Chapter 12: Earth, Moon, and the Sun

Let us first recall what Fig. 12.17 shows.

• The drawing represents the June solstice, when the Earth’s axis is tilted towards the Sun by about $$23.5^{\circ}$$.
• The circle that separates day from night (called the circle of illumination) is such that the entire region around the North Pole lies inside the daylight half, while the region around the South Pole lies completely inside the night half.

Now consider one complete rotation of the Earth (one full day):

1. North Pole
   Because the North Pole never crosses into the night half at this time of the year, it remains lit for the whole 24-hour period.
   Therefore, hours of sunlight at the North Pole = $$24\,\text{h}$$.
2. South Pole
   Conversely, the South Pole is entirely inside the dark half. As the Earth rotates, it never receives direct sunlight.
   Therefore, hours of sunlight at the South Pole = $$0\,\text{h}$$.

Thus, during the June solstice pictured in Fig. 12.17, the North Pole enjoys continuous daylight while the South Pole remains in continuous darkness.

Observation provided
Yesterday (Day 1) the constellation Orion was almost exactly overhead at 8 pm.

1. Why do constellations shift their overhead time?
The Earth has two simultaneously running motions:

• Rotation about its own axis: one solar day  = 24 h.
• Revolution around the Sun: one year  = 365 days.

Because the Earth advances a little (about $$1^{\circ}$$) along its orbit each day, it has to turn through slightly more than $$360^{\circ}$$ to bring the Sun back to the meridian. Hence the interval between two successive “noons” (a solar day) is longer than the interval between two successive alignments with a distant star (a sidereal day).

2. Length of a sidereal day

Actual values

$$\text{Solar day}=24\,\text{h}=1440\,\text{min}$$
$$\text{Sidereal day}=23\,\text{h}\;56\,\text{min}\;4\,\text{s}\approx1436.1\,\text{min}$$

Extra time the Sun needs each day:

$$1440\,\text{min}-1436.1\,\text{min}\approx3.9\,\text{min}\approx4\,\text{min}$$

Therefore every fixed star or constellation appears about 4 minutes earlier in the sky on each successive night.

3. Applying the 4-minute rule

Time Orion was overhead yesterday: $$8{:}00\,\text{pm}$$.

Shift next night: $$4\,\text{min}$$ earlier.

$$8{:}00\,\text{pm}-0{:}04=7{:}56\,\text{pm}$$

4. Conclusion
Padmashree will find Orion nearly overhead today at about 7 : 56 pm (roughly four minutes earlier).

Step 1 – Recall the daily shift of the night sky
Because the Earth goes once round the Sun in a year, the night-side of the Earth faces slightly different directions from one night to the next. A given star therefore rises about 4 minutes earlier every day.

Step 2 – Find how many days make the star come back to the same clock time
If the star has to rise at the same clock time (midnight) again, the total advance must add up to a full 24 h.

Time that has to be gained  = $$24 \text{ h}=24\times 60\;\text{min}=1440\;\text{min}$$

Daily advance  = $$4\;\text{min per day}$$

Number of days needed  = $$\dfrac{1440\;\text{min}}{4\;\text{min day}^{-1}} = 360\;\text{days}$$

Step 3 – Convert 360 days into a calendar date
Starting from 21 June, 360 days bring us to 16 June of the next calendar year (5 days earlier than the starting date).

Step 4 – Conclusion
Hence the same group of stars will again rise exactly at midnight about 360 days later, i.e. on ~16 June of the next year.

Concept 1 : The Earth is a sphere that rotates about its own axis

The rotation is from west to east and takes 24 hours for one complete turn. Because of this rotation

• the face that is turned towards the Sun is illuminated → day-time,
• the opposite face is in the Earth’s own shadow → night-time.

Concept 2 : Angular speed of rotation

The Earth sweeps $$360^{\circ}$$ in 24 h, so the angular speed is

$$\dfrac{360^{\circ}}{24\;\text{h}} = 15^{\circ}\;\text{per hour}.$$/p>

Concept 3 : Longitude decides local time

Every $$15^{\circ}$$ of longitude toward the east makes the local time 1 hour ahead; toward the west it is 1 hour behind.

Data for the two places

• India’s standard meridian: $$82.5^{\circ}\text{E}$$
• (One of) U.S.A. central meridian: about $$90^{\circ}\text{W}$$

The total longitudinal separation is

$$82.5^{\circ} + 90^{\circ} = 172.5^{\circ}.$$/p>

Time difference

$$\text{Time difference} = \dfrac{172.5^{\circ}}{15^{\circ}/\text{h}} = 11.5\;\text{h}\;\bigl(11\,\text{h}\,30\,\text{min}\bigr).$$

Thus local time in the U.S.A. is roughly 11–12 hours behind that in India. When it is about 10 a.m. in India, it will be

$$10\,\text{a.m.} - 11.5\,\text{h} \approx 10\,\text{p.m. (previous day)}$$

in the central United States, which is normal sleeping time.

Conclusion

Day and night do not occur everywhere on Earth at the same moment because the Earth keeps turning. Places lying toward the east (India) receive sunlight earlier, while places far to the west (U.S.A.) face away from the Sun at that instant, making it night there. This rotational effect produces the familiar time-zone differences that Abhay observed.

Concept recall — Why can looking at the Sun be dangerous?

• The Sun is an extremely strong source of visible and invisible (infra-red and ultraviolet) radiation.
• The lens of our eye focuses this intense light on a very small spot of the retina. Even a few seconds of direct exposure during an eclipse can burn that spot permanently (a condition called solar retinopathy).
• Therefore the only safe methods are those that reduce the intensity by many thousand times before the light enters our eyes.

Check each friend’s method

1. Ravikiran: He used a pair of certified solar-eclipse goggles (made with special Mylar or black polymer film).  These filters cut down the Sun’s brightness by a factor of about $$10^{5}$$, so they are safe.
2. Jyothi: She turned her back to the Sun and used a small mirror to project the Sun’s image onto a distant wall/screen.  The light never goes directly into her eyes, hence this indirect projection method is also safe.
3. Adithya: He looked straight at the Sun with his unaided eyes. Nothing reduced the Sun’s brightness, so this is unsafe and careless.
4. Aruna: She watched the eclipse through arrangements made by a planetarium. Such programmes always provide safe filters or projection devices, so her method is safe.

Conclusion

Only Adithya’s action involves direct viewing without protection, which can seriously damage eyesight.

The three empty circles in Fig. 12.18 represent the three bodies that take part in the Earth–Moon–Sun system. We decide which is which by comparing two obvious facts that you already know.

1. The source of light. One circle has arrows (or rays) coming out of it. The only real source of light in this trio is the Sun. Hence that largest, ray-emitting circle must be filled with the word Sun.
2. Relative sizes.
   The radii of the three bodies follow the order $$R_{\text{Sun}} \;\gg\; R_{\text{Earth}} \;>\; R_{\text{Moon}}.$$ Therefore:
   • the biggest remaining circle (after the Sun) stands for the Earth,
   • the smallest circle stands for the Moon.
   Because the Moon goes round the Earth, many textbook drawings also show the small circle close to (or orbiting) the medium one, which is another clue that the medium circle is the Earth.

Writing the names inside the circles, from the largest to the smallest, gives the correct labelling:

• Sun   →  Earth   →  Moon

Thus the figure should read “Sun” in the largest, ray-emitting circle, “Earth” in the medium circle, and “Moon” in the smallest circle.

Given fact to be explained — In a total solar eclipse the Moon, although far smaller than the Sun, hides the whole solar disc from an observer on the Earth.

To understand this we have to compare the apparent (angular) sizes of the two bodies as seen from Earth. An object of physical diameter $$D$$ situated at distance $$d$$ from the eye subtends an angular diameter $$\theta$$ that, for small angles, is

$$\theta \approx \dfrac{D}{d}\;,\qquad(\text{in radians}).$$

Step 1 : List the necessary data

Step 2 : Calculate the Sun’s apparent (angular) diameter

$$\theta_S = \dfrac{D_S}{d_S} = \dfrac{1.39\times10^6}{1.496\times10^8}\,\text{rad} \approx 0.0093\,\text{rad} \;\;(\approx 0.53^{\circ}).$$

Step 3 : Calculate the Moon’s apparent (angular) diameter

$$\theta_M = \dfrac{D_M}{d_M} = \dfrac{3.48\times10^3}{3.84\times10^5}\,\text{rad} \approx 0.0091\,\text{rad} \;\;(\approx 0.52^{\circ}).$$

Step 4 : Compare the two angles

The numerical results show that $$\theta_M$$ and $$\theta_S$$ are practically equal. Stated as a proportion

\[ \frac{D_M}{d_M} \;\approx\; \frac{D_S}{d_S}. \]

This means the Moon and the Sun appear almost the same size on the sky. Therefore, when the Moon happens to line up exactly between the Earth and the Sun, its disc can cover the Sun’s disc completely, producing a total solar eclipse.

Key idea in one sentence — The Sun is about 400 times larger in diameter than the Moon but it is also about 400 times farther away, so their apparent diameters are nearly equal, letting the nearer (and therefore apparently equally large) Moon hide the farther Sun.

Step 1 – Recall the reason for seasons.
The Earth’s axis is tilted by about $$23.5^{\circ}$$ with respect to the line perpendicular to its orbital plane. Because of this tilt, as the Earth goes around the Sun, each hemisphere is inclined either towards or away from the Sun at different times of the year. The hemisphere tilted towards the Sun receives more direct sunlight and has summer; the opposite hemisphere receives slanting rays and has winter.

Step 2 – Link months to hemispheres.
In the Northern Hemisphere (where India lies):

• June–August ⇒ summer
• September–November ⇒ autumn
• December–February ⇒ winter
• March–May ⇒ spring

• December–February ⇒ summer
• March–May ⇒ autumn
• June–August ⇒ winter
• September–November ⇒ spring

Step 3 – Locate Australia.
Australia is south of the equator, so it is in the Southern Hemisphere.

Step 4 – Determine the Australian season in December.
From Step 2, December belongs to the Southern Hemisphere’s summer period.

Step 5 – Conclude what clothes are needed.
Since December is summertime in Australia, the Indian cricket team will experience hot weather there. Therefore they should pack summer clothes (light cotton clothing, caps, sunglasses, etc.), not woollens.

(Optional diagram to draw): Show the Earth in its orbit with the axis tilted. Mark the position where the South Pole is inclined towards the Sun (December). Shade the Southern Hemisphere to indicate larger sunlight area and label it “summer in Australia”.

Step 1 · Recall how eclipses are produced

• A lunar eclipse happens when the Earth comes between the Sun and the Moon. The Moon then goes through Earth’s shadow.
• A solar eclipse happens when the Moon comes between the Sun and the Earth. A part of Earth then enters the Moon’s shadow.

In both cases two shadow zones are formed:

• Umbra – the central, completely dark cone.
• Penumbra – the outer, partially shaded region.

Step 2 · Compare the sizes of the bodies

For the same distance from the Sun we have

$$\text{Diameter of Earth} \approx 4\,\text{times the diameter of Moon}$$

Because the Earth is far larger than the Moon, its umbral cone is also much wider where the Moon crosses it, while the Moon’s umbral cone is much narrower where it meets Earth.

Step 3 · Understand the shadow cones geometrically

• Umbra of Earth at the Moon’s distance
  The width of Earth’s umbra at the distance of $$3.8\times10^5\,\text{km}$$ (the Moon’s orbit) is about $$\sim 9000\,\text{km}$$. This is wider than the Moon itself ($$\sim 3500\,\text{km}$$), so the whole Moon can fit inside the dark cone.
• Umbra of Moon at Earth’s surface
  Because the Moon is small, its shadow tapers to only about $$\sim 100\!–\!200\,\text{km}$$ wide when it reaches Earth. That is narrower than a typical country!

Step 4 · Consequences for observers on Earth

1. Lunar eclipse – Anyone located on the night-side hemisphere of Earth (roughly half the globe) can look up and see the Moon pass through Earth’s large umbra or penumbra. Therefore the event is visible from a very large part of Earth simultaneously.
2. Total solar eclipse – To see totality your position must fall inside the Moon’s tiny umbra. As Earth rotates, this spot sweeps out a narrow track a few thousand kilometres long but only about $$100$$ – $$200\,\text{km}$$ wide. Only the people situated along that thin path see a total eclipse; those just outside see only a partial eclipse or none at all.

Step 5 · Key idea in one sentence

The Earth’s shadow is huge at the Moon’s distance, while the Moon’s shadow is tiny at Earth’s surface; hence a lunar eclipse is visible from the whole night side of Earth but a total solar eclipse is restricted to a narrow strip.

(Suggestion for the textbook diagram): Draw two separate pictures to scale: (a) Earth casting a wide cone that completely covers the Moon; shade the night-side hemisphere of Earth to show where the eclipse can be seen. (b) Moon casting a very narrow cone that just touches a small patch of Earth; shade only that path to indicate totality.

Step 1 · Present situation
The Earth’s axis is inclined to the perpendicular drawn to the plane of its orbit (the ecliptic) by about $$\theta = 23.5^{\circ}$$. Because of this tilt, while the Earth revolves around the Sun, sometimes the Northern Hemisphere is turned towards the Sun (more heat → summer there, winter in the South) and six months later the reverse happens. These regular changes in the angle of sunlight give us the four seasons.

Step 2 · Assume the axis is not tilted
If the axis were exactly perpendicular to the orbital plane, then $$\theta = 0^{\circ}$$.

Step 3 · Consequences for sunlight

• The Sun’s rays would strike the Equator at $$90^{\circ}$$ every day of the year.
• The “circle of illumination” (the boundary between day and night) would always pass through both the North and South Poles.
• Hence every place on Earth would experience exactly 12 h of daylight and 12 h of night all year.
• The noon-time height of the Sun over any latitude $$L$$ would be constant, given by
  \[ h = 90^{\circ} - |L| \]
  and would never change with date.

Step 4 · Effect on temperature
Because the altitude of the Sun and the duration of daylight decide how much heat reaches the ground, a constant value of $$h$$ and a fixed 12-hour day mean each latitude would receive the same amount of solar energy every single day.

Step 5 · Final conclusion
With no periodic change in either the angle or the duration of sunlight, there would be no seasons at all. The Equator would stay permanently hot, the Poles permanently cold, and the intermediate latitudes would have an unchanging moderate climate. Spring, summer, autumn and winter would never occur.

Step 1 : Fix the scale
At the given scale,

• $$1\,\text{cm} \Longrightarrow 10\,\text{million km}$$

Step 2 : Convert the two radii into real distances

Step 3 : Find the difference between the two distances

$$152\,\text{million km}-147\,\text{million km}=5\,\text{million km}$$

Step 4 : Notice how small the difference is

The average Earth–Sun distance is

$$\frac{147+152}{2}=149.5\,\text{million km}$$

Percentage difference:

$$\frac{5}{149.5}\times100\%\approx3.3\%$$

What to draw

1. Mark a point to represent the Sun as the common centre.
2. With a compass, draw Circle 1 of radius $$14.7\,\text{cm}$$.
3. Without shifting the centre, draw Circle 2 of radius $$15.2\,\text{cm}$$.

The inner circle shows Earth at its closest (perihelion ≈ 147 million km), and the outer circle shows Earth at its farthest (aphelion ≈ 152 million km). The two circles almost overlap, visually demonstrating that Earth’s orbit is only slightly oval.

Chapter 12 — EARTH, MOON AND THE SUN

Worked Solution | NCERT Class 7 Science

Question. Suppose the tilt of the Earth's axis of rotation increases. Will it cause more extreme seasons? Find out if the tilt of Uranus is more than the Earth and about the seasons there. Write an interesting article for a newspaper or your school magazine about it.

1. Present tilt of the Earth

The Earth’s axis is not upright; it is tilted by about $$23.5^{\circ}$$ with respect to the perpendicular to its orbital plane (the ecliptic).

This tilt is called the axial tilt or obliquity. Because of it, the two hemispheres receive different amounts of sunlight at different times of the year, giving us the four seasons.

2. What if the tilt becomes larger?

Intensity of sunlight on a surface depends on the angle $$\theta$$ between the Sun’s rays and a line perpendicular to the ground. For the same beam of sunlight, the power received per unit area is

$$I = I_{0}\cos \theta$$

where $$I_{0}$$ is the intensity when rays fall straight down. A smaller $$\theta$$ (Sun high in the sky) makes $$\cos\theta$$ larger, so the ground heats up more.

• In summer the hemisphere tilted towards the Sun has a smaller $$\theta$$, so it becomes warm.
• In winter the same hemisphere is tilted away; $$\theta$$ is larger and it becomes cold.

If the tilt increased from $$23.5^{\circ}$$ to, say, $$30^{\circ}$$ or $$40^{\circ}$$, the difference between summer and winter $$\theta$$ values would grow. Therefore

\[\text{Bigger tilt }\Rightarrow\ \text{more extreme seasons (hotter summers, colder winters).}\]

The tropics would extend farther from the equator, and the polar circles would creep closer to it.

3. Uranus — the planet lying on its side

• Measured axial tilt of Uranus: $$97.86^{\circ}\;\text{(≈}98^{\circ}\text{)}$$.
• This is far greater than Earth’s $$23.5^{\circ}$$, in fact the planet is almost rolling around the Sun!
• One Uranian year = 84 Earth years. During about 42 years one pole points almost directly at the Sun and the other pole is in total darkness. After half a revolution the situation reverses.
• Result: the most bizarre seasons in the Solar System – each pole has a 42-year “day” followed by a 42-year “night”. Equatorial regions get sunlight in brief but rapid cycles during the planet’s 17-hour rotation.

4. Newspaper / School-magazine article

HEADLINE : Will Earth One Day Have 6-Month Days? – Lessons from Uranus

Have you ever complained about a sweltering summer afternoon or a freezing winter night? Be thankful you don’t live on Uranus! Scientists report that Earth’s gentle axial tilt of $$23.5^{\circ}$$ is just right for the moderate seasons we know. If our planet leaned over a little more, every June would sizzle and every December would bite.

Think of sunlight as a torch shining on a globe. When you tilt the globe further, the top half soaks up a more concentrated beam while the bottom half is left in the dark. Mathematically the warmth you feel is given by $$I = I_{0}\cos \theta$$. Make the tilt larger and the summer side turns the torch straight on – instant heatwave! Six months later the same place will catch the torch at a very slanted angle – deep-freeze.

Uranus takes this idea to the extreme. Lying almost on its side with a spectacular $$98^{\circ}$$ tilt, it gives each pole a 42-year long day followed by 42 years of night. Any school on Uranus would have two very, very long terms!

Could Earth’s tilt change? Yes – the Moon and the Sun gently wobble it between $$22.1^{\circ}$$ and $$24.5^{\circ}$$ every 41,000 years, but that is mild. A much bigger nudge from a giant asteroid could set Earth on a crazier path. Until then, enjoy the rhythm of four seasons – they could have been a lot wilder.

— Submitted by the Class 7 Science Club

Conclusion

• Increasing the Earth’s tilt would indeed make summers hotter and winters colder.
• Uranus, with an axial tilt of about $$98^{\circ}$$, proves the rule by experiencing the most extreme seasons in the Solar System.