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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

The weight of a star is usually balanced by two forces: the gravitational force, acting inward, and the force created by nuclear reactions, acting outward. Over a long period of time, the force due to nuclear reactions gets weaker, causing the gravitational collapse of the star and crushing atoms out of existence. Under such extreme conditions, protons and electrons are squeezed to form neutrons, giving birth to a neutron star. Neutron stars are massively heavy-a teaspoon of the substance of a neutron star would weigh 100 million metric tons on the Earth. a) Consider a neutron star whose mass is twice the mass of the Sun and whose radius is \(10.0 \mathrm{~km} .\) If it rotates with a period of \(1.00 \mathrm{~s}\), what is the speed of a point on the Equator of this star? Compare this speed with the speed of a point on the Earth's Equator. b) What is the value of \(g\) at the surface of this star? c) Compare the weight of a 1.00 -kg mass on the Earth with its weight on the neutron star. d) If a satellite is to circle \(10.0 \mathrm{~km}\) above the surface of such a neutron star, how many revolutions per minute will it make? e) What is the radius of the geostationary orbit for this neutron star?

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