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October 23, 2014

October 23, 2014

Posted by **Easha** on Tuesday, November 24, 2009 at 9:09pm.

time and is given in units of Watts (W) where 1 W = 1 Joule per second. The Sun's power

output is therefore 3.90*10^26 Watts. This is called the solar luminosity. As this energy travels

outward from the Sun it is spread (fairly evenly) over a spherical surface that increases in area

with distance from the Sun (recall the area of a sphere is proportional to its radius squared).

The energy received each second by an area of 1 m2 is called the energy flux.

a) Calculate the energy °ux from the Sun at a distance of 9:54 AU (that is, Saturn's mean

distance) from the Sun. How does this compare with the solar flux received at the Earth's

mean distance from the Sun?

b) Suppose we place a perfectly black sphere with radius 2600 km at a distance of 9:54 AU from

the Sun. How much direct solar energy does the sphere absorb in one second?

c) If the sphere in part (b) is in thermal equilibrium, what is its temperature?

d) Titan re°ects approximately 20% of the light that strikes its surface, and absorbs the rest.

Titan's radius is roughly 2600 km. How much energy from sunlight does Titan absorb each

second? What is the temperature of Titan if it has no internal heat sources of significance?

- astrophysics -
**MathMate**, Tuesday, November 24, 2009 at 9:56pmP = The Sun's power output is therefore 3.90*10^26 Watts.

D1 = distance of the earth from the sun

= 1 AU = 149,598,000,000 m

D2 = distance of Saturn from the sun

= 9.54 AU

Surface area, S, of a sphere of radius r

=4πr²

a.

For Saturn,

R2=9.54 AU

=9.54*149,598,000,000 m

S2=4πR2²

Solar flux

F2=P/S2

I get about 15 watts/m² for F2.

For earth, the flux would be 9.54² times higher.

b.

"perfectly black sphere" means all the electromagnetic energy that falls on it will be absorbed.

Cross sectional area of 2600 km sphere

A=π(2600000)² m²

Energy absorbed

=A*F2

For the equilibrium temperature, you will need the formulae from your course. The "black body surface temperature" generally requires the temperature of the surface of the sun and the sun's radius, which is not supplied. So I suspect a different method is used.

See also:

http://en.wikipedia.org/wiki/Black_body

- astrophysics -
**Easha**, Wednesday, November 25, 2009 at 4:42pmty so much for the help..for the 1st 1 i got the same answer...but y did u use pir^2 for the second 1 and 4pir^2 for the 1st 1??

- astrophysics -
**MathMate**, Wednesday, November 25, 2009 at 8:00pm4πr² is the surface area of an imaginary sphere in which the sun is at the centre. The energy output of the sun divided by this surface area represents the intensity of the sun's rays at the given radius.

For the second problem, we are to calculate the sun's energy absorbed by the sphere which has a cross sectional area of πr². This area is similar to the shape of the moon when we see it from a distance.

If that is still not clear, feel free to post again.

- astrophysics -
**Easha**, Wednesday, November 25, 2009 at 9:53pmyeah...i got it..ty sooo much...i can do the rest i guess..but i have any problem with any other questions...i'll post it..!...ty once agn!!:D

- astrophysics -
**Easha**, Wednesday, November 25, 2009 at 10:29pmsry to disturb u 1ce agn..but i had another problem related to galilean moons...i thot i would just post the question in case u can help me...i have done it...i just want to check my answer!!..

What is the largest angle that can separate Ganymede from Europa from the point of view of a

terrestrial astronomer who observes both moons to the west of Jupiter? Make the assumption

that Jupiter, Ganymede and Europa are aligned.

what i got is 2.08 degrees!!!

- astrophysics -
**MathMate**, Wednesday, November 25, 2009 at 11:26pmThe angle subtended by the moon from earth is 3474.8/384403 radians=0.00904 radians=0.52°!

You may have an error with units. Perhaps you calculated the number of minutes (and not degrees).

The angle subtended in radians, i.e. transverse distance divided by longitudinal distance, can be converted to minutes by dividing by the constant 0.000290888208666 and I got 2.18 minutes.

These are my data:

semi-major axis of Ganymede=1070412 km

semi-major axis of europa = 671034 km

Separation assuming they are aligned with Jupiter

= 1070412-671034=399,378 km

Mean distance of the earth from the sun

= 149,000,000 km

Mean distance of Jupiter from the sun

= 779,000,000 km

distance of the earth from Jupiter

= 779,000,000 - 149,000,000

= 630,000,000 km

Angle subtended by the two moons

= 399,378/630,000,000

= 0.000634 radians

= 0.000634/.000290888208666 minutes

= 2.18 minutes

I assume that the alignment of the moons with Jupiter occurs when it is perpendicular to the earth.

Ref:

http://en.wikipedia.org/wiki/Moons_of_Jupiter

http://www.windows.ucar.edu/tour/link=/kids_space/distance.html&fr=t

- astrophysics -
**Easha**, Wednesday, November 25, 2009 at 11:56pmwhat i did was..

distance between earth and jupiter= 4.20 AU= 6.30*10^8 km

distance between jupiter and europa= 6710000 km

distance between jupiter and ganymede= 1070000 km

then i found the angles subtended by jupiter and europa, x (from earth viewpoint) since earth and europa makes 90 deg angle on jupiter. Similarly angle made by jupiter and ganymede, y, on earth.

tan x = 671000/(6.3*10^8)

=> x= 0.0610 rad = 3.50 deg

tan y = 1070000/(6.3*10^8)

=> y= 0.0973 rad = 5.58 deg

Now these angles are subtracted and we get 2.08 deg (=124.8 arc minutes) as the largest angle separating ganymede from europa, as seen from earth.

- astrophysics -
**MathMate**, Thursday, November 26, 2009 at 9:53amYour method of calculations is correct and would have results that agree with mine except for two little things:

1.

"distance between jupiter and europa= 6710000 km "

The number has an extra zero (probably a transcription error for posting only). It does not change your answer because your subsequent calculations used the correct value.

2. The arctangent that you calcuated is already in degrees, so there is no need to convert from radians to degrees.

In fact, (0.0973-0.0610)=0.0363°=2.18 minutes of arc.

Check your calculator settings, and check how you can switch the setting from degrees to radians and vice versa.

- astrophysics -
**Easha**, Thursday, November 26, 2009 at 12:35pmyea...i figured that out earlier..my answer was in degree (as the calculator was in degree mode)...but i took the answer to be in radian then i converted that to degree...which is completely wrong...its 2.18arc mins now...ty sooo much for all ur help!!

- astrophysics -
**MathMate**, Thursday, November 26, 2009 at 1:35pmYou're very welcome!

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