The Sun
The Sun

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The Sun

2.6 End-of-course questions

Question 1

What is the approximate wavelength range (in metres) of microwaves?

Answer

The range is from about 1 m down to a little less than 0.001 m (see Figure 6).

Question 2

What are the main modes of energy transfer that participate in the transport of energy from the Sun's core to the Earth's surface?

Answer

Energy released by nuclear reactions in the Sun's core is initially carried away by radiation. Although repeatedly absorbed and re-emitted, it is radiation that carries the energy through the radiative zone. It is absorbed at the bottom of the convective zone, where it causes the convective flows that are largely responsible for transporting energy up to the photosphere. Radiation takes over again in the photosphere; the radiation emitted there escapes from the Sun and may transport energy to the Earth's surface where it can be absorbed.

Question 3

In a sentence or two, explain why the problem of the Sun's fuel puzzled scientists, and say how the discovery of nuclear reactions solved the puzzle.

Answer

Conventional fuels could maintain the Sun's output of light and heat for only a few thousand years, which is not nearly long enough to sustain the evolution of life on Earth over the millions of years deduced from fossil records. Nuclear reactions produce much more energy output for a given amount of fuel, enabling the Sun to emit a steady output over thousands of millions of years.

Question 4

The Sun is about 150 million km from the Earth. Use information from Section 4 about the angular diameters of the Sun and the Moon to calculate the Sun's approximate diameter. You should note the approximate relationship:

actual size = (angular size in degrees × distance) ÷ 57°.

(Do not spend more than a few minutes trying to answer this. There is a second way of tackling this question - see whether you can find it.)

Answer

There are two methods for answering this question.

Method 1

The expression given is:

actual size = (angular size in degrees × distance) ÷ 57.

This can be rewritten as:

diameter of Sun = (angular size of Sun in degrees × distance of Sun) ÷ 57.

The angular size of the Sun is about the same as that of the Moon, i.e. 0.5°, so:

diameter of Sun = (0.5 × 150 million km) ÷ 57 = 1.32 million km.

Method 2

The Sun's diameter is about 400 times the diameter of the Moon. Therefore, if the Moon is 3476 km in diameter, the Sun must have a diameter of about 400 × 3476 km = 1.39 million km. {This is not as accurate as in Method 1, because the diameter of the Sun is about 400 times the diameter of the Moon.}

Question 5

A large sunspot is observed to have an angular size one-twentieth that of the Sun. Taking the Sun's angular size to be 0.5°, what is the angular size of the sunspot? Express your answer in (a) arcmin and (b) arcsec.

Answer

(a) There are 60 arcmin in one degree (1°), so the Sun's angular size is 30 arcmin. The angular size of the sunspot is:

30 arcmin ÷ 20 = 1.5 arcmin

(b) 1.5 arcmin = 90 arcsec

S194_1

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