Kepler’s law – problems and solutions

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

Ar a dtugtar:

T = 1 year, r = 149.6 x 10⁶ km

Wanted : T.² /r³ = … ?

réiteach :

k = T² /r³ = 1² /(149.6x10⁶)³ = 1 / (3348071.9 x 10¹⁸) = 2.98×10^-25 bliain²/ km³

2. Universal constant (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?

Ar a dtugtar:

r of Earth = 149.6 x 10⁶ km

r of mercury = 57.9 x 10⁶ km

T of Earth = 1 year

Teastaíonn: T of mercury?

Réiteach:

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1. Fadhbanna agus réitigh dhlí imtharraingthe uilíoch Newton
2. Fórsaí imtharraingthe, fadhbanna meáchain, agus réitigh
3. Fadhbanna luasghéaraithe de bharr domhantarraingthe agus réitigh
4. Fadhbanna agus réitigh satailíte geoshioncrónacha
5. Fadhbanna agus réitigh dhlí Kepler, fadhbanna agus réitigh