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Chapter 12: Problem 76

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

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

Step by step solution

01

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.
02

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: \[\frac{d_{EM}^3}{d_{ES}^3} = 1\]
03

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.

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Most popular questions from this chapter

Two planets have the same mass, \(M .\) Each planet has a constant density, but the density of planet 2 is twice as high as that of planet \(1 .\) Identical objects of mass \(m\) are placed on the surfaces of the planets. What is the relationship of the gravitational potential energy, \(U_{1},\) on planet 1 to \(U_{2}\) on planet \(2 ?\) a) \(U_{1}=U_{2}\) b) \(U_{1}=\frac{1}{2} U_{2}\) c) \(U_{1}=2 U_{2}\) d) \(U_{1}=8 U_{2}\) e) \(U_{1}=0.794 U_{2}\)

Express algebraically the ratio of the gravitational force on the Moon due to the Earth to the gravitational force on the Moon due to the Sun. Why, since the ratio is so small, doesn't the Sun pull the Moon away from the Earth?

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