Luminosity Watch

Fross877
Badges: 10
Rep:
?
#1
Report Thread starter 8 months ago
#1
A sensor of cross-sectional area 4 x 10-4 mounted on a satellite orbiting the Earth is used to gather the EM radiation from the star Antares.
Antares is 550 light years from the Earth. The radiant power entering the sensor from Antares is 2.6 x 10-11W

i) Calculate the luminosity L of Antares.

L = P/A = 2.6x10-11/4x10-4 = 6.5 x 10-8

I dont understand why the markscheme has multiplied the radiant power by the distance to the star? Isn't the P of the equation the radiant power incident on the sensor?
0
reply
Agent007
Badges: 14
Rep:
?
#2
Report 8 months ago
#2
lmao literally just did this question paper. Baffled by the MS
0
reply
Fross877
Badges: 10
Rep:
?
#3
Report Thread starter 8 months ago
#3
(Original post by Agent007)
lmao literally just did this question paper. Baffled by the MS
rough paper xD
0
reply
Agent007
Badges: 14
Rep:
?
#4
Report 8 months ago
#4
Did some research

Apparently there's anothe Luminosity equation apart from the one we've learnt:

L = 4*p1*(d^2)*b where b is the brightness.

But b should be in Wm^-2 , but in the question they give you 2.6x10^-11 in W so im guessing that the divided by cross-sectional area to get the Wm^-2 unit.
Equation could be re-written as 4*pi*(d^2)*Power / Cross-sectional Area.

OCR are proper snakey, never heard of this equation.

Hope this helped
1
reply
Agent007
Badges: 14
Rep:
?
#5
Report 8 months ago
#5
(Original post by Agent007)
Did some research

Apparently there's anothe Luminosity equation apart from the one we've learnt:

L = 4*p1*(d^2)*b where b is the brightness.

But b should be in Wm^-2 , but in the question they give you 2.6x10^-11 in W so im guessing that the divided by cross-sectional area to get the Wm^-2 unit.
Equation could be re-written as 4*pi*(d^2)*Power / Cross-sectional Area.

OCR are proper snakey, never heard of this equation.

Hope this helped
Now that I think about it, by brightness they probably meant intensity? So it could be 4*pi*(d^2)*Intensity
0
reply
FP5
Badges: 12
Rep:
?
#6
Report 8 months ago
#6
Is the correct answer roughly 5.53 x 10^34 by any chance?

If this is the correct answer, I will show you my workings.
0
reply
FP5
Badges: 12
Rep:
?
#7
Report 8 months ago
#7
Scrap that, is it 6.5 x 10^14?
0
reply
Eimmanuel
  • Community Assistant
Badges: 12
Rep:
?
#8
Report 8 months ago
#8
(Original post by Fross877)
A sensor of cross-sectional area 4 x 10-4 mounted on a satellite orbiting the Earth is used to gather the EM radiation from the star Antares.
Antares is 550 light years from the Earth. The radiant power entering the sensor from Antares is 2.6 x 10-11W

i) Calculate the luminosity L of Antares.

L = P/A = 2.6x10-11/4x10-4 = 6.5 x 10-8

I dont understand why the markscheme has multiplied the radiant power by the distance to the star? Isn't the P of the equation the radiant power incident on the sensor?
I follow the definition of luminosity from Wikipedia
"In astronomy, luminosity is the total amount of energy emitted per unit time by a star, galaxy, or other astronomical object. As a term for energy emitted per unit time, luminosity is synonymous with power."

We can solve the problem using standard ratio of finding the luminosity or based on intensity formula.

I assume the cross sectional area to 4 × 10−4 m because no unit is stated in the question.

Method 1: Ratio
For a cross sectional area of 4 × 10−4 m, the power is 2.6 × 10−11 W.

Since Antares is 550 light years from the Earth, the total spherical surface area is


 4 \pi R^2 = 4 \pi (550 \times 9.4607 \times 10^{15} )^2

The power corresponding to this area is


 \dfrac{ 4 \pi (550 \times 9.4607 \times 10^{15} )^2  \text{m}^2}{4 \times 10^{-4} \text{m}^2 } \times 2.6 \times 10^{-11} \text{W} = 2.21 \times 10^{31} \text{W}


Method 2 : intensity formula
You had already determined the intensity to be 6.5 × 10−8 W m−2. Extended to find the total power of Antares which is the luminosity of Antares,

  I = \dfrac{ P_{total}}{4 \pi R^2 } => P_{total} = 4 \pi R^2 \times I = 4 \pi (550 \times 9.4607 \times 10^{15} )^2 \times  6.5 \times 10^{-8} = 2.21 \times 10^{31} \text{W}

I hope it helps.
1
reply
Fross877
Badges: 10
Rep:
?
#9
Report Thread starter 8 months ago
#9
(Original post by FP5)
Scrap that, is it 6.5 x 10^14?
The answer was 2.2 x 10^31
0
reply
Fross877
Badges: 10
Rep:
?
#10
Report Thread starter 8 months ago
#10
(Original post by Eimmanuel)
I follow the definition of luminosity from Wikipedia
"In astronomy, luminosity is the total amount of energy emitted per unit time by a star, galaxy, or other astronomical object. As a term for energy emitted per unit time, luminosity is synonymous with power."

We can solve the problem using standard ratio of finding the luminosity or based on intensity formula.

I assume the cross sectional area to 4 × 10−4 m because no unit is stated in the question.

Method 1: Ratio
For a cross sectional area of 4 × 10−4 m, the power is 2.6 × 10−11 W.

Since Antares is 550 light years from the Earth, the total spherical surface area is



 4 \pi R^2 = 4 \pi (550 \times 9.4607 \times 10^{15} )^2



The power corresponding to this area is



 \dfrac{ 4 \pi (550 \times 9.4607 \times 10^{15} )^2  \text{m}^2}{4 \times 10^{-4} \text{m}^2 } \times 2.6 \times 10^{-11} \text{W} = 2.21 \times 10^{31} \text{W}




Method 2 : intensity formula
You had already determined the intensity to be 6.5 × 10−8 W m−2. Extended to find the total power of Antares which is the luminosity of Antares,

  I = \dfrac{ P_{total}}{4 \pi R^2 } => P_{total} = 4 \pi R^2 \times I = 4 \pi (550 \times 9.4607 \times 10^{15} )^2 \times  6.5 \times 10^{-8} = 2.21 \times 10^{31} \text{W}

I hope it helps.
Thank you for the help
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Edge Hill University
    Undergraduate and Postgraduate - Campus Tour Undergraduate
    Mon, 18 Feb '19
  • University of the Arts London
    MA Innovation Management Open Day Postgraduate
    Mon, 18 Feb '19
  • University of Roehampton
    Department of Media, Culture and Language; School of Education; Business School Undergraduate
    Tue, 19 Feb '19

Do you give blood?

Yes (46)
9.22%
I used to but I don't now (14)
2.81%
No, but I want to start (177)
35.47%
No, I am unable to (117)
23.45%
No, I chose not to (145)
29.06%

Watched Threads

View All