Orbits and Kepler's Laws Worksheet Generator
Kepler's third law, Newtonian orbital period and speed. No AI, no credits.
Grades 8 – 12No account needed
Kepler's third law in AU and years, Newtonian orbital period and speed using sourced constants and planetary data, with the source shown. Every answer is computed by code. No AI, so it never uses a generation credit.
What students practice
- Earth and Space
- Orbits
- Kepler's laws
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Name: ____________________Date: ____________
Orbits and Kepler's Laws
1. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Jupiter, which has a semi-major axis of 5.203 AU.
2. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Mars, which has a semi-major axis of 1.524 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
3. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Uranus, which has a semi-major axis of 19.191 AU.
4. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Saturn, which has a semi-major axis of 9.537 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
5. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Venus, which has a semi-major axis of 0.723 AU.
6. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Jupiter, which has a semi-major axis of 5.203 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
7. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Earth, which has a semi-major axis of 1 AU.
8. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Jupiter, which has a semi-major axis of 5.203 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
9. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Uranus, which has a semi-major axis of 19.191 AU.
10. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Mars, which has a semi-major axis of 1.524 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
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Answer Key — Orbits and Kepler's Laws
1. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Jupiter, which has a semi-major axis of 5.203 AU.
Solution:
Kepler's third law: T² = a³
T = a^(3/2) = 5.203^(3/2)
T = 11.87 years
Answer: 11.87 years
2. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Mars, which has a semi-major axis of 1.524 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
Solution:
a = 1.524 AU = 2.280e+11 m
T = 2π × √(a³ / (G × M))
T = 2π × √(2.280e+11³ / (6.6743e-11 × 1.989e+30))
T = 59365545 s = 1.88 years
Answer: 1.88 years
3. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Uranus, which has a semi-major axis of 19.191 AU.
Solution:
Kepler's third law: T² = a³
T = a^(3/2) = 19.191^(3/2)
T = 84.07 years
Answer: 84.07 years
4. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Saturn, which has a semi-major axis of 9.537 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
Solution:
a = 9.537 AU = 1.427e+12 m
T = 2π × √(a³ / (G × M))
T = 2π × √(1.427e+12³ / (6.6743e-11 × 1.989e+30))
T = 929340105 s = 29.45 years
Answer: 29.45 years
5. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Venus, which has a semi-major axis of 0.723 AU.
Solution:
Kepler's third law: T² = a³
T = a^(3/2) = 0.723^(3/2)
T = 0.61 years
Answer: 0.61 years
6. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Jupiter, which has a semi-major axis of 5.203 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
Solution:
a = 5.203 AU = 7.784e+11 m
T = 2π × √(a³ / (G × M))
T = 2π × √(7.784e+11³ / (6.6743e-11 × 1.989e+30))
T = 374487861 s = 11.87 years
Answer: 11.87 years
7. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Earth, which has a semi-major axis of 1 AU.
Solution:
Kepler's third law: T² = a³
T = a^(3/2) = 1^(3/2)
T = 1.00 years
Answer: 1.00 years
8. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Jupiter, which has a semi-major axis of 5.203 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
Solution:
a = 5.203 AU = 7.784e+11 m
T = 2π × √(a³ / (G × M))
T = 2π × √(7.784e+11³ / (6.6743e-11 × 1.989e+30))
T = 374487861 s = 11.87 years
Answer: 11.87 years
9. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Uranus, which has a semi-major axis of 19.191 AU.
Solution:
Kepler's third law: T² = a³
T = a^(3/2) = 19.191^(3/2)
T = 84.07 years
Answer: 84.07 years
10. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Mars, which has a semi-major axis of 1.524 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
Solution:
a = 1.524 AU = 2.280e+11 m
T = 2π × √(a³ / (G × M))
T = 2π × √(2.280e+11³ / (6.6743e-11 × 1.989e+30))
T = 59365545 s = 1.88 years
Answer: 1.88 years
Answers computed by code. No AI, no credits.
Assumptions and Conventions
Assumptions:
- Kepler's third law: T² = a³ (T in years, a in AU).
- Newton's form: T = 2π√(a³/(GM)).
- Orbital speed (circular): v = √(GM/a).
- 1 AU = 1.496 × 10¹¹ m.
Conventions:
- G = 6.67430 × 10⁻¹¹ N⋅m²/kg² (NIST 2018).
- M_sun = 1.989 × 10³⁰ kg.
- Planetary data from NASA Planetary Fact Sheet.
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Every answer is computed by code and verified by a second independent method. No AI is used.
What grades is the Orbits and Kepler's Laws for?
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Name: ____________________Date: ____________
Orbits and Kepler's Laws
1. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Neptune, which has a semi-major axis of 30.07 AU.
2. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Earth, which has a semi-major axis of 1 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
3. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Mars, which has a semi-major axis of 1.524 AU.
4. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Uranus, which has a semi-major axis of 19.191 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
5. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Jupiter, which has a semi-major axis of 5.203 AU.
6. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Earth, which has a semi-major axis of 1 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
7. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Mercury, which has a semi-major axis of 0.387 AU.
8. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Uranus, which has a semi-major axis of 19.191 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
9. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Neptune, which has a semi-major axis of 30.07 AU.
10. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Uranus, which has a semi-major axis of 19.191 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
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Answer Key — Orbits and Kepler's Laws
1. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Neptune, which has a semi-major axis of 30.07 AU.
Solution:
Kepler's third law: T² = a³
T = a^(3/2) = 30.07^(3/2)
T = 164.89 years
Answer: 164.89 years
2. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Earth, which has a semi-major axis of 1 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
Solution:
a = 1 AU = 1.496e+11 m
T = 2π × √(a³ / (G × M))
T = 2π × √(1.496e+11³ / (6.6743e-11 × 1.989e+30))
T = 31554188 s = 1.00 years
Answer: 1.00 years
3. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Mars, which has a semi-major axis of 1.524 AU.
Solution:
Kepler's third law: T² = a³
T = a^(3/2) = 1.524^(3/2)
T = 1.88 years
Answer: 1.88 years
4. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Uranus, which has a semi-major axis of 19.191 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
Solution:
a = 19.191 AU = 2.871e+12 m
T = 2π × √(a³ / (G × M))
T = 2π × √(2.871e+12³ / (6.6743e-11 × 1.989e+30))
T = 2652793309 s = 84.06 years
Answer: 84.06 years
5. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Jupiter, which has a semi-major axis of 5.203 AU.
Solution:
Kepler's third law: T² = a³
T = a^(3/2) = 5.203^(3/2)
T = 11.87 years
Answer: 11.87 years
6. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Earth, which has a semi-major axis of 1 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
Solution:
a = 1 AU = 1.496e+11 m
T = 2π × √(a³ / (G × M))
T = 2π × √(1.496e+11³ / (6.6743e-11 × 1.989e+30))
T = 31554188 s = 1.00 years
Answer: 1.00 years
7. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Mercury, which has a semi-major axis of 0.387 AU.
Solution:
Kepler's third law: T² = a³
T = a^(3/2) = 0.387^(3/2)
T = 0.24 years
Answer: 0.24 years
8. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Uranus, which has a semi-major axis of 19.191 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
Solution:
a = 19.191 AU = 2.871e+12 m
T = 2π × √(a³ / (G × M))
T = 2π × √(2.871e+12³ / (6.6743e-11 × 1.989e+30))
T = 2652793309 s = 84.06 years
Answer: 84.06 years
9. Using Kepler's third law (T² = a³, where T is in years and a is in AU), find the orbital period of Neptune, which has a semi-major axis of 30.07 AU.
Solution:
Kepler's third law: T² = a³
T = a^(3/2) = 30.07^(3/2)
T = 164.89 years
Answer: 164.89 years
10. Using Newton's form of Kepler's third law (T = 2π√(a³/GM)), find the orbital period of Uranus, which has a semi-major axis of 19.191 AU. (M_sun = 1.989e+30 kg, G = 6.6743e-11 N⋅m²/kg².)
Solution:
a = 19.191 AU = 2.871e+12 m
T = 2π × √(a³ / (G × M))
T = 2π × √(2.871e+12³ / (6.6743e-11 × 1.989e+30))
T = 2652793309 s = 84.06 years
Answer: 84.06 years
Answers computed by code. No AI, no credits.
Assumptions and Conventions
Assumptions:
- Kepler's third law: T² = a³ (T in years, a in AU).
- Newton's form: T = 2π√(a³/(GM)).
- Orbital speed (circular): v = √(GM/a).
- 1 AU = 1.496 × 10¹¹ m.
Conventions:
- G = 6.67430 × 10⁻¹¹ N⋅m²/kg² (NIST 2018).
- M_sun = 1.989 × 10³⁰ kg.
- Planetary data from NASA Planetary Fact Sheet.
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Orbits and Kepler's Laws - General earth and space - Mixed orbital mechanics
Kepler's third law, Newtonian orbital period and speed. No AI, no credits.