Question

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 1 AU would have an orbital period of about __________ year(s).

Answer 1

Step-by-step explanation:

Orbital radius,
`r=1\ AU=1.496* 10^(11)\ m`

According to Kepler's law :

`T^2\propto r^3`

`T^2=(4\pi^2r^3)/(GM)`

Where

M is the mass of sun,
`M=1.98* 10^(30)\ kg`

`T^2=(4\pi^2* (1.496* 10^(11))^3)/(6.67* 10^(-11)* 1.98* 10^(30))`

`T=\sqrt{1.0008* 10^(15)}`

T = 31635423.18 s

or

T = 1.0003 years

So, a solar-system planet that has an orbital radius of 1 AU would have an orbital period of about 1.0003 year(s).