Kepler’s law – problems and solutions

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

Zodziwika:

T = 1 year, r = 149.6 x 10⁶ km

anafuna :T² /r³ = … ?

Anakonza :

k = T² /r³ = 1² / (149.6 x 10⁶)³ = 1 / (3348071.9 x 10¹⁸) = 2.98 x 10^-25 chaka²/km³

2. Chokhazikika cha chilengedwe chonse (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?

Zodziwika:

r of Earth = 149.6 x 10⁶ km

r of mercury = 57.9 x 10⁶ km

T of Earth = 1 year

SE basi: T of mercury?

Yankho:

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1. Lamulo la Newton la mavuto a mphamvu yokoka padziko lonse lapansi ndi mayankho ake
2. Mphamvu yokoka, mavuto olemera, ndi mayankho
3. Kuthamanga chifukwa cha mavuto ndi mphamvu yokoka
4. Mavuto ndi mayankho a satellite ya geosynchronous
5. Mavuto ndi mayankho a malamulo a Kepler