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(II) Two satellites orbit Earth at altitudes of \(5000 \mathrm{~km}\) and \(15,000 \mathrm{~km} .\) Which satellite is faster, and by what factor?

Short Answer

Expert verified
The satellite at 5000 km is faster by a factor of about 1.37.

Step by step solution

01

Understand the Problem

We have two satellites orbiting Earth. One is at an altitude of 5000 km, and the other is at 15000 km. We need to determine which satellite is faster and by what factor.
02

Review Orbital Speed Formula

The orbital speed of a satellite is given by the formula: \( v = \sqrt{\frac{GM}{R}} \), where \( G \) is the gravitational constant, \( M \) is the mass of Earth, and \( R \) is the distance from the center of Earth to the satellite. \( R = R_{earth} + h \), where \( R_{earth} \approx 6371 \ km \) and \( h \) is the altitude of the satellite.
03

Calculate Orbital Radius for Satellite 1

For the satellite at 5000 km altitude: \( R_1 = 6371 \ km + 5000 \ km = 11371 \ km \). This is the total distance from the center of Earth.
04

Calculate Orbital Radius for Satellite 2

For the satellite at 15000 km altitude: \( R_2 = 6371 \ km + 15000 \ km = 21371 \ km \). This is also measured from Earth's center.
05

Calculate Orbital Speeds

Using the formula \( v = \sqrt{\frac{GM}{R}} \), calculate the speeds:- Satellite 1 speed: \( v_1 = \sqrt{\frac{GM}{11371}} \).- Satellite 2 speed: \( v_2 = \sqrt{\frac{GM}{21371}} \).Notice that \( v_1 > v_2 \) because the speed decreases with increasing radius.
06

Calculate the Speed Ratio

We find the speed factor by calculating \( \frac{v_1}{v_2} = \frac{\sqrt{\frac{GM}{11371}}}{\sqrt{\frac{GM}{21371}}} = \sqrt{\frac{21371}{11371}} \).
07

Simplify and Result

Simplify \( \sqrt{\frac{21371}{11371}} \) to determine the factor by which the speed of satellite 1 exceeds that of satellite 2. This yields approximately \( \sqrt{1.88} \approx 1.37 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Constant
The gravitational constant, denoted as \( G \), is a fundamental constant in physics that appears in Newton's law of universal gravitation. It represents the strength of the gravitational force between two masses. The value of \( G \) is approximately \( 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \). This constant is crucial for calculating gravitational forces not only on Earth but throughout the universe.

When computing the orbital speed of a satellite around Earth, \( G \) is used in conjunction with the mass of Earth \( M \) in the formula \( v = \sqrt{\frac{GM}{R}} \). Here, \( v \) is the orbital speed and \( R \) is the distance from the Earth's center to the satellite. The simplified expression highlights how \( G \) directly influences the speed, demonstrating how gravity holds the satellite in its orbital path.

Understanding \( G \) helps us appreciate how universal gravitational interactions are foundational to celestial mechanics, from the movement of planets to the flight path of satellites.
Satellite Altitude
Satellite altitude refers to the height above the Earth's surface at which a satellite orbits. It is measured in kilometers or miles and plays a pivotal role in determining the satellite's orbital speed as well as its function.

The formula \( R = R_{earth} + h \) combines the Earth's radius \( R_{earth} \) with the satellite's altitude \( h \) to calculate the total orbital radius \( R \), which is crucial for finding the orbital speed. For instance:
  • For a satellite at 5000 km altitude, \( R = 6371 \text{ km} + 5000 \text{ km} = 11371 \text{ km} \).
  • For another at 15000 km altitude, \( R = 6371 \text{ km} + 15000 \text{ km} = 21371 \text{ km} \).
As the altitude increases, the total orbital radius increases, which in turn affects the speed of the satellite. Higher altitude satellites have larger orbital paths and thus travel slower, as illustrated by the exercise where a satellite at 5000 km is faster than one at 15000 km.

Understanding satellite altitude is essential for different applications, like communication, where closer satellites provide quicker transmission.
Orbital Mechanics
Orbital mechanics is the study of the motions of satellites and celestial bodies under the influence of gravitational forces. Central to this field is the concept of orbital speed, which dictates how quickly a satellite travels around a planet.

The primary formula used to understand these speeds is \( v = \sqrt{\frac{GM}{R}} \), which incorporates the gravitational constant \( G \), the mass of the central body (e.g., Earth), and the orbital radius \( R \). This formula shows the inverse relationship between speed and radius: as the satellite altitude, and thus the radius, increases, the required orbital speed decreases. This is due to the lesser gravitational pull experienced at higher distances.

In practical terms, mastering orbital mechanics allows for precise satellite positioning, essential for everything from weather forecasting to GPS navigation. Engineers use these principles to ensure satellites maintain their intended orbits and operational functionality.

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Most popular questions from this chapter

Jupiter is about 320 times as massive as the Earth. Thus, it has been claimed that a person would be crushed by the force of gravity on a planet the size of Jupiter since people can't survive more than a few \(g\) 's. Calculate the number of \(g\) 's a person would experience at the equator of such a planet. Use the following data for Jupiter: mass \(=1.9 \times 10^{27} \mathrm{~kg}, \quad\) equatorial radius \(=7.1 \times 10^{4} \mathrm{~km}\) rotation period \(=9 \mathrm{hr} 55 \mathrm{~min} .\) Take the centripetal acceleration into account.

(II) Determine the mass of the Sun using the known value for the period of the Earth and its distance from the Sun. [Hint: The force on the Earth due to the Sun is related to the centripetal acceleration of the Earth.] Compare your answer to that obtained using Kepler's laws, Example \(6-9 .\)

(II) Two identical point masses, each of mass \(M,\) always remain separated by a distance of 2\(R .\) A third mass \(m\) is then placed a distance \(x\) along the perpendicular bisector of the original two masses, as shown in Fig. \(26 .\) Show that the gravitational force on the third mass is directed inward along the perpendicular bisector and has a magnitude of $$F=\frac{2 G M m x}{\left(x^{2}+R^{2}\right)^{\frac{3}{2}}}$$

(II) Planet \(A\) and planet \(B\) are in circular orbits around a distant star. Planet \(A\) is 9.0 times farther from the star than is planet B. What is the ratio of their speeds \(v_{A} / v_{B} ?\)

(II) The accompanying table shows the data for the mean distances of planets (except Pluto) from the Sun in our solar system, and their periods of revolution about the Sun. $$ \begin{array}{lcc} \hline \text { Planet } & \text { Mean Distance (AU) } & \text { Period (Years) } \\ \hline \text { Mercury } & 0.387 & 0.241 \\ \text { Venus } & 0.723 & 0.615 \\ \text { Earth } & 1.000 & 1.000 \\ \text { Mars } & 1.524 & 1.881 \\ \text { Jupiter } & 5.203 & 11.88 \\ \text { Saturn } & 9.539 & 29.46 \\ \text { Uranus } & 19.18 & 84.01 \\ \text { Neptune } & 30.06 & 164.8 \\ \hline \end{array} $$ (a) Graph the square of the periods as a function of the cube of the average distances, and find the best-fit straight line. (b) If the period of Pluto is 247.7 years, estimate the mean distance of Pluto from the Sun from the best- fit line.

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