00:01
According to kepler's third law of planetary motion, according to the law, it states that t square is directly proportional to r cube, where t represents the orbital period of celestial object.
00:35
And here, r represents the orbital radius of the object.
00:48
And here, the equation for orbital radius is equal to g into m into t squared by 4 pi square raised to 1 by 3, where g means universal gravitational constant and the value of g is equal to 6 .67 into 10 power minus 11 newton meter square.
01:22
Per kilogram square.
01:25
M represents the mass of sun and the value of mass is equal to 1 .99 into 10 power 30 kilogram.
01:37
Then we have got t which implies the time period of the celestial object.
01:45
Here the celestial object is an asteroid and the value of time period is given in the question as 6 years.
01:54
We know that 1 year.
01:56
Means 3 .16 into 10 power 7 seconds and here earth unit means we have to find the answer in astronomical unit and one astronomical unit is equal to 1 .5 into 10 power 11 meter so these two conversions we have to follow as per the question therefore here the orbital radius of the object is equal to g into m into t square here value of g is 6 .67 into 10 power minus 11...