Need help with three questions

[Ronnie67]

Hi guys I just wanted to pick your brains sort of speak, it would be most appreciated if you could help. here are the questions.

a)  The luminosity of the Sun is 4*10^33 erg/s, and its radius is 7*10^10 cm.

 You are tasked with building a solar power plant in the Arizona desert, using solar panels with 10% efficiency. How large an area (km^2) must your solar panels cover to match the power output of a large nuclear powerplant (about a GigaWatt)? Please enter your answer in units of km^2

 The Keck telescope on Mauna Kea has an angular resolution on Earth of half an arcsecond.

How far away (in meters) could you read ("resolve the letters of") a book with 3 mm square type, using the Keck telescope on Earth?

c)  In space, the angular resolution of the Keck telescope is govererned by the diffraction limit.

How far away could you read the same book, using the Keck telescope in space? Please express your answers in units of meters.

These questions are from a free astronomy topic from coursera, anyone can join for free, its fun but some of the questions have placed me in a stump, I cannot move forward until I complete these three.

Ronnie

[bob1957]

Ronnie,

a. First you will need to calculate the flux (watts/km2) which falls on the Earth's surface from the Sun. You have the Sun's luminosity as 4.0E33 erg/s which is equal to 4.0E26 Watts. Using the distance from the Sun to the Earth of about 1.5E8 km calculate the surface area of a sphere using this distance as the radius. Divide the power from the Sun by the surface area and this will give you the flux at the Earth's surface. You know you need 1 Gigawatt (1E10 Watts) output power from the solar array, but remember it's only 10% efficient so you need more power going into the array, i.e. 1E10 Watts. Now divide this required input power to the array by the flux falling onto it and it should give you the area of the array in square kilometres.

b. Here you can you use the small angle approximation which basically says that the angle (in radians) formed by a small object at a very large distance can be calculated by dividing the size of the object by it's distance (all in metres). So if you know the angle (in this case the angular resolution) you can rearrange it to find the distance in metres by dividing the size of the object (in metres) by the angle (in radians). But you will have to convert the angular resolution of 0.5 arcseconds into radians. If you are not familiar with this multiply the number of degrees by pi and divide by 180. Remember there are 60 minutes in a degree and 60 seconds in a minute so one degree equals 1/3,600 arcseconds.

c. For diffraction limited optics you will need to calculate the angular diffraction. For this you will need the diameter of the optics in this case the Keck which I believe is 10 metres diameter and also the wavelength of light which you can approximate to 500 nm (5E-7 metres). The angular diffraction is calculated by dividing the wavelength of light by the diameter of the optics and multiplying by 1.22. This gives you the angular diffraction in radians. You can then use the small angle approximation outlined in b. to work out the distance using the angular diffraction you have just calculated and the size of the object.

I hope all of the above helps.

Cheers, Bob

Edited September 4, 2015 by bob1957

[Ronnie67]

Cheers Bob for the help, will give it a go tomorrow, its a bit late to try now, just finished work so brain power is down lol

Ronnie

[ollypenrice]

For question 1 you'd also need to factor in the angle of the panel relative to the sun because the insolation varies as the angle of incidence changes. I suppose you could assume a multi-panel array with each panel equatorially mounted to track the sun.

Olly

[bob1957]

Good point Olly. The panel area decreases as the angle towards the vertical increases as the Sun passes overhead, unless as you say the panels can track the movement of the Sun for optimum area.

[bob1957]

Good point Olly. The panel area decreases as the angle towards the vertical increases as the Sun passes overhead, unless as you say the panels can track the movement of the Sun for optimum area.

[bob1957]

Seems I've developed a stutter!

[Brodie3857]

What's the answer for (b)

[robin_astro]

For (a), if this was a real situation and not a typical simplistic course question, then you would also need to consider the solar spectral energy density, the atmospheric transmission and the efficiency of the solar cells with wavelength. (Not sure why they have given you the diameter of the sun, except that  with this and the luminosity,  you could estimate the temperature and hence the spectral energy density, though I doubt that is what they intended.

Cheers

Robin

EDIT: Holy thread resurrection batman!

Edited May 22, 2019 by robin_astro