Solar Luminosity: Calculating Sphere of Light Illuminating Earth's Orbit

  • Thread starter Thread starter newsflash
  • Start date Start date
  • Tags Tags
    Luminosity Solar
AI Thread Summary
The discussion focuses on calculating the surface area of the sphere of light illuminated by the sun, specifically regarding the Earth's orbit. The formula used is S = 4(3.14) r^2, with r being the mean distance from the sun to the Earth, approximately 149,597,829 kilometers. The initial calculation yields a surface area of about 2.81 x 10^15 km^2. A key point emphasized is the importance of converting the radius into meters before squaring to avoid complications with unit conversion. This calculation is crucial for understanding solar luminosity in relation to Earth's orbit.
newsflash
Messages
1
Reaction score
0

Homework Statement


The formula for the surface area of a sphere is 4(3.14) r2, where r is the mean distance from the sun to the earth. Determine the sphere of light (square meters) illuminated by the sun the size of the Earth's orbit.

Homework Equations



astronomical Unit-149,597,829 k

The Attempt at a Solution


my answer is 2.8122922(17)
no idea
 
Last edited:
Physics news on Phys.org
Welcome to the Physics Forum, newsflash!

S = 4(3.14) r^2
= 4*3.14*(149,597,829 km)^2
= 2.81 x 10^15 km^2
If you need that in square meters, I suggest you change the r value into meters before squaring. It is tricky changing units on something after it is squared.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top