Kepler’s law – problems and solutions

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

Known :

T = 1 year, r = 149.6 x 106 km

Wanted : T2 / r3 = … ?

Solution :

k = T2 / r3 = 12 / (149.6 x 106)3 = 1 / (3348071.9 x 1018) = 2.98 x 10-25 year2/km3

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2. Universal constant (G) = 6.67 x 10-11 N.m2/kg2 and Sun’s 1.99 x 1030 kg.

Kepler's law – problems and solutions 1

Kepler's law – problems and solutions 2

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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 10

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6 km. The period of Earth’s revolutions is 1 year, what is the period of Mercury’s revolution?

Known :

r of Earth = 149.6 x 106 km

r of mercury = 57.9 x 106 km

T of Earth = 1 year

Wanted: T of mercury?

Solution :

Kepler's law – problems and solutions 3

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