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Ch 09: Work and Kinetic Energy
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 09: Work and Kinetic EnergyProblema 70b
Capitolo 9, Problema 70b

A 12 kg weather rocket generates a thrust of 200 N. The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 550 N/m, is anchored to the ground. After the engine is ignited, what is the rocket’s speed when the spring has stretched 40 cm?

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Step 1: Understand the problem. A geosynchronous orbit means the satellite has a period equal to the rotational period of the Earth, which is 24 hours. However, we will verify this by using Kepler's Third Law and the given altitude of the orbit.
Step 2: Write down Kepler's Third Law for circular orbits: T2 = 4π2 r3GM, where T is the orbital period, r is the orbital radius, G is the gravitational constant, and M is the mass of the Earth.
Step 3: Calculate the orbital radius r. The orbital radius is the sum of the Earth's radius and the altitude of the satellite. Use the Earth's radius (approximately 6.37 x 10^6 m) and the given altitude (3.58 x 10^7 m) to find r: r = REarth + h.
Step 4: Rearrange Kepler's Third Law to solve for T: T = 4π2 r3GM. Substitute the values for G (6.674 x 10^-11 N·m²/kg²), M (5.972 x 10^24 kg), and r (calculated in Step 3).
Step 5: Simplify the expression to find T. Ensure the units are consistent (meters, kilograms, seconds) throughout the calculation. The result should confirm that the period is approximately 24 hours, as expected for a geosynchronous orbit.

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Geosynchronous Orbit

A geosynchronous orbit is a circular orbit around the Earth where a satellite's orbital period matches the Earth's rotation period, approximately 24 hours. This allows the satellite to remain fixed over a specific point on the equator, making it ideal for communication purposes. The altitude of such an orbit is about 35,786 kilometers (22,236 miles) above sea level.
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Geosynchronous Orbits

Orbital Period

The orbital period is the time it takes for a satellite to complete one full orbit around a celestial body. For geosynchronous satellites, this period is equal to the rotational period of the Earth, which is approximately 24 hours. This synchronization ensures that the satellite appears stationary relative to the Earth's surface.
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Geosynchronous Orbits

Centripetal Force

Centripetal force is the net force required to keep an object moving in a circular path, directed towards the center of the circle. In the context of satellites, gravitational force provides the necessary centripetal force to maintain the satellite's orbit. The balance between gravitational force and the satellite's inertia is crucial for stable orbiting.
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Intro to Centripetal Forces
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Domanda del libro di testo

A 12 kg weather rocket generates a thrust of 200 N. The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 550 N/m, is anchored to the ground. Initially, before the engine is ignited, the rocket sits at rest on top of the spring. How much is the spring compressed?

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Domanda del libro di testo

A gardener pushes a 12 kg lawnmower whose handle is tilted up 37° above horizontal. The lawnmower's coefficient of rolling friction is 0.15. How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.2 m/s? Assume his push is parallel to the handle.

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Domanda del libro di testo

Draw a pictorial representation.

T−(1500kg)(9.8m/s2)=(1500kg)(1.0m/s2)T-(1500\,\(\text{kg}\))(9.8\,\(\text{m/s}\)^2)=(1500\,\(\text{kg}\))\(\left\)(1.0\,\(\text{m/s}\)^2\(\right\))

P=T(2.0m/s)P=T(2.0\,\(\text{m/s}\))

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Domanda del libro di testo

Finish the solution of the problem.

T−(1500kg)(9.8m/s2)=(1500kg)(1.0m/s2)T-(1500\,\(\text{kg}\))(9.8\,\(\text{m/s}\)^2)=(1500\,\(\text{kg}\))\(\left\)(1.0\,\(\text{m/s}\)^2\(\right\))

P=T(2.0m/s)P=T(2.0\,\(\text{m/s}\))

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