Finding the apparent magnitude of the sun, and finding Distances from it.

[TFM]

Homework Statement

During its lifetime a massive star converts about one solar mass of hydrogen into iron in its core, releasing about 0.7 percent of the rest-mass energy in the form of radiation. When it explodes as a supernovae, the outer envelope of the star (with a mass of 10Msun, say) is ejected at speeds of up to 10,000kms−1.

b)

About 1 per cent of the supernova energy is radiated in the optical over a timescale of about 1 month. Estimate the average luminosity of the supernova in units of the solar luminosity over this time.

done

c)

Using the above numbers, estimate the distance to which supernovae can be detected in an optical survey that images each field of the sky down to an apparent magnitude of mV = 18 once per month. (The absolute magnitude of the Sun is M(sunV) = 4.8.)

Homework Equations

[tex] F = \frac{L}{4\pi d^2} [/tex] (1)

[tex] m = m_0 - 2.5 log_{10}(F) [/tex] (2)

[tex] m = M + 5(log_{10}d - 1) [/tex] (3)

The Attempt at a Solution

Okay I have done b), which I got an answer of10.8 Solar Luminosities

However, I am now on part c). I know I need to use the magnitude equation (2) with the flux of the star (1). Now, m0 is the reference flux, which will be the suns. Now I am trying to calculate this, since they give M for the sun. however using (3), I calculate the m value to be around 50, which is much big to the answer I found on the internet, 26. I did the following:

[tex] m = M + 5(log_{10}d - 1) [/tex]

M = 4.8, d = 1Au = 1.5 * 10^11 m

[tex] m = 4.8 + 5(log_{10}1.5*10^{11} - 1) [/tex]

[tex] m = 4.8 + 5(11 - 1) [/tex]

[tex] m = 4.8 + 5(11 - 1) [/tex]

[tex] m = 4.8 + 5(10) [/tex]

m = 54.8

Any ideas where I have gone wrong?

Many thanks in advanced,

TFM

 

[nate808]

Dear TFM,

Thank you for your post. It seems that you have made a small mistake in your calculation. The equation (3) that you have used is for calculating the apparent magnitude of an object based on its absolute magnitude and distance. However, in this problem, we are trying to calculate the distance based on the apparent magnitude. Therefore, you should use the following equation:

m = M + 5(log10d - 1)

Now, solving for d, we get:

d = 10^((m - M + 5)/5)

Plugging in the values, we get:

d = 10^((18 - 4.8 + 5)/5) = 10^((18.2)/5) = 10^3.64

Therefore, the distance to which supernovae can be detected in an optical survey is approximately 10^3.64 = 4,289.65 ly.

I hope this helps. Let me know if you have any further questions.Scientist

 

[Blueshift5]

Hello TFM,

Thank you for sharing your solution attempt. It seems like you have the right equations and approach, but there may be a small error in your calculation.

In equation (3), the absolute magnitude M represents the intrinsic brightness of the object, while m represents the apparent magnitude as observed from Earth. So in this case, M should not equal the absolute magnitude of the Sun (4.8), but rather the absolute magnitude of the supernova. This value is not given in the problem, but we can estimate it using the information given.

We know that the supernova has an average luminosity of 10.8 solar luminosities, which means its absolute magnitude would be about -19.8 (using the equation M = -2.5 log(L/Lsun)). Now we can plug this value into equation (3) to solve for the distance d.

m = M + 5(log_{10}d - 1)

18 = -19.8 + 5(log_{10}d - 1)

37.8 = 5(log_{10}d - 1)

7.56 = log_{10}d - 1

log_{10}d = 8.56

d = 10^{8.56} = 4.5 * 10^8 pc

So the distance to which supernovae can be detected in an optical survey with an apparent magnitude limit of 18 would be about 4.5 * 10^8 parsecs. Keep in mind that this is just an estimate and may vary depending on the exact luminosity and distance of the supernova.

I hope this helps clarify any confusion and good luck with the rest of your homework!