# Kepler’s law – problems and solutions

1. The Earth’s distance from the Sun is 149.6 x 10^{6} km and period of Earth’s revolution is 1 year. Calculate T^{2} / r^{3}

__Known :__

T = 1 year, r = 149.6 x 10^{6} km

__Wanted __: T^{2} / r^{3} = … ?

__Solution __:

k = T^{2} / r^{3 }= 1^{2} / (149.6 x 10^{6})^{3} = 1 / (3348071.9 x 10^{18}) = 2.98 x 10^{-25 }year^{2}/km^{3}

2. Universal constant (G) = 6.67 x 10^{-11} N.m^{2}/kg^{2 }and Sun’s 1.99 x 10^{30} kg.

3. The mean distance of Earth from the Sun is 149.6 x 10^{6} km and the mean distance of Mercury from the Sun is 57.9 x 10

^{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 10^{6} km

r of mercury = 57.9 x 10^{6} km

T of Earth = 1 year

__Wanted:__ T of mercury?

__Solution :__

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