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The moon has a period of 27.3 days (earth days) and its average distance from the earth is 3.8x10^5 km. If a new International Space Station (ISS) is set in orbit at 1.2x10^3 km average distance from earth, what is the period of the new ISS?

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Final answer:

The period of an orbiting object can be calculated using Kepler's third law. By setting up a proportion with the known period and distance of the Moon's orbit, we can solve for the period of the new International Space Station at its given distance. The period of the ISS is approximately 0.09 days.

Step-by-step explanation:

To calculate the period of the new International Space Station (ISS) orbiting at an average distance of 1.2x10^3 km from Earth, we can use Kepler's third law. This law states that the period of an orbit is related to the radius of the orbit. We already know the period of the Moon's orbit (27.3 days) and its average distance from Earth (3.8x10^5 km). We can set up a proportion using the Moon's period and distance, and then solve for the period of the ISS:

27.3 days / 3.8x10^5 km = x days / 1.2x10^3 km

Cross multiplying, we get:

27.3 days * 1.2x10^3 km = 3.8x10^5 km * x days

Simplifying, we find that the period of the ISS is x = (27.3 days * 1.2x10^3 km) / (3.8x10^5 km) = 0.0864 days, or approximately 0.09 days.

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