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Chapter 6: Problem 20

A \(2150 \mathrm{~kg}\) satellite used in a cellular telephone network is in a circular orbit at a height of \(780 \mathrm{~km}\) above the surface of the earth. What is the gravitational force on the satellite? What fraction is this force of the satellite's weight at the surface of the earth?

Short Answer

Expert verified
Gravitational force on satellite: ~16067 N, fraction: ~76.1% of Earth's surface weight.

Step by step solution

01

Calculate Gravitational Force

The gravitational force (\( F \)) on the satellite can be calculated using the formula:\[ F = \frac{G \cdot M \cdot m}{r^2} \]where \( G = 6.674 \times 10^{-11} \, \mathrm{Nm^2/kg^2} \) (gravitational constant), \( M = 5.972 \times 10^{24} \, \mathrm{kg} \) (mass of Earth), and \( m = 2150 \, \mathrm{kg} \) (mass of satellite).The distance \( r \) from the center of the Earth to the satellite is the sum of the Earth's radius \( R = 6371 \mathrm{~km} \, (or \, 6.371 \times 10^6 \mathrm{~m}) \) and the height above the surface \( h = 780 \mathrm{~km} \, (or \, 780 \times 10^3 \mathrm{~m}) \).Therefore, \( r = R + h = 6.371 \times 10^6 \mathrm{~m} + 780 \times 10^3 \mathrm{~m} = 7.151 \times 10^6 \mathrm{~m} \).Now, substituting these values into the equation:\[ F = \frac{6.674 \times 10^{-11} \cdot 5.972 \times 10^{24} \cdot 2150}{(7.151 \times 10^6)^2} \]\[ F \approx 16067 \mathrm{~N} \].
02

Calculate Satellite Weight on Earth's Surface

The weight of the satellite on the Earth's surface is given by\( W = m \cdot g \)where \( g = 9.81 \, \mathrm{m/s^2} \) is the acceleration due to gravity on Earth's surface.\( m = 2150 \, \mathrm{kg} \), therefore\[ W = 2150 \cdot 9.81 = 21081.5 \, \mathrm{N} \].
03

Calculate the Fraction of the Gravitational Force

The fraction of the gravitational force on the satellite compared to its weight on the Earth's surface is calculated by dividing the force in orbit by the weight on the surface:\[ \text{Fraction} = \frac{F}{W} \]Substituting the values from the previous steps:\[ \text{Fraction} = \frac{16067}{21081.5} \approx 0.761 \]This means the gravitational force in orbit is about 76.1% of its weight at Earth's surface.

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Key Concepts

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

Satellite Motion
Satellites, like the one used in a cellular telephone network, move in what we call "satellite motion." Their movement happens because of the gravitational pull exerted by Earth. This pull keeps them in a constant path or orbit around our planet. When a satellite orbits Earth, it balances between two forces: gravity pulling it towards Earth and its tendency to move in a straight line into space. This constant pull and resistance keep it moving in a circle or elliptical path.
When a satellite is in motion, it doesn't lose contact or fall back to Earth unless something disrupts its path or momentum. Their speed and the gravitational pull are what keep satellites from floating away into space or crashing back to Earth.
Orbits
An orbit is the path that an object takes as it moves around another larger object, usually due to gravity. The satellite in our problem is in a circular orbit, which is a special type of orbit characterized by a constant distance from the center of the Earth throughout its motion. Circular orbits are the simplest form of orbits where the speed and altitude remain the same.
The height of the orbit, such as 780 km above the Earth's surface as given, determines the satellite's period and speed. The further a satellite is from Earth, the slower it moves along its orbit. This is because the gravitational force decreases with distance, allowing satellites farther away to move at slower speeds and still maintain their path. However, if a satellite moves to a lower orbit, closer to Earth, it must move faster to counteract the stronger gravitational pull.
Newton's Law of Universal Gravitation
Newton's Law of Universal Gravitation is a fundamental law that tells us how bodies are attracted to each other by gravity. It states that every two masses in the universe attract each other with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
The formula \[ F = \frac{G \cdot M \cdot m}{r^2} \] utilizes the gravitational constant \( G \), the masses \( M \) and \( m \), and the distance \( r \). This calculation tells us the gravitational force acting on the satellite. It highlights why objects like our satellite, orbiting at high altitudes, experience less gravitational force than when they are closer to Earth's surface.
Weight and Mass
Weight and mass are related but distinct concepts. Mass is a measure of how much matter is in an object, and it stays constant regardless of location. In the satellite's case, its mass is 2150 kg.
Weight, however, is the measure of the force exerted by the gravitational field on that mass. On Earth's surface, this is calculated with the formula \( W = m \cdot g \), where \( g \) is the acceleration due to gravity. As the satellite moves into orbit, the gravitational pull (and hence its weight) decreases, since weight depends on the distance from Earth's center. The satellite does not "weigh" the same as on Earth, because weight changes with gravity, while mass remains consistent.
Understanding this distinction helps explain why objects feel lighter in space鈥攍ess gravitational force acts upon them, although their mass remains unchanged.

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Most popular questions from this chapter

The asteroid 243 Ida has a mass of about \(4.0 \times 10^{16} \mathrm{~kg}\) and an average radius of about \(16 \mathrm{~km}\) (it's not spherical, but you can assume it is). (a) Calculate the acceleration of gravity on 243 Ida. (b) What would an astronaut whose earth weight is \(650 \mathrm{~N}\) weigh on 243 Ida? (c) If you dropped a rock from a height of \(1.0 \mathrm{~m}\) on 243 Ida, how long would it take for the rock to reach the ground? (d) If you can jump \(60 \mathrm{~cm}\) straight up on earth, how high could you jump on 243 Ida? (Assume the asteroid's gravity doesn't weaken significantly over the distance of your jump.)

I A flat (unbanked) curve on a highway has a radius of \(220 \mathrm{~m}\). A car successfully rounds the curve at a speed of \(35 \mathrm{~m} / \mathrm{s}\) but is on the verge of skidding out. (a) If the coefficient of static friction between the car's tires and the road surface were reduced by a factor of \(2,\) with what maximum speed could the car round the curve? (b) Suppose the coefficient of friction were increased by a factor of 2; what would be the maximum speed?

A \(50.0 \mathrm{~kg}\) stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. (a) If the plane's speed at the lowest point of the circle is \(9.50 \mathrm{~m} / \mathrm{s}\) what should the minimum radius of the circle be in order for the cen- tripetal acceleration at this point not to exceed \(4.00 \mathrm{~g} ?\) (b) What is the apparent weight of the pilot at the lowest point of the pullout?

Apparent weightlessness in a satellite. You have probably seen films of astronauts floating freely in orbiting satellites. People often think the astronauts are weightless because they are free of the gravity of the earth. Let us see if that explanation is correct. (a) Typically, such satellites orbit around \(400 \mathrm{~km}\) above the surface of the earth. If an astronaut weighs \(750 \mathrm{~N}\) on the ground, what will he weigh if he is \(400 \mathrm{~km}\) above the surface? (b) Draw the orbit of the satellite in part (a) to scale on a sketch of the earth. (c) In light of your answers to parts (a) and (b), are the astronauts weightless because gravity is so weak? Why are they apparently weightless?

If an object's weight is \(W\) on the earth, what would be its weight (in terms of \(W\) ) if the earth had (a) twice its present mass but was the same size, (b) half its present radius but the same mass, (c) half its present radius and half its present mass, (d) twice its present radius and twice its present mass?
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