Question

The asteroid ceres lies at an average distance of 414 million kilometers from the sun. The period of revolution of ceres around the sun is approximately

Answer 1

Answer:

Answer is 1.156x10^23 sec

Step-by-step explanation:

According to kepler's law, the square of the period of revolution is proportional to the cube of the distance within planets.

At a distance r of 414 million kilometers I.e 414x10^6 km, period calculation is shown in the image below.

Answer 2

T2 = 1680,4 days

Step-by-step explanation:

Kepplers law:

`(T^(2) )/(a^(3) ) = constant`

For Earth:

T1 = 365 days ; a1 = 149 597 870 700 m

For Ceres:

T2 = ? days ; a1 =
`414*10^(9) m`

Then:

`(T1^(2) )/(a1^(3) ) = (T2^(2) )/(a2^(3) ) ----> T2 = T1*\sqrt{(a2^(3))/(a1^(3))}`

Replacing values:

T2 = 1680,4 days