Question

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

Answer 1

Step-by-step explanation:

We have,

Semimajor axis is
`4* 10^(12)\ m`

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

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

G is universal gravitational constant

M is solar mass

Plugging all the values,

`T^2=(4\pi ^2)/(6.67* 10^(-11)* 1.98* 10^(30))* (4* 10^(12))^3\\\\T=\sqrt{(4\pi^(2))/(6.67*10^(-11)*1.98*10^(30))*(4*10^(12))^(3)}\\\\T=4.37* 10^9\ s`

Since,

`1\ s=3.17* 10^(-8)\ \text{years}\\\\4.37* 10^9\ s=4.37\cdot10^(9)\cdot3.17\cdot10^(-8)\\\\4.37* 10^9\ s=138.52\ \text{years}`

So, the orbital period of a dwarf planet is 138.52 years.