Kepler’s law – problems and solutions

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

Mōhiotia:

T = 1 year, r = 149.6 x 10⁶ km

hiahia :T² /r³ = … ?

otinga :

k = T² /r³ = 1² / (149.6 x 10⁶)³ = 1 / (3348071.9 x 10¹⁸) = 2.98 x 10^-25 tau²/km³

2. Pūmau whānui (G) = 6.67 x 10^-11 Nm²/ kg² and Sun’s 1.99 x 10³⁰ kg

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?

Mōhiotia:

r of Earth = 149.6 x 10⁶ km

r of mercury = 57.9 x 10⁶ km

T of Earth = 1 year

Hiahia: T of mercury?

Rongoā:

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1. Ngā raruraru me ngā otinga o te ture a Newton mō te kaha ā-ao
2. Ngā kaha ā-papa, ngā raruraru taumaha, me ngā otinga
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4. Ngā raruraru me ngā otinga o te amiorangi tukutahi-whenua
5. Ngā raruraru me ngā otinga o te ture a Kepler ngā raruraru me ngā otinga