Question

A satellite is placed between the Earth and the Moon, along a straight line that connects their centers of mass. The satellite has an orbital period around the Earth that is the same as that of the Moon, 27.3 days. How far away from the Earth should this satellite be placed?

Short answer

Answer: The satellite should be placed 384,400 km away from the Earth.

Step-by-step solution

Understand Kepler's third law

Kepler's third law states that the ratio of the squares of the orbital periods of two satellites is equal to the ratio of the cubes of the semi-major axes of their orbits. Mathematically, this can be written as: \[\frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3}\] where \(T_1\) and \(T_2\) are the respective orbital periods, and \(a_1\) and \(a_2\) are the respective semi-major axes (average distances from the satellite to the planet). For this problem, we know that the orbital period of the Moon is 27.3 days, and the satellite's orbital period is also 27.3 days. Therefore, they have the same orbital period ratio.

Set up proportion using Kepler's third law

Let the distance between the Earth and the Moon be represented by \(d_{EM}\) and the distance between the Earth and the satellite by \(d_{ES}\). Using Kepler's third law, we can write the following proportion: \[\frac{d_{EM}^3}{d_{ES}^3} = \frac{T_M^2}{T_S^2}\] We know that the orbital period of the Moon is equal to the satellite's orbital period, so we can substitute this information into our proportion: \[\frac{d_{EM}^3}{d_{ES}^3} = \frac{27.3^2}{27.3^2}\] Since the values are the same, we can cancel the orbital period terms on both sides: \[\frac{d_{EM}^3}{d_{ES}^3} = 1\]

Solve for the distance between the Earth and the satellite

To solve for the distance between the Earth and the satellite, we can isolate \(d_{ES}\) on one side of the equation: \[\frac{d_{EM}^3}{d_{ES}^3} = 1\] \[d_{EM}^3 = d_{ES}^3\] Now, we will take the cube root of both sides: \[d_{EM} = d_{ES}\] Given that the average distance between the Earth and the Moon is approximately 384,400 km, the distance between the Earth and the satellite should also be: \[d_{ES} = 384,400 \, km\] Thus, the satellite should be placed 384,400 km away from the Earth.

Key concepts

Orbital Period

The orbital period is the time it takes for a satellite or celestial body to complete one full orbit around a planet, like Earth. This concept is pivotal in understanding many aspects of our solar system. Specifically, the orbital period indicates how long it will take for a celestial object to return to the same position in its orbit.

If you're observing the Moon, for example, its orbital period around the Earth is about 27.3 days. This means every 27.3 days, the Moon completes a full circle around our planet. This measure of time applies to any satellite, not just celestial bodies like moons and planets.

For designing satellite systems, knowing the orbital period helps determine how long the satellite will be visible from a particular point on Earth. In our problem, the satellite set to orbit with the same period as the Moon (27.3 days) allows us to use specific mathematical relationships, such as Kepler's third law, to find other orbital characteristics like distance.

Semi-Major Axis

The semi-major axis is a key concept in orbital mechanics. It represents half of the longest diameter of an elliptical orbit. In simpler terms, it's the average distance between the satellite and the celestial body it orbits. This distance is crucial in determining other orbital parameters.

According to Kepler's third law, if you know the semi-major axis, you can determine the orbital period. The law states that the square of the orbital period is proportional to the cube of the semi-major axis. The relationship shows:

• Larger semi-major axis = longer orbital period.
• Smaller semi-major axis = shorter orbital period.

In our case, since the satellite's orbital period is the same as the Moon's, the semi-major axis of the satellite's orbit is also identical to that of the Moon. This realization enables precise predictions about the satellite's behavior and location, as these are not independent measures but intricately linked through Kepler's frameworks.

Satellite Distance Calculation

Calculating the precise distance from Earth to a satellite involves understanding the interrelationship outlined in Kepler's third law. In this problem, we have a satellite with an orbital period that matches the Moon's. This information is crucial for determining its distance from Earth.

The mathematical expression from Kepler’s law \[ \frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3} \] helps us set up a proportion where the orbital periods and semi-major axes of the satellite and the Moon are equated. Given the periods are the same, the equation simplifies considerably, indicating the semi-major axes must also be identical. Thus, the satellite's distance from Earth equals the Moon's average distance, about 384,400 km.

This calculation shows how interconnected these terms are, emphasizing the utility of principles like Kepler’s third law in space operations and the placement of satellites.