HS-ESS1-4NGSS*ScienceEarth and Space ScienceGrades 9-12
HS-ESS1-4: Predicting Orbits with Kepler's Laws and Newton's Gravity
In plain English: HS-ESS1-4 is the NGSS high school performance expectation that asks students to use mathematical or computational representations to predict the motion of orbiting objects in the solar system. Students apply Kepler's laws and Newton's law of gravitation to planets, moons and satellites, one pair of bodies at a time. It is usually taught in Earth and Space Science or Physics.
Use mathematical or computational representations to predict the motion of orbiting objects in the solar system.
Clarification Statement: Emphasis is on Newtonian gravitational laws governing orbital motions, which apply to human-made satellites as well as planets and moons.
Assessment Boundary: Mathematical representations for the gravitational attraction of bodies and Kepler's Laws of orbital motions should not deal with more than two bodies, nor involve calculus.
Next Generation Science Standards · Disciplinary Core Idea: Earth's Place in the Universe · Official standard
Students use equations and spreadsheets to predict where and how fast objects in the solar system move. They apply Kepler's three laws (elliptical orbits with the Sun at one focus, equal areas in equal times, and T² proportional to a³) and Newton's law of universal gravitation, F = Gm1m2/r². Setting the gravitational force equal to the force needed for a circular orbit gives the orbital speed v = √(GM/r) and the period T = 2π√(r³/GM), which students use to predict the motion of planets, moons and human-made satellites and then compare their predictions with observed values.
The NGSS Clarification Statement says: "Emphasis is on Newtonian gravitational laws governing orbital motions, which apply to human-made satellites as well as planets and moons." The Assessment Boundary says: "Mathematical representations for the gravitational attraction of bodies and Kepler's Laws of orbital motions should not deal with more than two bodies, nor involve calculus." Every problem on this page involves one central body and one orbiting body and uses algebra only. The page uses these standard values: G = 6.674 × 10⁻¹¹ N·m²/kg², Earth: M = 5.97 × 10²⁴ kg and R = 6,371 km, Sun: M = 1.99 × 10³⁰ kg, 1 AU = 1.496 × 10¹¹ m. Planet and moon orbits use published average values, rounded; any invented data are labeled as invented.
Learning Objectives
By the end of this lesson, students will be able to:
Use Kepler's third law, T² = a³ (years and AU), to predict the period or orbit size of an object orbiting the Sun
Use Kepler's first and second laws to describe an elliptical orbit and predict where an object moves fastest and slowest
Calculate the gravitational force between two bodies with F = Gm1m2/r² and predict how it changes with mass and distance
Predict the speed and period of a satellite or moon in a circular orbit with v = √(GM/r) and T = 2π√(r³/GM), and use a moon's orbit to find the mass of its planet
Build a spreadsheet or table that turns orbit data into predictions and compare the predictions with observed values
Prior Knowledge Required
Students should already be comfortable with:
The role of gravity in the motions of the solar system MS-ESS1-2
Scientific notation and unit conversion, including km to m and days to seconds
Rearranging formulas to solve for one quantity HSA.CED.A.4
Square roots, cube roots and fractional exponents such as a1.5 on a calculator
Project a table with the average distance from the Sun and the orbital period of Mercury, Venus, Earth and Mars, without any formula.
Warm-Up Prompt
"Mars is about 1.5 times as far from the Sun as Earth, but its year is almost 2 Earth years long. If the period only grew in proportion to the distance, what would Mars's period be? What does the real value tell you about how period depends on distance?"
Students notice that the period grows faster than the distance: Mars, 1.52 times as far from the Sun, takes 1.88 years, not 1.52. Record their guesses about why (longer path, slower speed) and return to them after Kepler's third law.
Direct Instruction15-20 minutes
Two ways to represent an orbit. Kepler's laws describe orbits: (1) each planet moves on an ellipse with the Sun at one focus; (2) the line from the Sun to the planet sweeps out equal areas in equal times, so the planet moves fastest at perihelion and slowest at aphelion; (3) T² is proportional to a³, and for objects orbiting the Sun T² = a³ when T is in years and a in AU. Newton explained these laws with gravity: F = Gm1m2/r². For a circular orbit, gravity supplies exactly the force needed to keep the object turning, GMm/r² = mv²/r, so v = √(GM/r) and T = 2πr/v = 2π√(r³/GM). The orbiting object's mass cancels. Use Diagram 1 for the third law and Diagram 2 for the first two laws. For satellites, remind students that r is measured from the center of Earth: r = REarth + altitude.
Kepler's third law: predicting a planet's year
Mars orbits the Sun at an average distance of 1.52 AU. Predict its orbital period.
Result: T² = a³ = 1.52³ = 3.51, so T = √3.51 ≈ 1.87 years. The observed value is 1.88 years.
Newton's law of gravitation: the Earth-Moon pull
The Moon (7.35 × 10²² kg) is on average 3.84 × 10⁸ m from Earth (5.97 × 10²⁴ kg). Find the gravitational force between them.
Result: F = (6.674 × 10⁻¹¹)(5.97 × 10²⁴)(7.35 × 10²²) ÷ (3.84 × 10⁸)² ≈ 1.99 × 10²⁰ N, the same size on each body.
Satellite speed and period (circular orbit)
The International Space Station orbits about 420 km above Earth's surface. Predict its speed and period.
Result: r = 6,371 km + 420 km = 6.791 × 10⁶ m. v = √(GM/r) = √(3.98 × 10¹⁴ ÷ 6.791 × 10⁶) ≈ 7.66 km/s. T = 2πr/v ≈ 5,570 s ≈ 93 minutes.
Kepler's first and second laws: a comet's elliptical orbit
Halley's Comet comes within 0.59 AU of the Sun at perihelion and reaches 35.1 AU at aphelion. Predict its period and compare its speeds at the two ends of the orbit.
Result: a = (0.59 + 35.1) ÷ 2 ≈ 17.8 AU, so T = 17.81.5 ≈ 75 years. At the two ends, speed × distance is the same, so vperihelion ÷ vaphelion = 35.1 ÷ 0.59 ≈ 59.
Designing an orbit: the geostationary satellite
A communications satellite must circle Earth once per sidereal day (86,164 s) so it stays above one point on the equator. Predict the orbital radius and altitude.
Result: r = ∛(GMT²/4π²) = ∛(3.98 × 10¹⁴ × 86,164² ÷ 4π²) ≈ 4.22 × 10⁷ m, so the altitude is about 42,160 km - 6,371 km ≈ 35,800 km.
After each example, ask: "Which two bodies are we modeling, which one is the central body, and which quantity did the representation let us predict?" Point out that Kepler's T² = a³ in years and AU is Newton's T² = 4π²a³/(GM) with the Sun's mass built in, so it does not work for moons or satellites.
Guided Practice15 minutes
Pairs test Kepler's third law with the published orbit data below (rounded standard values). They add a column for T² ÷ a³, then plot T against a and compare with Diagram 1.
Guided practice data: planets from Mercury to Saturn
Planet
a (AU)
T (years)
Mercury
0.387
0.241
Venus
0.723
0.615
Earth
1.000
1
Mars
1.524
1.881
Jupiter
5.203
11.86
Saturn
9.537
29.46
Compute T² ÷ a³ for each planet. Every value comes out between 0.99 and 1.01, which supports T² = a³.
Plot T against a. Explain why the points lie on a curve that bends upward rather than on a straight line.
Predict the period of Uranus, which orbits at 19.19 AU (about 84 years), and compare with the observed 84.0 years.
Write one sentence that states what the representation predicts and one sentence that names its limit (it applies only to objects orbiting the Sun).
Independent Practice10-15 minutes
Each student predicts the orbit of Phobos, the larger moon of Mars, using a two-body model of Mars (mass 6.42 × 10²³ kg) and Phobos. Phobos orbits 9,376 km from the center of Mars. Students find the period with T = 2π√(r³/GM) and convert it to hours.
Expected result: GM = 4.28 × 10¹³ m³/s², T ≈ 27,560 s ≈ 7.65 hours, close to the observed 7.66 hours. Ask students why Phobos rises in the west as seen from Mars (it goes around faster than Mars spins, once every 24.6 hours).
Closure10 minutes
Exit ticket: "Satellite X circles Earth at twice the orbital radius of satellite Y. Predict how many times longer X takes to complete one orbit, and name the law you used." (T ∝ r1.5, so 21.5 ≈ 2.83 times as long.) Return to the warm-up and ask students to explain the Mars period with T² = a³.
Differentiation Strategies
For Struggling Students
Give a spreadsheet with the constants already entered and the columns labeled (r in m, GM, v, T in s, T in minutes) so students focus on what each column predicts
Practice Kepler's third law with perfect cubes first (a = 4 AU, 9 AU) before using decimal distances
Provide a checklist for satellite problems: add Earth's radius, convert km to m, and convert seconds to minutes at the end
For Advanced Students
Show that T² = 4π²a³/(GM) reduces to T² = a³ in years and AU by substituting the Sun's mass, and explain why the constant differs for Jupiter's moons
Use the two-body form T² = 4π²a³/(G(M + m)) to explain why the Moon's predicted period improves when the Moon's own mass is included
Design a spreadsheet that takes any altitude and returns speed and period, then find the altitude that gives exactly 12 hours
Assessment Guidance
What to Look For
Check that students measure orbital radius from the center of the central body, convert km to m and days to seconds, and use T² = a³ only for objects orbiting the Sun. Look for predictions that are compared with an observed value and a stated reason for any gap. Listen for students who think a heavier satellite must move faster at the same height, or that there is no gravity where astronauts orbit.
02
Classroom Activities
3 Activities
1
Weigh Jupiter with Its Moons
25 minPairs
Pairs build a spreadsheet that uses the orbits of Jupiter's four large moons to test Newton's form of Kepler's third law and to calculate the mass of Jupiter. Each moon and Jupiter form their own two-body system.
Data (published values, rounded)
Io: a = 421,700 km, T = 1.769 days
Europa: a = 671,034 km, T = 3.551 days
Ganymede: a = 1,070,412 km, T = 7.155 days
Callisto: a = 1,882,709 km, T = 16.69 days
Procedure
Enter a in meters and T in seconds (multiply days by 86,400).
Add a column for a³ ÷ T². It comes out near 3.21 × 10¹⁵ m³/s² for every moon.
Add a column for M = 4π²a³ ÷ (GT²). Every moon gives about 1.90 × 10²⁷ kg.
Use the average value of a³ ÷ T² to predict the period of a small moon at 181,400 km from Jupiter's center (about 0.50 days), then compare with the observed 0.498 days for Amalthea.
Discussion Questions
All four moons give the same mass for Jupiter to within 0.03%. What does that agreement say about the model?
Why can this method find the mass of Jupiter but not the mass of any of its moons?
Why would using T² = a³ (years and AU) give a wrong answer for these moons?
Modification for Distance Learning
Share the spreadsheet template online. Students fill in the formulas at home and post their value of Jupiter's mass with one sentence about how close the four moons agree.
2
String-and-Pin Ellipses: Testing the First and Second Laws
20 minPairs
Pairs draw ellipses with two push pins and a 30 cm loop of string, then use the drawings to predict how an object's speed changes around its orbit.
Procedure
Push two pins into the cardboard 6 cm apart. Loop the string around them, pull it tight with a pencil and trace the ellipse. One pin stands for the Sun.
Measure the long axis (2a) and the closest and farthest distances from the Sun pin. Compute e = (distance between pins) ÷ (long axis).
Move the pins to 12 cm apart and repeat with the same loop.
Safety: push the pins in with the flat of the thumb, keep the points facing down into the cardboard, and return the pins to the teacher at the end.
Expected Results
With a 30 cm loop, the long axis is 30 cm minus the pin distance. Pins 6 cm apart: 2a = 24 cm, e = 0.25, closest 9 cm and farthest 15 cm, so the speed at the closest point is 15 ÷ 9 ≈ 1.7 times the speed at the farthest point. Pins 12 cm apart: 2a = 18 cm, e ≈ 0.67, closest 3 cm and farthest 15 cm, a speed ratio of 5.
Discussion Questions
Which ellipse is closer to Earth's nearly circular orbit (e = 0.017)? Which is closer to a comet's orbit?
Shade a thin wedge near the far end and a wide wedge near the Sun pin with about the same area. Which one takes the object longer to sweep out, if either?
What happens to the ellipse as the pins move together? What orbit shape does that describe?
Challenge Variation
Students choose a pin distance that gives the orbit of Mercury (e = 0.21) and predict its closest and farthest distances as fractions of the long axis.
3
Mission Control: Predict, Then Compare
25 minGroups of 3-4
Each station describes one real orbiting object with a single central body. Groups predict the requested quantity with the representation they choose, then open an envelope with the observed value and explain any difference. Groups rotate every 6 minutes.
Stations
Station 1, Hubble Space Telescope: altitude about 540 km. Predict the period (about 95.3 minutes; observed about 95.4).
Station 2, a GPS satellite: orbital radius 26,560 km from Earth's center. Predict the period (about 11.97 hours, so two orbits per day).
Station 3, the Moon: average distance 384,400 km. Predict the period with T = 2π√(r³/GM) (about 27.46 days) and compare with the observed 27.32 days.
Station 4, the dwarf planet Pluto: period 248 years. Predict its average distance from the Sun with T² = a³ (about 39.5 AU).
Discussion Questions
Station 3 is the only prediction that misses by more than 0.3%. Why? (Hint: the Moon's mass is about 1.2% of Earth's, and the two-body form T² = 4π²r³/(G(M + m)) fixes most of the gap.)
Which stations used Kepler's T² = a³ and which needed Newton's form? What decides?
A student used the altitude instead of the orbital radius at Station 1. How far off would the period be?
Modification for Distance Learning
Post each station as a slide with a hidden answer. Groups work in breakout rooms and reveal the observed value only after they submit their prediction.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Kepler's Third Law for the Planets
Orbital period against average distance for the six planets known before the telescope (published values), drawn to scale. Every point lies on the curve T = a1.5, which is Kepler's third law T² = a³ with T in years and a in AU. The curve bends upward: each step outward adds more time to the period than the step before.
Diagram 2: Kepler's First and Second Laws on One Orbit
An orbit with eccentricity 0.6, drawn to scale, with the Sun at one focus (first law). The two shaded wedges are swept out in equal times and have equal areas (second law). The farthest point is 4 times as far from the Sun as the closest point, so the speed at the closest point is 4 times the speed at the farthest point.
04
Homework Assignment
~30 min
HS-ESS1-4 Homework: Predicting Orbits
Directions: Each problem involves one central body and one orbiting body. Use G = 6.674 × 10⁻¹¹ N·m²/kg², Earth: M = 5.97 × 10²⁴ kg and R = 6,371 km, Sun: M = 1.99 × 10³⁰ kg, 1 AU = 1.496 × 10¹¹ m. Show each calculation with units, and end every problem with one sentence that states your prediction and compares it with an observed value when one is given.
Part 1: Kepler's Laws (Problems 1-2)
The dwarf planet Ceres orbits the Sun at an average distance of 2.77 AU. (a) Predict its orbital period. (b) An asteroid takes 5.20 years to orbit the Sun. Predict its average distance from the Sun. (c) Which of the two objects moves faster along its orbit? Explain with Kepler's third law.
A comet comes within 0.5 AU of the Sun and travels out to 9.5 AU. (a) Find the semi-major axis and the eccentricity of its orbit. (b) Predict its period. (c) How many times faster does it move at perihelion than at aphelion? (d) Sketch the orbit and mark where the comet spends most of its time.
Part 2: Newton's Gravitation and Circular Orbits (Problems 3-4)
Find the gravitational force between the Sun and Earth, which are 1.00 AU apart. Then predict the force if Earth were moved to 2.00 AU, and explain your answer with the inverse-square law.
An Earth-observation satellite orbits 2,000 km above the surface. (a) Predict its orbital speed in km/s. (b) Predict its period in minutes. (c) Compare it with the International Space Station at 420 km: which one takes longer to go around, and why?
Part 3: Using Data and Evaluating Claims (Problems 5-6)
The table gives published orbits of four moons of Saturn: Tethys (a = 294,619 km, T = 1.888 days), Dione (377,396 km, 2.737 days), Rhea (527,108 km, 4.518 days) and Titan (1,221,870 km, 15.945 days). (a) Build a table or spreadsheet that computes M = 4π²a³/(GT²) for each moon. (b) Do the results support Newton's model? (c) State the mass of Saturn.
A student claims: "A 1,500 kg satellite must orbit faster than a 150 kg satellite to stay at the same 800 km altitude, because gravity pulls harder on it." Predict the speed and period of each satellite, and write a claim, evidence and reasoning response to the student.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
Choice of Representation
Uses T² = a³ only for objects orbiting the Sun and Newton's forms for moons and satellites, with the correct central mass
Correct representation with one wrong input, such as altitude instead of orbital radius
Calculation and Units
Converts km to m and days to s, and every prediction has the right size and units
Correct method with one unit or arithmetic error
Prediction and Comparison
States each prediction clearly and compares it with an observed value or a claim, explaining any gap
Prediction stated but not compared or not explained
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Every question involves two bodies, one orbiting the other. Use G = 6.674 × 10⁻¹¹ N·m²/kg², Earth: M = 5.97 × 10²⁴ kg and R = 6,371 km, Sun: M = 1.99 × 10³⁰ kg, 1 AU = 1.496 × 10¹¹ m. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
According to Kepler's first law, what is the shape of a planet's orbit and where is the Sun?
Answer: D
Kepler's first law says each orbit is an ellipse with the Sun at one focus, not at the center (choice B). Many orbits are close to circles, but choice A leaves out the ellipse and the off-center Sun. Choice C describes no stable orbit.
Question 2 of 20 · Multiple Choice
An asteroid orbits the Sun at an average distance of 4.0 AU. Using T² = a³, what is its period?
Answer: B
T² = 4.0³ = 64, so T = 8.0 years. Choice A squares the distance instead of cubing it and does not take the root (T = a²). Choice C assumes the period grows in proportion to distance, and choice D uses a2/3, which is the inverse relationship.
Question 3 of 20 · Multiple Choice
The distance between two bodies is tripled and their masses stay the same. What happens to the gravitational force between them?
Answer: A
F = Gm1m2/r², so tripling r divides the force by 3² = 9. Choice B forgets that the distance is squared. Choices C and D reverse the relationship: a larger distance gives a smaller force.
Question 4 of 20 · Multiple Choice
A space probe's mass doubles when it docks with a second probe, and its distance from Mars does not change. What happens to the gravitational force between Mars and the probe?
Answer: C
The force is proportional to each mass, so doubling one mass doubles F. Choice D squares the factor, which applies to distance, not mass. Choice B confuses the force with the orbital speed, which does not depend on the probe's mass.
Question 5 of 20 · Multiple Choice
By Kepler's second law, where in its elliptical orbit does a planet move fastest?
Answer: B
Equal areas in equal times means the planet must cover more of its path when it is close to the Sun, so it moves fastest at perihelion. Choice A reverses this. Choice D is true only for a perfectly circular orbit.
Question 6 of 20 · Multiple Choice
A small body takes 27 years to orbit the Sun. What is its average distance from the Sun?
Answer: D
a³ = T² = 27² = 729, so a = ∛729 = 9.0 AU. Choice A takes the cube root of 27 without squaring first. Choice B is √27, which uses T = a² instead of T² = a³. Choice C squares 9 again.
Question 7 of 20 · Multiple Choice
Two satellites orbit Earth at the same altitude in circular orbits. One has a mass of 500 kg and the other 5,000 kg. How do their orbital speeds compare?
Answer: C
v = √(GM/r) uses only the mass of Earth and the orbital radius: the satellite's own mass cancels because a larger mass feels a larger force but also needs a larger force to turn. Choices A and D assume the force grows without the needed force growing too.
Question 8 of 20 · Multiple Choice
A weather satellite is in a circular orbit 1,000 km above Earth's surface. What is its predicted period?
Answer: A
r = 6,371 + 1,000 = 7,371 km = 7.371 × 10⁶ m. T = 2π√(r³/GM) = 2π√((7.371 × 10⁶)³ ÷ 3.98 × 10¹⁴) ≈ 6,300 s ≈ 105 minutes. Choice D uses r = 1.0 × 10⁶ m, the altitude instead of the distance from Earth's center. Choice C is the period of a geostationary satellite, much higher up.
Question 9 of 20 · Multiple Choice
A comet moves at 41 km/s at perihelion, 1.0 AU from the Sun. Its aphelion is 20 AU from the Sun. Predict its speed at aphelion.
Answer: D
At the two ends of the orbit, speed × distance is the same: 41 × 1.0 = v × 20, so v = 2.05 km/s. Choice A multiplies instead of dividing. Choice C divides by 20² as if speed followed an inverse-square law, and choice B only halves the speed.
Question 10 of 20 · Multiple Choice
Four students report an average distance and a period for the same imaginary object orbiting the Sun. Which pair agrees with Kepler's third law?
Answer: C
T = 1.81.5 = √5.832 ≈ 2.4 years, so T² ≈ a³. Choice A uses T = a², choice B uses T = a³ without taking the square root, and choice D uses T = √a.
Question 11 of 20 · Multiple Choice
A 1,000 kg satellite orbits at a distance of 2 Earth radii from Earth's center. About how large is Earth's gravitational force on it?
Answer: C
At the surface the force is GMm/R² ≈ 9,820 N. At 2R the force is 1/2² = 1/4 of that, about 2,450 N. Choice A ignores the distance, choice B divides by 2 instead of 4, and choice D divides by 8.
Question 12 of 20 · Multiple Choice
An astronomer measures the orbital radius and period of a small moon of Neptune. Which quantity can she calculate from those two measurements alone, using T² = 4π²a³/(GM)?
Answer: A
In the formula, M is the mass of the central body, so the orbit gives the mass of Neptune. The moon's mass cancels out of the orbit (choice B), which is also why its size and density (choices C and D) cannot be found this way.
Question 13 of 20 · Multiple Choice
Mission planners raise a satellite from a low circular orbit to a higher circular orbit. How do its speed and period change?
Answer: B
v = √(GM/r) gets smaller as r grows, and T = 2πr/v grows because the path is longer and the speed is lower. Choice A reflects the idea that a higher orbit needs more speed, but a smaller speed is needed there because gravity is weaker.
Question 14 of 20 · Multiple Choice
An orbit around the Sun has its closest point at 2.0 AU and its farthest point at 6.0 AU. What is its eccentricity?
Answer: D
The semi-major axis is (2.0 + 6.0) ÷ 2 = 4.0 AU, and e = (farthest - closest) ÷ (farthest + closest) = 4.0 ÷ 8.0 = 0.50. Choice B gives the semi-major axis, and choice A divides the closest distance by the farthest. An eccentricity must be less than 1 for a closed orbit.
Question 15 of 20 · Short Answer
The dwarf planet Eris orbits the Sun at an average distance of 67.9 AU. Predict its orbital period and explain which representation you used and why it applies.
T² = a³ applies because Eris orbits the Sun and its mass is tiny next to the Sun's. T = 67.91.5 = √(67.9³) ≈ 560 years (the observed value is about 559 years).
Question 16 of 20 · Short Answer
Jupiter (1.90 × 10²⁷ kg) orbits 5.20 AU from the Sun. Find the gravitational force between Jupiter and the Sun, and compare it with the Earth-Moon force of about 1.99 × 10²⁰ N from the worked examples.
r = 5.20 × 1.496 × 10¹¹ = 7.78 × 10¹¹ m. F = (6.674 × 10⁻¹¹)(1.99 × 10³⁰)(1.90 × 10²⁷) ÷ (7.78 × 10¹¹)² ≈ 4.2 × 10²³ N, about 2,100 times the Earth-Moon force. Both the Sun and Jupiter are far more massive than Earth and the Moon, which outweighs the much larger distance.
Question 17 of 20 · Short Answer
Describe a spreadsheet (a computational representation) that predicts the period of an Earth satellite for any altitude. Name the columns and the formula in each.
Columns: altitude (km); r (m) = (altitude + 6,371) × 1,000; GM = 3.98 × 10¹⁴; T (s) = 2π√(r³/GM); T (min) = T ÷ 60. Filling the altitude column with 200, 400, 600 km and so on gives a prediction for every row at once. A good answer also names one check, such as comparing one row with a known satellite.
Question 18 of 20 · Short Answer
Use v = √(GM/r) with the Sun's mass to predict Earth's orbital speed. Then check it with the distance Earth travels in one year (365.25 days).
v = √(6.674 × 10⁻¹¹ × 1.99 × 10³⁰ ÷ 1.496 × 10¹¹) ≈ 29.8 km/s. Check: 2π × 1.496 × 10¹¹ m ÷ 3.156 × 10⁷ s ≈ 29.8 km/s. The two representations agree, which supports the circular-orbit model for Earth (e = 0.017).
Question 19 of 20 · Short Answer
Titania, a moon of Uranus, orbits 435,910 km from the planet's center with a period of 8.706 days. Use Newton's form of Kepler's third law to find the mass of Uranus.
a = 4.3591 × 10⁸ m, T = 8.706 × 86,400 = 7.522 × 10⁵ s. M = 4π²a³ ÷ (GT²) ≈ 8.7 × 10²⁵ kg, about 14.5 times the mass of Earth.
Question 20 of 20 · Short Answer
A student says: "Astronauts float inside a space station 400 km up because there is no gravity there." Use Newton's law of gravitation to evaluate the claim.
At r = 6,771 km, g = GM/r² = 3.98 × 10¹⁴ ÷ (6.771 × 10⁶)² ≈ 8.7 m/s², about 89% of the surface value. Gravity is strong there; the astronauts float because they and the station fall around Earth together in the same orbit, so the floor does not push on them.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HS-ESS1-4 mean?
HS-ESS1-4 asks students to use math or a computational tool such as a spreadsheet to predict how planets, moons and satellites move in their orbits. In practice that means Kepler's laws and Newton's law of gravitation applied to one orbiting body and one central body. The code stands for High School, Earth and Space Science, core idea 1 (Earth's Place in the Universe), performance expectation 4.
Is HS-ESS1-4 taught in Earth science or physics?
Both are common. It is usually taught in an Earth and Space Science or astronomy course, and many Physics courses cover the same Newtonian gravity in their gravitation unit. The math needed is algebra with square roots, cube roots and scientific notation.
What is the assessment boundary for HS-ESS1-4?
The official Assessment Boundary limits the math to two bodies and rules out calculus. A problem may pair the Sun with a planet, a planet with a moon, or Earth with a satellite, but not three bodies pulling on each other at once. Deriving Kepler's laws with calculus, orbital transfers that need calculus, and three-body problems are beyond this standard.
What are Kepler's three laws in simple terms?
Orbits are ellipses with the Sun at one focus; a planet sweeps out equal areas in equal times, so it speeds up near the Sun; and the square of the period is proportional to the cube of the average distance. For planets, T² = a³ when T is in years and a in astronomical units.
How does Newton's law of gravitation connect to Kepler's third law?
Newton showed that the pull of gravity, F = GMm/r², explains Kepler's third law. For a circular orbit, gravity supplies the force needed to keep the object turning, which gives T² = 4π²r³/(GM). With the Sun's mass, years and AU, the constant becomes 1, so Kepler's version is a special case of Newton's.
Why doesn't a satellite's mass change its orbit?
A heavier satellite feels a stronger pull, but it also needs a proportionally stronger pull to curve its path, so its mass cancels out. The orbital speed v = √(GM/r) depends only on the central body's mass and the orbital radius. That is why a tiny bolt and the space station can share the same orbit.
What mistakes do students make with orbit calculations?
A frequent one is using a satellite's altitude as r instead of adding the planet's radius. Others are leaving distances in km while G is in m, forgetting to convert days to seconds, and using T² = a³ for moons of other planets, where the years-and-AU constant does not apply.
What counts as a computational representation for HS-ESS1-4?
A spreadsheet or short program that turns inputs (distance, mass, altitude) into predictions (period, speed, force) counts. Students might fill a column of altitudes and let formulas compute the period for each, or compare predicted and observed periods for a set of moons. The key is that the tool makes predictions students can check against data.
How can students check an orbit prediction?
Compare it with published values for real objects: planet periods, the periods of moons, or the roughly 90-minute periods of satellites in low Earth orbit. Space agencies publish the orbits of many satellites, and a prediction within a percent or two supports the model. Larger gaps are worth discussing, such as ignoring the mass of the orbiting body when it is not small.
What comes after HS-ESS1-4?
HS-PS2-4 uses Newton's law of gravitation alongside Coulomb's law to predict forces between objects, and HS-ESS1-6 uses evidence from the solar system to reconstruct Earth's formation. In later physics courses, students add energy methods and more than two bodies to their orbit models.
07
Related Standards
5 standards
These standards connect to HS-ESS1-4: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
MS-ESS1-2Prerequisite
Model the role of gravity in the motions within galaxies and the solar system
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