× Didn't find what you were looking for? Ask a question
Top Posters
Since Sunday
y
2
m
2
m
2
u
2
m
2
B
2
M
2
e
2
k
2
N
2
y
2
m
2
New Topic  
LenChan64 LenChan64
wrote...
11 years ago
I heard about it a while ago, I think you needed the object's orbital radius (semi-major axis) and the solar flux of Earth (1367?), but I forgot how the equation goes. Thanks.
Read 351 times
2 Replies

Related Topics

Replies
wrote...
11 years ago
Flux(R) = (3.827*10^26 W)/(4*pi*R^2 m^2)

Express R in meters, and the flux will be in watts/m^2.
wrote...
11 years ago
The answer given by the first responder gets solar flux by assuming that the power output of the sun is already known.

The method you wanted was, to use the known solar flux at earth's orbit,
and use that to find the power of the sun?

The total area of the sphere that is as far from the sun as the earth is
would be
4 pi (150,000,000,000 m)^2,
so if the known solar flux at earth's orbit is 1367 W/m^2,
then the total power of the sun must be

(1367 W/m^2) (4 pi) (150,000,000,000 m)^2 = 3.9 x 10^26 Watts.

The value is perhaps really closer to 3.8 x 10^26,
but the error may come from squaring the approximate 150 million km.
New Topic      
Explore
Post your homework questions and get free online help from our incredible volunteers
  509 People Browsing
Related Images
  
 193
  
 269
  
 1028
Your Opinion
What's your favorite coffee beverage?
Votes: 305