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A large Asteroid in the asteroid belt has a mean distance from the Sun of 5.743 X 10m. Find the period of this object in Earth days.

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User Zignd
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Final answer:

The period of an asteroid in Earth days can be determined using Kepler's third law and converting the mean distance from the Sun to astronomical units (AU). The period is approximately 0.0075 Earth days.

Step-by-step explanation:

The period of revolution of an object around the Sun can be determined using Kepler's third law, which states that the square of the period is proportional to the cube of the semi-major axis of the orbit. To find the period in Earth days, we need to convert the given mean distance from the Sun in meters to astronomical units (AU). 1 AU is equal to the mean distance from the Earth to the Sun, which is about 150 million kilometers or 93 million miles.

Converting 5.743 X 10^9 meters to AU: 5.743 X 10^9 meters = (5.743 X 10^9) / (1.5 X 10^11) AU = 0.03829 AU.

Now, we can use Kepler's third law to find the period in Earth's days. Let's use the formula: T^2 = a^3, where T is the period in Earth days and a is the semi-major axis in AU.

T^2 = (0.03829)^3

T^2 = 0.000056059

T = sqrt(0.000056059)

T = 0.0075 Earth days

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User Jonty Morris
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