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Astronomy & motions: Earth Science Regents Practice

Practice Astronomy & motions for the Earth Science Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 10 · 1 pt

An observer in New York State measures the altitude of Polaris (the North Star). That altitude is approximately equal to the observer's

Worked solution
$\text{altitude of Polaris} \approx \text{latitude}$ A key Earth Science relationship.
Answer: \text{latitude}

Because Polaris sits nearly above Earth's axis, its angle above the horizon equals the observer's latitude in the Northern Hemisphere.

Question 1 · 2 pts

An ellipse is drawn to model a planet's orbit. The distance between its two foci is $2.0$ cm and the length of its major axis is $8.0$ cm. Using the equation on the ESRT, what is the eccentricity of this orbit?

Worked solution
$e = \dfrac{\text{distance between foci}}{\text{length of major axis}}$ ESRT eccentricity equation
$e = \dfrac{2.0\ \text{cm}}{8.0\ \text{cm}}$ Substitute the given lengths
$e = 0.25$ Eccentricity is unitless and ranges from 0 to 1
Answer: 0.25

Eccentricity measures how elongated an ellipse is and equals the distance between the foci divided by the length of the major axis, with values near 0 being nearly circular.

Question 2 · 2 pts

An observer in New York State measures the altitude of Polaris (the North Star) to be $43^\circ$ above the northern horizon. What is the observer's approximate latitude?

Worked solution
$\text{altitude of Polaris} = \text{observer's latitude}$ Key relationship for Northern Hemisphere observers
$43^\circ\ \text{altitude} \Rightarrow 43^\circ\,\text{N latitude}$ Apply the relationship directly
Answer: 43 degrees N

Because Polaris sits nearly directly above the North Pole, its angle above the northern horizon equals the observer's north latitude.

Question 3 · 2 pts

Which motion of Earth is chiefly responsible for the apparent daily rising of the sun in the east and setting in the west?

Worked solution
$\text{rotation: } 360^\circ \text{ in } 24\ \text{h}$ One full spin produces day and night
$\text{apparent sun motion} = 15^\circ/\text{h, east to west}$ Opposite to Earth's west-to-east spin
Answer: Earth's rotation on its axis

Earth's daily west-to-east rotation makes the sun and stars appear to move from east to west, producing the cycle of day and night.

Question 9 · 2 pts

The Sun appears to move across the sky from east to west at an average rate of about $15°$ per hour, an apparent motion produced by Earth's rotation on its axis. Approximately how many degrees will the Sun appear to move during a $3$-hour period?

Worked solution
$\text{rate} = 15°\ \text{per hour}$ Earth rotates 360 degrees in 24 hours, so 360/24 = 15 degrees per hour.
$\text{motion} = 15°/\text{hr} \times 3\ \text{hr} = 45°$ Multiply the apparent rate by the elapsed time.
Answer: $45°$

The Sun's apparent east-to-west motion is caused by Earth's west-to-east rotation; at $15°$ per hour it sweeps through $45°$ in $3$ hours.

Question 14 · 2 pts

Light from distant galaxies is observed to be shifted toward the red (longer-wavelength) end of the visible spectrum, and more distant galaxies show greater red shifts. This Doppler observation is the main evidence that

Worked solution
$\text{red shift} \Rightarrow \text{source receding (wavelength lengthened)}$ The Doppler effect lengthens the wavelength of a source moving away.
$\text{greater distance} \Rightarrow \text{greater red shift} \Rightarrow \text{faster recession}$
$\therefore \text{space between galaxies is expanding}$ This supports the Big Bang model.
Answer: The universe is expanding, with galaxies moving away from one another.

A red shift means a light source is moving away; because nearly all galaxies are red-shifted and more distant ones more strongly, the universe is expanding, which supports the Big Bang theory.

Question 1 · 2 pts

An observer standing at a location in the Northern Hemisphere measures the altitude of the star Polaris above the northern horizon to be $43°$. What is the observer's approximate latitude?

Worked solution
$In the Northern Hemisphere, the altitude of Polaris above the horizon equals the observer's latitude.$ This is because Polaris lies almost directly above Earth's North Pole, along the extension of Earth's axis.
$Altitude of Polaris = $43°$, so latitude = $43°N$.$
Answer: $43°N$

Because Polaris sits nearly on the extension of Earth's rotational axis, its altitude above the horizon as seen by an observer always equals that observer's latitude in the Northern Hemisphere.

Question 13 · 2 pts

Which motion of Earth is primarily responsible for the apparent daily rising and setting of the Sun?

Worked solution
$Earth spins on its axis once approximately every 24 hours.$ Rotation period
$This spin causes different locations on Earth to face toward or away from the Sun over the course of a day.$
$The result is the observed daily (diurnal) rising in the east and setting in the west of the Sun.$
Answer: Earth's rotation on its axis

The Sun's apparent daily east-to-west path across the sky is caused by Earth's rotation on its axis, not by its yearly revolution or axial tilt, which instead cause seasonal changes.

Question 26 · 2 pts

An observer in New York State measures the altitude of Polaris above the northern horizon to be 43 degrees. What is the observer's approximate latitude?

Worked solution
$For an observer in the Northern Hemisphere, the altitude of Polaris above the horizon is equal to the observer's latitude.$ Relationship from the Earth Science Reference Tables
$Altitude of Polaris = 43°, so latitude = 43°N$
Answer: 43°N

Because Polaris lies almost directly above Earth's North Pole, its angle of altitude as seen by an observer equals that observer's latitude in the Northern Hemisphere.

Question 39 · 2 pts

An observer standing at a location with a latitude of 42°N measures the altitude of the star Polaris above the northern horizon on a clear night. What is the approximate altitude of Polaris for this observer?

Worked solution
$\text{Altitude of Polaris} = \text{observer's latitude}$ Fundamental relationship on the ESRT
$\text{Latitude} = 42^{\circ}\text{N}$
$\text{Altitude of Polaris} = 42^{\circ}$
Answer: 42°

Because Polaris lies almost directly above Earth's North Pole along the extended axis of rotation, the angle of altitude at which an observer in the Northern Hemisphere sees Polaris is numerically equal to that observer's latitude.

Question 67 · 4 pts · constructed response

An observer standing on Earth's surface measures the altitude (angle above the horizon) of the star Polaris to be 43 degrees above the northern horizon. State the latitude of the observer. Explain the relationship between the altitude of Polaris and an observer's latitude that supports your answer.

Model answer: The observer's latitude is 43 degrees North.
Worked solution
$Polaris is located almost directly above Earth's North Pole, along Earth's axis of rotation.$ geometry of Polaris
$The altitude of Polaris above the horizon, in degrees, is always equal to the observer's latitude in the Northern Hemisphere.$
$Altitude of Polaris = 43 degrees, so latitude = 43 degrees N.$
Answer: The observer's latitude is 43 degrees North.

Because Polaris lies almost exactly along Earth's rotational axis, the angle you must look up from the horizon to see it exactly equals your latitude, a fact used by navigators for centuries; a 43 degree altitude therefore places the observer at 43 degrees N latitude.

Question 75 · 4 pts · constructed response

An observer in the Northern Hemisphere measures the altitude of Polaris above the northern horizon to be $43°$. State the observer's latitude and explain the relationship between the altitude of Polaris and an observer's latitude in the Northern Hemisphere.

Model answer: The observer's latitude is 43°N. In the Northern Hemisphere, the altitude of Polaris (in degrees above the horizon) is numerically equal to the observer's latitude, because Polaris lies nearly along the extension of Earth's rotational axis.
Worked solution
$Rule: altitude of Polaris = observer's latitude (Northern Hemisphere)$ General relationship
$Given altitude = 43°$ Apply the rule
$Latitude = 43°N$
Answer: The observer's latitude is 43°N. In the Northern Hemisphere, the altitude of Polaris (in degrees above the horizon) is numerically equal to the observer's latitude, because Polaris lies nearly along the extension of Earth's rotational axis.

Polaris is located almost directly above Earth's North Pole along the extension of the rotational axis, so its angle above the horizon as seen by an observer equals that observer's latitude in the Northern Hemisphere.

Question 1 · 2 pts

Which piece of evidence provides direct proof that Earth rotates on its axis?

Worked solution
$Seasonal change in visible constellations and changing daylight length are caused by Earth's revolution and axial tilt, not rotation.$ eliminate revolution-based evidence
$Solar eclipse timing depends on the Moon's orbital motion, not Earth's spin.$
$A Foucault pendulum keeps swinging in a fixed plane relative to space; as Earth turns beneath it, the swing plane appears to shift, directly demonstrating rotation.$
Answer: The plane of swing of a Foucault pendulum appears to rotate over the course of a day

A Foucault pendulum is the classic direct proof of Earth's rotation because its plane of swing stays fixed in an inertial frame while the ground (and observer) turns beneath it, unlike other seasonal or orbital effects which reflect revolution rather than spin.

Question 13 · 2 pts

An observer determines that the altitude of Polaris above the northern horizon is $42°$. What is the observer's latitude?

Worked solution
$Polaris lies nearly directly above Earth's North Pole, along the extension of Earth's axis of rotation.$ geometry
$The angle of altitude of Polaris above the horizon equals the observer's latitude on Earth.$
$Given altitude of Polaris $= 42°$, the observer's latitude $= 42°N$.$
Answer: $42°N$

Because Polaris is positioned almost exactly along Earth's rotational axis, the angle between the horizon and Polaris (its altitude) is numerically equal to the observer's latitude in the Northern Hemisphere. A common error is subtracting from 90 degrees, which does not apply to this relationship.

Question 25 · 2 pts

Earth rotates on its axis once approximately every 24 hours. Because of this rotation, the Sun, Moon, and stars appear to move across the sky each day, with the Sun rising in the east and setting in the west. In which direction does the Sun appear to move across the sky as seen by most observers on Earth's surface?

Worked solution
$Earth's actual rotation direction is west to east (counterclockwise as viewed from above the North Pole).$
$This real motion of the observer causes celestial objects to appear to move in the opposite direction across the sky.$
$Therefore the Sun appears to move east to west, rising in the east and setting in the west.$
Answer: East to west

Earth spins west to east, so from the surface the Sun, Moon, and stars all appear to travel east to west, which is why the Sun rises in the east and sets in the west every day.

Question 37 · 2 pts

Which observation provides direct evidence that Earth rotates on its axis?

Worked solution
$A Foucault pendulum swings in a fixed plane relative to space.$ The pendulum itself does not change its swing direction.
$Because Earth rotates beneath the pendulum, an observer on the ground sees the swing plane appear to shift direction over time.$
$Seasons and changing daylight length are caused by Earth's revolution and axis tilt, not its rotation.$ Eliminates B and C
$Orbital shape (elliptical path) relates to revolution, not rotation.$ Eliminates D
Answer: The apparent path of a Foucault pendulum appears to slowly rotate over the course of a day

Rotation is Earth spinning on its axis once every 24 hours. The Foucault pendulum is the classic proof of rotation because the pendulum's swing plane stays fixed in space while Earth turns underneath it, making the plane appear to shift to an observer on the ground.

Question 51 · 4 pts · constructed response

The Sun has an apparent magnitude of -26.8 and an actual magnitude (absolute magnitude) of 4.8, giving it a luminosity of 1 (on the Sun-luminosity scale) and a surface temperature of 5,700 K. A student compares the Sun to Star X, which has a surface temperature of 5,700 K and a luminosity of 100 (100 times the Sun's luminosity). Identify the classification of Star X as a main-sequence star, giant, or supergiant, based on its position relative to the Sun on the temperature-luminosity relationship, and explain your reasoning using the fact that both stars have the same surface temperature.

Model answer: Star X is a giant (or supergiant) star. Since Star X has the same surface temperature as the Sun but is far more luminous (100 times), it must have a much larger surface area to radiate that much more energy at the same temperature, meaning it is a much larger star than a main-sequence star of that temperature.
Worked solution
$Compare surface temperatures: Sun = 5,700 K; Star X = 5,700 K (equal)$ Same temperature means same color/spectral class
$Compare luminosities: Sun = 1; Star X = 100$ Star X radiates 100 times more energy per second
$Luminosity depends on both temperature and surface area$ L is proportional to (surface area) x (T^4)
$Since temperature is equal but luminosity is much greater, Star X must have a much greater surface area (radius)$ Larger radius stars at the same temperature are classified as giants or supergiants
$Conclusion: Star X is a giant star$ It lies above the main sequence on the Hertzsprung-Russell type relationship
Answer: Star X is a giant (or supergiant) star. Since Star X has the same surface temperature as the Sun but is far more luminous (100 times), it must have a much larger surface area to radiate that much more energy at the same temperature, meaning it is a much larger star than a main-sequence star of that temperature.

On the temperature-luminosity relationship used in Earth Science, stars of equal surface temperature but far greater luminosity must have far greater surface area, since luminosity increases with both temperature and size; this places large, highly luminous stars of a given temperature in the giant or supergiant category rather than on the main sequence.

Question 65 · 4 pts · constructed response

An observer standing in New York State measures the altitude of the star Polaris above the northern horizon to be 42 degrees. (a) Determine the observer's latitude. (b) Explain the general relationship between the altitude of Polaris above the horizon and an observer's latitude in the Northern Hemisphere.

Model answer: The observer's latitude is 42 degrees North. Because Polaris lies almost directly above Earth's North Pole (along Earth's rotational axis), the altitude of Polaris above the northern horizon, in degrees, is equal to the observer's latitude in the Northern Hemisphere.
Worked solution
$\text{Altitude of Polaris} = \text{observer's latitude (Northern Hemisphere)}$
$\text{Altitude given} = 42^{\circ} \Rightarrow \text{latitude} = 42^{\circ}\text{N}$
Answer: The observer's latitude is 42 degrees North. Because Polaris lies almost directly above Earth's North Pole (along Earth's rotational axis), the altitude of Polaris above the northern horizon, in degrees, is equal to the observer's latitude in the Northern Hemisphere.

Polaris sits almost directly above the North Pole, so a sighting of Polaris's altitude above the horizon directly gives the observer's latitude anywhere in the Northern Hemisphere.

Question 75 · 4 pts · constructed response

On March 21 (the spring equinox), an observer at a location in the Northern Hemisphere measures the Sun's altitude at solar noon to be 50 degrees above the horizon. Calculate the latitude of the observer. Show your work.

Model answer: The observer's latitude is 40 degrees North.
Worked solution
$\text{On an equinox, the Sun's declination is } 0^\circ.$ Sun is directly over the equator
$\text{Noon altitude} = 90^\circ - \text{latitude} + \text{declination}$
$50^\circ = 90^\circ - \text{latitude} + 0^\circ$
$\text{latitude} = 90^\circ - 50^\circ = 40^\circ \text{N}$
Answer: The observer's latitude is 40 degrees North.

At either equinox the Sun's rays strike the equator directly, so the noon altitude and latitude are related by a simple complement (90 degrees minus altitude equals latitude), letting the latitude be found directly from the measured Sun angle.

Question 1 · 2 pts

An observer in the Northern Hemisphere measures the altitude of Polaris above the northern horizon to be $42°$. What is the observer's approximate latitude?

Worked solution
$The altitude of Polaris above the horizon equals the observer's latitude in the Northern Hemisphere.$ Basic relationship from the ESRT star charts
$Altitude of Polaris = $42°$$ given
$Therefore latitude = $42°N$$
Answer: $42°N$

Because Polaris sits almost exactly above the North Pole, its altitude angle above the horizon as seen by an observer is numerically equal to that observer's latitude. This is a standard navigation and astronomy fact tested on the Earth Science Regents.

Question 42 · 2 pts

The apparent daily path of the Sun, Moon, and stars across the sky, moving from east to west, is a direct result of which motion of Earth?

Worked solution
$Earth spins west to east on its axis once every 24 hours.$ Rotation
$Because the observer on the surface is turning eastward, distant celestial objects appear to move westward across the sky each day.$
$Revolution around the Sun takes about 365.25 days and instead explains the yearly cycle of seasons and constellation changes, not the daily east-to-west motion.$
Answer: Earth's rotation on its axis

Daily (diurnal) motion of celestial objects is caused by Earth's rotation, completed once every 24 hours, while revolution around the Sun is responsible for much slower, yearly changes such as seasons.

Question 59 · 4 pts · constructed response

Earth revolves around the Sun in a nearly circular orbit with an average orbital speed of 30 kilometers per second. The circumference of Earth's orbit is approximately $9.4 \times 10^{8}$ kilometers. Using the relationship distance = speed $\times$ time, calculate the time, in days, required for Earth to complete one full revolution around the Sun. Show all work, including unit conversions.

Model answer: Approximately 363 days (about $3.13 \times 10^{7}$ seconds), which is close to Earth's actual orbital period of 365.25 days.
Worked solution
$time = distance \div speed$ rearrange distance = speed × time
$time = \dfrac{9.4 \times 10^{8}\ km}{30\ km/s} \approx 3.13 \times 10^{7}\ s$
$3.13 \times 10^{7}\ s \div 86{,}400\ s/day \approx 363\ days$ 86,400 seconds in one day
Answer: Approximately 363 days (about $3.13 \times 10^{7}$ seconds), which is close to Earth's actual orbital period of 365.25 days.

Since speed equals distance divided by time, time equals distance divided by speed. Dividing Earth's orbital circumference by its average orbital speed, then converting seconds to days using 86,400 seconds per day, gives an orbital period close to one year.

Question 22 · 1 pt

An elliptical orbit has a major axis of $20\ \mathrm{cm}$ and a distance between foci of $5\ \mathrm{cm}$. What is its eccentricity? (eccentricity $=$ distance between foci $\div$ length of major axis)

Worked solution
$e = \frac{d_{\text{foci}}}{L_{\text{major axis}}}$ Formula from the Earth Science Reference Tables.
$= \frac{5}{20} = 0.25$
Answer: e = 0.25

Eccentricity measures how 'stretched' an ellipse is: distance between foci divided by the major-axis length. 0 is a circle.

Question 4 · 2 pts

The sun appears to move across the sky at a steady rate due to Earth's rotation. An observer notes that the sun is $30^\circ$ west of the noon meridian. Approximately how much time has passed since solar noon?

Worked solution
$\text{rate} = \dfrac{360^\circ}{24\ \text{h}} = 15^\circ/\text{h}$ Apparent rate of the sun's motion
$t = \dfrac{\text{angle}}{\text{rate}} = \dfrac{30^\circ}{15^\circ/\text{h}}$ Divide angular distance by rate
$t = 2.0\ \text{h}$
Answer: 2.0 hours

The sun appears to move 15 degrees each hour, so the angular distance from the noon meridian divided by 15 deg/h gives the time elapsed since solar noon.

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