Kepler txoj cai - teeb meem thiab kev daws teeb meem

1. The Earth’s deb from the Sun is 149.6 x 10⁶ km and period of Earth’s revolution is 1 year. Calculate T² / r ib³

Paub:

T = 1 year, r = 149.6 x 10⁶ km

xav : Twm² / r ib³ = … ?

tshuaj :

k = T² / r ib³ = 1² / (149.6 x 10⁶)³ = 1 / (3348071.9 x 10¹⁸) = 2.98 x 10^-25 xyoo²/km³

2. Qhov tsis hloov pauv thoob ntiaj teb (G) = 6.67 x 10^-11 Nm²/ kg² and Sun’s 1.99 x 10³⁰ kg ua.

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?

Paub:

r of Earth = 149.6 x 10⁶ km

r of mercury = 57.9 x 10⁶ km

T of Earth = 1 year

SE pob: T of mercury?

Tshuaj:

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