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Chapter 10: Problem 23

A satellite is in a circular orbit about Earth moving at a speed of \(v\). In order to escape the planet, the satellite must attain a speed of (SSM) A. \(v\). B. \(\sqrt{2} v\) C. \(2 v\). D. \(v / 2\). E. \(\frac{v}{\sqrt{2}}\)

Short Answer

Expert verified
The correct answer is B. \(\sqrt{2} v\).

Step by step solution

01

Derive the Formula for the Escape Velocity

Escape velocity is given by \(v_{esc} = \sqrt{2GM\over r}\), where G is the gravitational constant, M is the mass of the earth and r is the distance of the satellite from the center of the earth.
02

Derive the Formula for Circular Orbit Speed

The speed of a satellite in circular orbit \(v_0\) is given by \(v_0 = \sqrt{GM\over r}\). This is derived from setting the gravitational force equal to the centripetal force.
03

Compare Escape Velocity and Orbital Speed

Comparing the two formulas, it is seen that \(v_{esc}\) is \(\sqrt{2}\) times \(v\). Meaning, the escape speed of the satellite is \(\sqrt{2}\) times its current speed.

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Key Concepts

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

Circular Orbit
A circular orbit is when a satellite moves around Earth (or any celestial body) in a circle. This means that the speed and direction of the satellite remain constant. The satellite is in perfect balance. The gravitational pull of Earth draws the satellite inwards, but at the same time, its forward speed keeps it from falling into the planet.
For an object to maintain a circular orbit, two forces must balance out:
  • Gravity: Pulls the satellite toward the center of the Earth.
  • Centripetal force: Keeps the satellite moving along its circular path.
Understanding circular orbits helps us grasp why satellites don't just fall to Earth or drift off into space.
Gravitational Constant
The gravitational constant, denoted by 'G', plays a vital role in calculating gravitational forces. Its value is approximately \(6.674 \times 10^{-11} \, \text{m}^3/(\text{kg} \cdot \text{s}^2)\). This constant appears in Newton's law of universal gravitation, which states that any two masses attract each other with a force proportional to their masses and inversely proportional to the square of the distance between their centers.
In our context, G helps us figure out the gravitational force between Earth and a satellite, making it possible to determine the satellite's speed and how much energy is needed to escape Earth's pull.
Centripetal Force
Centripetal force is a key player in keeping satellites in their orbits. It's the "center-seeking" force that ensures an object follows a curved path. When a satellite is in circular motion, the centripetal force required to maintain its orbit comes from Earth's gravity.
Here's how it works:
  • The force pulls the satellite toward Earth, ensuring it doesn't fly off.
  • It's what gives the satellite its circular path.
This force can be calculated using the formula:\[ F_c = \frac{mv^2}{r} \] where:
  • \( m \) is the mass of the satellite,
  • \( v \) is the velocity of the satellite,
  • \( r \) is the radius of the orbit.
Satellite Motion
Satellite motion is all about balance. It's a dance between speed and gravity. Satellites need to move fast enough to stay in orbit but not so fast that they drift away from the planet. This nuanced relationship is governed by several factors:
  • Initial speed: Determines if a satellite stays in orbit or continues into outer space.
  • Gravity: Constantly pulls the satellite towards the planet, acting as a central force.
  • Mass and distance: Determine the gravitational pull, influencing how fast a satellite needs to move to maintain its orbit.
For a satellite to leave its orbit and go into space, it must reach escape velocity. This means exceeding its current circular orbit speed by a factor of \(\sqrt{2}\), as explored earlier in the escape velocity concept.

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Most popular questions from this chapter

A satellite is to be raised from one circular orbit to one further from Earth's surface. What will happen to its period?

Two satellites having equal masses are in circular orbits around Earth. Satellite A has a smaller orbital radius than satellite B. Which statement is true? SSM A. Satellite A has more kinetic energy, less potential energy, and less mechanical energy (potential energy plus kinetic energy) than satellite B. B. Satellite A has less kinetic energy, less potential energy, and less mechanical energy (potential energy plus kinetic energy) than satellite B. C. Satellite A has more kinetic energy, more potential energy, and less mechanical energy (potential energy plus kinetic energy) than satellite \(\mathrm{B}\). D. Satellite A and satellite B have the same amount of mechanical energy (potential energy plus kinetic energy). E. Satellite A and satellite B have the same amount of kinetic energy and no potential energy because they are in motion.

Spherical storage tanks are used to hold natural gas. A storage facility contains a row of five such tanks, each with a volume of \(5000 \mathrm{~m}^{3}\). Determine the net gravitational force on the sphere at the end of the row of tanks due to the other tanks if they are filled with natural gas (natural gas' density is \(0.8 \mathrm{~kg} / \mathrm{m}^{3}\) ) and constructed from steel (steel's density is \(8000 \mathrm{~kg} / \mathrm{m}^{3}\) ) that is \(10 \mathrm{~cm}\) thick. The tanks are \(75 \mathrm{~m}\) apart (this distance is the closest one from the right edge of one tank to the left edge of the adjacent one).

Much attention has been devoted to the exact numerical power in Newton's law of universal gravitation \(\left(F \propto r^{-2}\right)\). Some theorists have investigated whether the dependence might be slightly larger or smaller than 2 . What would be the significance (what impact would there be) if the power was not exactly 2?

A satellite and the International Space Station have the same mass and are going around Earth in concentric orbits. The distance of the satellite from Earth's center is twice that of the International Space Station's distance. What is the ratio of the centripetal force acting on the satellite compared to that acting on the International Space Station? A. \(\frac{1}{4}\) B. \(\frac{1}{2}\) C. 1 D. 2 E. 4
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