00:01
In this problem, we have been given that there is a dwarf planet and this revolves around the sun provided that this planet is at a distance of 6 into 10 raised to 12 meters.
00:16
So that's the distance of this planet from the sun.
00:20
And the diameter of this planet is observed to be 300 kilometers.
00:25
So the radius of the planet represented by capital r, that will be equal to 150.
00:33
Kilometers that's 150 into 10 raise to 3 meters and now we need to determine the orbital period of this planet around the sun so we need to determine the t here so we can observe that the force between these this planet and the sun that's the gravitational force which is g into mass of sun into mass of the planet divided by square of the distance and that's equal to mass of the planet into r into omega square.
01:09
So from here we can observe that mass of the planet that gets cut and we can get the value of omega here as root of g into mass of sun divided by rq.
01:24
So from here we take the mass of sun and we put the values here.
01:31
So approximately mass of sun is 2 into 10 raise to 30 kilograms.
01:35
So from here we can replace omega with 2 pi by t and get the value of the time period as 2 pi times root of r cubed divided by g into mass of the sun.
01:53
So here we put the value to get the time period so that will be 2 pi into root of cube of the distance r.
02:01
So that will be 6 into 10 raise to 12 cube and divided by g times m.
02:10
G is the universal gravitational constant that's 6 .67 into 10 raise to minus 11 into m here is the mass of the sun that's 2 into 10 raise to 30...