Lex Kepleri – problemata et solutiones

1. The Earth’s spatium from the Sun is 149.6 x 106 km and period of Earth’s revolution is 1 year. Calculate T2 /r3

Notum:

T = 1 year, r = 149.6 x 106 km

voluit :T2 /r3 = … ?

Solutio :

k = T2 /r3 = 12 / (149.6 × 10)6)3 = 1 / (3348071.9 × 10)18) = 2.98 × 10-25 anno2/ km3

Vide quoque  Energia electrica in circuitibus condensatorum – problemata et solutiones

2. Constans universalis (G) = 6.67 × 10-11 Nm2/kg2 and Sun’s 1.99 x 1030 kg.

Kepler's law – problems and solutions 1

Kepler's law – problems and solutions 2

Vide quoque  Telescopia astronomica – problemata et solutiones

3. The mean distance of Earth from the Sun is 149.6 x 106 km and the mean distance of Mercury from the Sun is 57.9 x 106 km. The period of Earth’s revolutions is 1 year, what is the period of Mercury’s revolution?

Notum:

r of Earth = 149.6 x 106 km

r of mercury = 57.9 x 106 km

T of Earth = 1 year

SE busca: T of mercury?

solution:

Kepler's law – problems and solutions 3

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