is 550 nm, as that is the average wavelength of visible light), we get about 2.796 * 10^-7 radians. To find the spatial resolution, multiply the angular resolution by the distance to the object. When we do that for this problem, we get about 0.156 m, or about 15 cm.

If you have a question about Remote Sensing, ask it in the event page or PM me.

Sorry about that guys. Correct, your turn Xuax.

What is the peak wavelength of a blackbody at temperature 310 K?

Xuax wrote:What is the peak wavelength of a blackbody at temperature 310 K?

You can use Wein's law so you can do:

0.29/310K = 0.000935483871 nm. That's in the gamma ray range I think.

hearthstone224 wrote:

Xuax wrote:What is the peak wavelength of a blackbody at temperature 310 K?

You can use Wein's law so you can do:

0.29/310K = 0.000935483871 nm. That's in the gamma ray range I think.

Er, either your units out constant are incorrect I think. No way the wavelength is that low at 310K.

Private Wang Fire wrote:

hearthstone224 wrote:

Xuax wrote:What is the peak wavelength of a blackbody at temperature 310 K?

You can use Wein's law so you can do:

0.29/310K = 0.000935483871 nm. That's in the gamma ray range I think.

Er, either your units out constant are incorrect I think. No way the wavelength is that low at 310K.

If that was in centimeters, it would be correct. Remember, Wien's displacement constant is 2.8977729 × 10^-3 m K. That divided by 310 K gives 9.34765452e-6 m, which in nm is 9347.6 nm.