Question

Part A. What would be the luminosity of the Sun if its surface temperature were 3900 K and its radius were 1.5 AU ?

Part B. What would be the luminosity of the Sun if its surface temperature were 3900 K and its radius were 7.0 AU ?

*I don't understand how to calculate this out. Instructor didn't really cover this.

Answer 1

When speaking of the Luminosity, it can best be described as the total amount of light that will be emitted from the Surface of the sun, in a given period of time. the Luminotsity of the star, or the Sun (as in this question) can be calculated using different equations (in genral termed as the Luminosity Equation).

if we have to calculate the Luminosity odf the interstellar star,then it is calcualetd on the basis of comparison between the Luminosity of the Sun (Radius and surface temperature ratio).

but here in thiscase, we arer only to calculate the Luminosity of the Sun's surface, so we consider using simple formula.

A) First we have - Radius = 1.5 AU, and the ST (Surface Temp.) = 3900 K.

so, the L = R2T4

L = (1.5)2(3900)4 = 2.25 x (39 x 102)4 = 2.25 x (23,13,441 x 108)= 52,05,242 x 10^8. K.A.U.

B) Second we have - Radius = 7 AU, and Temp = 3900 K

usig the same formula we have (7)2 x (39 x 102)4 = 49 x 23,13,441 x 108 = 1,13,35 x 108 K. AU.

Answer 2

Part A: Luminosity at 3900 K and 1.5 AU

1. Convert radius to meters:

   $$R=1.5\text{AU}=1.5\times 1.496\times 10^{11}\text{m}=2.244\times 10^{11}\text{m}$$

2. Plug into the formula:

   $$L=4\pi (2.244\times 10^{11}{)}^2\times (5.67\times 10^{-8})\times (3900{)}^4$$

3. Simplify step-by-step:

• Calculate $R^2$:

  $$(2.244\times 10^{11}{)}^2=5.036\times 10^{22}{\text{m}}^2$$

• Calculate $T^4$:

  $$(3900{)}^4=2.31\times 10^{14}{\text{K}}^4$$

• Multiply all terms:

  $$L=4\pi \times 5.036\times 10^{22}\times 5.67\times 10^{-8}\times 2.31\times 10^{14}$$$$L\approx 8.23\times 10^{30}\text{W}$$

4. Compare to the Sun's current luminosity:

   $$\frac{L}{L_{\odot }}=\frac{8.23\times 10^{30}}{3.828\times 10^{26}}\approx 21,500$$

5. Answer: The Sun would be ~21,500 times more luminous than its current state.

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Part B: Luminosity at 3900 K and 7.0 AU

1. Convert radius to meters:

   $$R=7.0\text{AU}=7.0\times 1.496\times 10^{11}=1.047\times 10^{12}\text{m}$$

2. Plug into the formula:

   $$L=4\pi (1.047\times 10^{12}{)}^2\times (5.67\times 10^{-8})\times (3900{)}^4$$

3. Simplify step-by-step:

• $R^2$:

  $$(1.047\times 10^{12}{)}^2=1.096\times 10^{24}{\text{m}}^2$$

• $T^4$: Same as Part A ($2.31\times 10^{14}{\text{K}}^4$).

• Multiply all terms:

  $$L=4\pi \times 1.096\times 10^{24}\times 5.67\times 10^{-8}\times 2.31\times 10^{14}$$$$L\approx 1.79\times 10^{32}\text{W}$$

4. Compare to the Sun's current luminosity:

   $$\frac{L}{L_{\odot }}=\frac{1.79\times 10^{32}}{3.828\times 10^{26}}\approx 467,000$$

5. Answer: The Sun would be ~467,000 times more luminous than its current state.

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