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Alex787 [66]
3 years ago
15

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha

t the emissivity eee is equal to 1 for these surfaces.
Required:
a. Find the radius RRigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7 x 10^31 W and has a surface temperature of 11,000 K.
b. Find the radius RProcyonB of the star Procyon B, which radiates energy at a rate of 2.1 x 10^23 W and has a surface temperature of 10,000 K. Assume both stars are spherical. Use σ=5.67 x 10−8^ W/m^2*K^4 for the Stefan-Boltzmann constant.
Physics
1 answer:
belka [17]3 years ago
8 0

Given that,

Energy H=2.7\times10^{31}\ W

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

H=Ae\sigma T^4

A=\dfrac{H}{e\sigma T^4}

4\pi R^2=\dfrac{H}{e\sigma T^4}

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}

R=5.0\times10^{10}\ m

(b). Given that,

Radiates energy H=2.1\times10^{23}\ W

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}

R=5.42\times10^{6}\ m

Hence, (a). The radius of the star is 5.0\times10^{10}\ m

(b). The radius of the star is 5.42\times10^{6}\ m

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From the case we know that:

  1. The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
  2. The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
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Please refer to the image below.

We know from the case, that:

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We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

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Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

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Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

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