A satellite is placed in orbit, with a radius that is half the radius of the Moon's orbit. Find the period of the satellite in units of the period of the Moon. (sketch and analyze the situation, show all working so I can understand how you arrived at your answer).

http://www.jiskha.com/display.cgi?id=1356618182

To determine the period of the satellite in relation to the period of the Moon, we need to first understand the relationship between the orbital radius and the period of a satellite.

The period of an object in orbit around another object can be calculated using Kepler's third law of planetary motion, which states that the square of the period of an orbiting object is proportional to the cube of its semi-major axis (orbital radius).

Let's analyze the situation step by step:

1. Sketch the situation: Draw a circle to represent the Moon's orbit. Within that circle, draw a smaller circle to represent the satellite's orbit. Label the radius of the Moon's orbit as R and the radius of the satellite's orbit as r, where r = R/2.

Moon's Orbit (R)
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| |
| |
| Satellite |
| Orbit |
| |
| |
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2. Analyze the situation: The satellite's radius is stated to be half the radius of the Moon's orbit, so r = R/2.

3. Use Kepler's third law: According to Kepler's third law, the ratio of the periods of two objects in orbit around the same central object is equal to the square root of the ratio of their semi-major axes.

T₁ / T₂ = √(R₁³ / R₂³)

Here, T₁ is the period of the satellite, T₂ is the period of the Moon, R₁ is the radius of the satellite's orbit, and R₂ is the radius of the Moon's orbit.

4. Substitute the given values: In this case, we know that R₁ = r = R/2 and R₂ = R (radius of the Moon's orbit).

T₁ / T₂ = √((R/2)³ / R³)

5. Simplify the expression:

T₁ / T₂ = √(1/8)

Since the square root of 1/8 is (1/2) * √2, we can rewrite the equation as:

T₁ / T₂ = (1/2) * √2

Therefore, the period of the satellite (T₁) in relation to the period of the Moon (T₂) is given by (1/2) * √2, or approximately 0.707.

This means that the period of the satellite is approximately 0.707 times the period of the Moon.