Question

Find the period of revolution for the planet Mercury, whose average distance from the Sun is 5.79 x 1010 m.

Answer 1

T = 7.61*10^6 s

Step-by-step explanation:

In order to calculate the Mercury's period. in its orbit around the sun, you take into account on the Kepler's law. You use the following formula:

`T=\sqrt{(4\pi^2r^3)/(GM_s)}` (1)

T: period of Mercury

r: distance between Mercury and Sun

Ms: mass of the sun = 1.98*10^30 kg

G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2

You replace the values of all parameters in the equation (1):

`T=\sqrt{(4\pi^2(5.79*10^(10)m)^3)/((6.674*10^(-11)m^3kg^(-1)s^(-2))(1.98*10^(30)kg))}\\\\T=7.61*10^6s*(1h)/(3600s)*(1d)/(24h)=88.13\ days`

The period of Mercury is 7.61*10^6 s, which is approximately 88.13 Earth's days