/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 A \(600-\mathrm{kg}\) satellite ... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(600-\mathrm{kg}\) satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. Find (a) the satellite's orbital speed, (b) the period of its revolution, and (c) the gravitational force acting on it.

Short Answer

Expert verified
The orbital speed of the satellite is approximately \(5574.9\) m/s. The period of its revolution is around \(7200\) seconds, and the gravitational force acting on it is roughly \(1183.5\) Newtons.

Step by step solution

01

Determine the Satellite's Orbital Speed

First, use the given mass of the earth: \( M = 5.98 \times 10^{24}\) kg and the given radius of the earth \( R_e = 6.37\times 10^6 \) m. As the Satellite is at the same height above Earth as the Earth's radius, the radius of the orbit \( r = 2R_e \) or \( r = 2\times 6.37 \times 10^6 = 1.274\times10^7 \) m. Use the orbital speed formula \( v = \sqrt{\frac{G M}{r}} \) where \( G = 6.67\times 10^{-11} \) m^3 kg^-1 s^-2. Substituting these values yields \( v = \sqrt{\frac{6.67\times 10^{-11} \times 5.98\times10^{24}}{1.274\times10^7}} \approx 5574.9 \) m/s.
02

Calculate the Period of Revolution

Use the equation \( T = \frac{2 \pi r}{v} \). Substituting the values from the above steps yields \( T = \frac{2 \pi \times 1.274\times10^7}{5574.9} \approx 7200 \) s. So, the period of the Satellite's revolution is approximately 7200 seconds.
03

Find the Gravitational Force on the Satellite

Lastly, use Newton's Law of Universal Gravitation: \( F = \frac{GMm}{r^2} \). Replacing the known values yields \( F = \frac{6.67\times 10^{-11} \times 5.98\times10^{24} \times 600}{(1.274\times10^7)^2} \approx 1183.5 \) N. Here, the gravitational force acting upon the satellite is approximately 1183.5 Newtons.

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Key Concepts

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

Orbital Speed
When a satellite orbits the Earth, it must travel at a certain velocity to maintain its circular path. This velocity is known as the orbital speed. It is the speed at which the satellite must move so that the force of gravity on the satellite is equal to the centripetal (central seeking) force that is required to keep the satellite in its orbit. The balance of these forces allows the satellite to remain in a stable orbit without flying off into space or falling back to Earth.

The formula to calculate the orbital speed (\( v \)) is given by:\[ v = \small{\sqrt{\frac{G M}{r}}} \]where:\[ G \]is the gravitational constant (\( 6.67 \times 10^{-11} \) N\(\cdot\)m²/kg²),\[ M \]is the mass of the Earth, and\[ r \]is the distance from the Earth's center to the object. Thus, for our satellite, the orbital speed is a crucial factor in determining the success of its mission.
Period of Revolution
The period of revolution, commonly denoted as (\( T \)), is the time it takes for a satellite to complete one full orbit around the Earth. This is an important factor for satellites, including those used for communication or observing the Earth. It affects how often the satellite passes over the same point on Earth's surface and determines the pattern and frequency of data collection or signal transmission.

The equation to find the period is:\[ T = \frac{2 \pi r}{v} \]where:\[ 2\pi r \]provides the circumference of the satellite's orbit, and\[ v \]is the orbital speed. For many satellites, like the one in our example, the period of revolution is a constant value as long as the orbit remains circular and unaltered by outside forces.
Gravitational Force
Gravitational force is the attractive force between two bodies due to their masses. It is an essential concept in orbital mechanics as it keeps the satellite in orbit around the Earth. According to Newton's law of gravitation, this force is directly proportional to the product of the masses of the two objects, and inversely proportional to the square of the distance between their centers.

If we consider Earth and a satellite, the force (\( F \)) can be expressed with the formula:\[ F = \frac{GMm}{r^2} \]where:\[ G \]is the gravitational constant,\[ M \]and\[ m \]are the masses of the Earth and the satellite respectively, and\[ r \]is the distance between the centers of the Earth and the satellite. The gravitational force not only dictates the required orbital speed for circular motion but also affects the satellite's period of revolution.
Newton's Law of Universal Gravitation
Sir Isaac Newton formulated the Law of Universal Gravitation, which is fundamental to understanding why planets, stars, and satellites behave as they do in space. This law states that every point mass attracts every other point mass in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

The law can be written as:\[ F = \frac{GMm}{r^2} \]where:\[ F \]is the force of gravity,\[ G \]is the gravitational constant,\[ M \]and\[ m \]are the masses of two objects, and\[ r \]is the distance between the centers of the two masses. This fundamental principle allows us to calculate the motion of celestial bodies and understand the dynamics of galaxies, solar systems, and even the trajectories of spacecraft.

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