According to Kepler's third law:
- A) Equal to its perihelion distance from the Sun in A.U.
- B) Inversely proportional to its mass in kilograms.
- C) Equal to the fourth power of its average temperature in degrees Kelvin.
- D) Proportional to the cube of its semimajor axis in A.U.
- E) Equal to the square of its aphelion distance in A.U.
- B) Inversely proportional to its mass in kilograms.
- C) Equal to the fourth power of its average temperature in degrees Kelvin.
- D) Proportional to the cube of its semimajor axis in A.U.
- E) Equal to the square of its aphelion distance in A.U.
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pampaaaa
about 5 years ago
Kepler’s third law of motion, also known as law of harmonies, says that orbital period of a plant (P) is related to size of planet’s orbit (a). The square of period is directly proportion with cube of orbit size.
P(^2) ∝ a(^3)
Using Newton laws of motion and gravity, this equation transform into:
P(^2) = a(^3) 4 π(^2)/[G (M1 + M2)]
Where M1 and M2 are the masses of two orbiting planets. Kepler assumed the size of orbit is measured in astronomical units (AU). Solving problem with Kepler laws is explained in lecture slides.
P(^2) ∝ a(^3)
Using Newton laws of motion and gravity, this equation transform into:
P(^2) = a(^3) 4 π(^2)/[G (M1 + M2)]
Where M1 and M2 are the masses of two orbiting planets. Kepler assumed the size of orbit is measured in astronomical units (AU). Solving problem with Kepler laws is explained in lecture slides.
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