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Brian M.
Experienced teacher in astronomy
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Astrophysics
TutorMe
Question:

Imagine yourself sitting in a room with a 100 W light bulb located about 2 $$\mathrm{m}$$ away. (a) What is the flux (in $$\mathrm{erg}\;\mathrm{cm}^2\;\mathrm{s}^{-1}$$) of the light bulb? (b) How far away would the Sun have to be to have the same flux? (c) What is the absolute bolometric magnitude of the light bulb?

Brian M.
Answer:

(a) Flux is $$f = \frac{L}{4\pi r^2},$$ where $$L$$ is luminosity and $$r$$ is radius. In this case, luminosity is $$L = 100 W \cdot \frac{10^7\;\mathrm{erg}}{\mathrm{J}} = 10^9\;\mathrm{erg}\;\mathrm{s}^{-1}$$, and the radius is $$r = 2\;\mathrm{m} \cdot \frac{100\;\mathrm{cm}}{\mathrm{m}} = 200\;\mathrm{cm}$$. The flux is then $$f = 2\times10^{3}\;\mathrm{erg}\;\mathrm{cm}^2\;\mathrm{s}^{-1}$$. (b) First set the fluxes equal to one another: $$f_\odot = f \rightarrow \frac{L_\odot}{4\pi r_\odot^2} = \frac{L}{4\pi r^2}.$$ Now solve for the distance to the Sun: $$r_\odot = r \sqrt{\frac{L_\odot}{L}}.$$ The luminosity of the Sun is $$L_\odot = 2\times10^{33}\;\mathrm{erg}\;\mathrm{s}^{-1}.$$ Plugging in the numbers, this results in $$r_\odot = 3\times10^{14}\;\mathrm{cm} \approx 20\;\mathrm{au}$$, so if you were on Uranus the Sun would appear about as bright as the 100 W light bulb in your room. (c) Bolometric magnitude based upon that of the Sun is $$M_{bol} - M_{\odot,bol} = -2.5 \log \frac{L}{L_\odot},$$ where $$M_{bol}$$ is the absolute bolometric magnitude and $$M_{\odot,bol} = 4.74$$ is the absolute bolometric magnitude of the Sun. Plugging in the numbers, this results in $$M_{bol} = 70.$$

C++ Programming
TutorMe
Question:

If you create a class foo for which you wish to overload the + operator to work with a built-in type, say float. How can you create an operator that allows both $$\mathrm{foo} = \mathrm{foo} + 12.5$$ and $$\mathrm{foo} = 12.5 + \mathrm{foo}$$?

Brian M.
Answer:

The overload operator can be declared as a method of the class foo in the usual way, i.e. class foo { // ... define other components of foo public: inline foo operator +(float i_fValue) const { foo cRet; //perform desired operation between the cRet instance and i_fValue return cRet; } }; This definition allows $$\mathrm{foo} = \mathrm{foo} + 12.5$$. The operator is declared as const because the contents of the instance on which it is called are not modified,. To allow the other method, $$\mathrm{foo} = 12.5 + \mathrm{foo}$$, we need to override the operator of the base type (i.e. that of float), with a right hand operator of type foo. For this operator, we need to explicitly specify that the operator is binary. To ensure consistency and minimize opportunity for duplication errors, the operator should use the method defined within foo. foo operator +(float i_fValue, foo i_cFoo) { return i_cFoo + i_fValue; }

Astronomy
TutorMe
Question:

(a) Consider two stars that have the same spectral type. What can you say about the temperature of those stars? (b) One star of type K5 has a luminosity of 16\% that of the Sun; another has a luminosity 32$$\times$$ (3200%) that of the Sun, and a third has a luminosity 16,000$$\times$$ that of the Sun. (i)What relation are you aware of between the temperature and the luminosity of stars? (ii)How can you explain the difference between the stars? (iii)What is the quantitative relation between the relevant physical parameter of these stars?

Brian M.
Answer:

(a) The temperature of the stars are all the same. Spectral type is a proxy for temperature. (b) (i) The luminosity and temperature are related by the Stefan-Boltzmann law. Specifically, $$L \propto T^4 R^2$$, where $$L$$ is the luminosity, $$T$$ is the temperature, and $$R$$ is the radius of the star. (b) (ii) The stars that are brighter must be bigger in radius to explain the higher luminosity. (b) (iii) To find these relations, we know that the luminosity is proportional to the square of the radius and the 4th power of the temperature. If we take the ratio of these, then the ratio of luminosities is related to the ratio of temperatures and ratio of radii. Since the temperatures are the same, the ratio of temperatures is 1. We are left with the ratio of luminosities and the ratio of radii: $$\frac{L_2}{L_1} = \frac{R_2^2}{R_1^2},$$ where the subscript (1 or 2) refers to a particular star. We can take the square root of both sides so that we can find the ratio of radii: $$\sqrt{\frac{L_2}{L_1}} = \sqrt{\frac{R_2^2}{R_1^2}} = \frac{R_2}{R_1},$$ Then using the values above: $$\frac{R_2}{R_1} = \sqrt{\frac{32}{0.16}} \sim 14,$$ and $$\frac{R_3}{R_1} = \sqrt{\frac{16000}{0.16}} \sim 316,$$ so the star 32$$\times$$ the brightness of the Sun is 14 times larger than the star that is 16\% the brightness of the Sun. The star that is 16,000$$\times$$ the brightness of the Sun is about 300 times the radius of the the dimmest star.

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