Question

Suppose a small planet is discovered that is 14 times far from the sun as the Earth. The distance of earth from the sun $(1.5\times {10}^{11}m)$. Use Kepler's law of harmonies to predict the orbital period of such a planet. GIVEN: $\frac{T^2}{R^3}=2.97\times {10}^{-19}{s}^2/m^3$

• A. 52.4 years
• B. 60 years
• C. 79.8 years
• D. 36.2 year

Answer 1

The correct option is A 52.4 years

Use Kepler's third law:
$\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}$
Rearranging to solve for $T_p$:
$(T_p{)}^2=\left(\frac{T_e^2}{R_e^3}\right)\times (R_p{)}^3$
or $(T_p{)}^2=(T_e{)}^2\times {\left(\frac{R_p}{R_e}\right)}^{3}where\frac{R_p}{R_e}=14$
so $(T_p{)}^2=(T_e{)}^2\times (14{)}^{3}where{T}_e=1yr$
$(T_p{)}^2=(1yr{)}^2\times (14{)}^3=2744y{r}^2$
$T_p=\sqrt{\left(2744y{r}^2\right)}$
$T_{planet}=52.4year$