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

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

Answer:

T2 = 1680,4 days

Explanation:

Kepplers law:

$\frac{T^{2} }{a^{3} } = constant$

For Earth:

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

For Ceres:

T2 = ? days ; a1 = $414*10^{9} m$

Then:

$\frac{T1^{2} }{a1^{3} } = \frac{T2^{2} }{a2^{3} } ----> T2 = T1*\sqrt{\frac{a2^{3}}{a1^{3}}}$

Replacing values:

T2 = 1680,4 days