/*! 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 1 A 400 -kg satellite is placed in... [FREE SOLUTION] | 91Ó°ÊÓ

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A 400 -kg satellite is placed in a circular orbit \(6394 \mathrm{km}\) above the surface of the earth. At this elevation, the acceleration of gravity is \(4.09 \mathrm{m} / \mathrm{s}^{2} .\) Knowing that its orbital speed is \(20000 \mathrm{km} / \mathrm{h}\), determine the kinetic energy of the satellite.

Short Answer

Expert verified
The kinetic energy is approximately \(6.1739 \times 10^9 \text{ J}\).

Step by step solution

01

Identify Known Values

Firstly, gather all the information given in the problem. The mass of the satellite is 400 kg, the gravitational acceleration at the satellite's orbit is \(4.09 \text{ m/s}^2\), and its orbital speed is \(20000 \text{ km/h}\). Convert the speed from km/h to m/s for consistency in SI units.
02

Convert Speed Units

Convert the speed from kilometers per hour to meters per second. Use the conversion factor: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). Thus, \(20000 \text{ km/h} = \frac{20000}{3.6} \text{ m/s}\).
03

Apply Kinetic Energy Formula

Use the formula for kinetic energy: \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass of the satellite and \(v\) is its speed. Substitute the mass (400 kg) and converted speed into the formula to find the kinetic energy.
04

Perform Calculations

First, calculate the converted speed: \( \frac{20000}{3.6} = 5555.56 \text{ m/s} \). Next, compute the kinetic energy: \( KE = \frac{1}{2} \times 400 \times (5555.56)^2 \). This results in a kinetic energy of approximately \(6.1739 \times 10^9 \text{ J}\).

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Key Concepts

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

Orbital Mechanics
In the realm of physics, orbital mechanics is a fascinating subject dealing with the motion of objects in space. It primarily revolves around the gravitational forces that govern the movement of celestial bodies, like satellites or planets.
When we place a satellite in a circular orbit around Earth, as in this exercise, we are essentially balancing two forces:
  • The gravitational pull of the Earth which tends to draw the satellite towards the surface.
  • The centripetal force needed to keep the satellite moving in a circle at a constant speed.
Orbital mechanics involves determining the trajectory and velocity needed for this balance. In our exercise, the satellite is set in a specific path with a tailored speed of 20000 km/h. This speed ensures that the gravitational pull is perfectly balanced with the tendency of the satellite to move along a straight line. This balance is crucial for maintaining a stable orbit, which is what keeps our satellites in the sky, allowing technologies like GPS and communication systems to function seamlessly.
Gravity
Gravity is a fundamental force in the universe that not only pulls objects towards one another but also dictates the motion of planets and satellites. In the context of satellites, gravity acts as the anchoring force, keeping them locked in their orbits.
As specified in the exercise, the acceleration of gravity at the altitude of the satellite is 4.09 m/s². This is different from the standard gravity of 9.81 m/s² experienced on Earth's surface because gravity weakens with distance from the center of the Earth.
The gravitational pull on the satellite determines two key factors:
  • The altitude at which the satellite will orbit. Higher altitudes experience weaker gravitational forces.
  • The required velocity for an orbit to be circular and stable. Faster velocities at lower altitudes or slower ones at higher altitudes are needed to maintain the balance.
In this way, gravity plays a pivotal role in the satellite’s behavior, orchestrating a delicate dance that allows it to continuously circumnavigate the Earth.
Unit Conversion
Effective problem-solving in physics often hinges on the correct handling of units. Understanding and performing unit conversions is essential, especially when working with different systems of measurement.
In this exercise, the satellite's speed is initially given in kilometers per hour (km/h), which must be converted to meters per second (m/s) to use within the kinetic energy formula, which uses SI units.
  • The conversion factor we use is: 1 km/h = 1/3.6 m/s.
  • So, to convert 20000 km/h to m/s, we divide by 3.6.
This yields a speed of approximately 5555.56 m/s.
Unit conversion ensures that all terms in an equation are compatible, allowing for accurate calculations and better comprehension of the physical quantities being measured. Without correct conversions, the results could be misleading or erroneous, thus complicating the understanding of the topics like kinetic energy and orbital mechanics.

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Most popular questions from this chapter

The Mars Pathfinder spacecraft used large airbags to cushion its impact with the planet's surface when landing. Assuming the space-craft had an impact velocity of \(18.5 \mathrm{m} / \mathrm{s}\) at an angle of \(45^{\circ}\) with respect to the horizontal, the coefficient of restitution is 0.85 and neglecting friction, determine \((a)\) the height of the first bounce, (b) the length of the first bounce. (Acceleration of gravity on Mars \(=3.73 \mathrm{m} / \mathrm{s}^{2}\) )

Collar \(A\) has a mass of \(3 \mathrm{kg}\) and is attached to a spring of constant \(1200 \mathrm{N} / \mathrm{m}\) and of undeformed length equal to \(0.5 \mathrm{m}\). The system is set in motion with \(r=0.3 \mathrm{m}, v_{\theta}=2 \mathrm{m} / \mathrm{s}\), and \(v_{r}=0 .\) Neglecting the mass of the rod and the effect of friction, determine \((a)\) the maximum distance between the origin and the collar, \((b)\) the corresponding speed. (Hint: Solve the equation obtained for \(r\) by trial and error.)

Packages in an automobile parts supply house are transported to the loading dock by pushing them along on a roller track with very little friction. At the instant shown, packages B and C are at rest, and package A has a velocity of 2 m/s. Knowing that the coefficient of restitution between the packages is 0.3, determine (a) the velocity of package C after A hits B and B hits C, (b) the velocity of A after it hits B for the second time.

A ballistic pendulum is used to measure the speed of high-speed projectiles. A \(6-\mathrm{g}\) bullet \(A\) is fired into a \(1-\mathrm{kg}\) wood block \(B\) suspended by a cord with a length of \(l=2.2 \mathrm{m}\). The block then swings through a maximum angle of \(\theta=60^{\circ} .\) Determine \((a)\) the initial speed of the bullet \(v_{0},(b)\) the impulse imparted by the bullet on the block, \((c)\) the force on the cord immediately after the impact.

A golf ball struck on earth rises to a maximum height of 60 m and hits the ground 230 m away. How high will the same golf ball travel on the moon if the magnitude and direction of its velocity are the same as they were on earth immediately after the ball was hit? Assume that the ball is hit and lands at the same elevation in both cases and that the effect of the atmosphere on the earth is neglected, so that the trajectory in both cases is a parabola. The acceleration of gravity on the moon is 0.165 times that on earth.

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