Question

The dwarf planet praamzius is estimated to have a diameter of about 300km and orbits the sun at a distance of 6.4E12m . What is it’s orbital period in years?

Answer 1

Answer:

Time period of the planet is

$T_2 = 278.7 years$

Explanation:

As we know by Kepler's law of time period

square of time period of planet is proportional to the cube of its orbital radius

$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

here we know for Earth

$T_1 = 1 year$

$R_1 = 1.5 \times 10^{11} m$

now we have

$\frac{1^2}{T_2^2} = \frac{(1.5 \times 10^{11})^3}{(6.4 \times 10^{12})^3}$

$T_2^2 = 7.77 \times 10^4$

$T_2 = 278.7 years$