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A 6200 -kg satellite is in a circular earth orbit that has a radius of \(3.3 \times 10^{7} \mathrm{~m}\). A net external force must act on the satellite to make it change to a circular orbit that has a radius of \(7.0 \times 10^{6} \mathrm{~m}\). What work must the net external force do?

Short Answer

Expert verified
Work done is approximately \(5.54 \times 10^{10} \text{J}\).

Step by step solution

01

Understanding the Problem

The problem involves calculating the work required to change the satellite's orbit from one circular path to another. We use the concept of gravitational potential energy to find the work done by the external force.
02

Gravitational Potential Energy Calculation

First, calculate the initial gravitational potential energy (U_initial) of the satellite in its original orbit and the final gravitational potential energy (U_final) in the new orbit. The formula for gravitational potential energy is \( U = -\frac{G \, M \, m}{r} \), where \( G = 6.674 \, \times \, 10^{-11} \, \text{Nm}^2/\text{kg}^2 \) is the gravitational constant, \( M = 5.972 \, \times \, 10^{24} \, \text{kg} \) is Earth's mass, \( m = 6200 \, \text{kg} \) is the satellite's mass, and \( r \) is the orbit radius.
03

Calculate Initial Potential Energy

Substitute the initial radius, \(3.3 \times 10^7 \, \text{m}\), into the gravitational potential energy formula: \[ U_{\text{initial}} = -\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24} \times 6200}{3.3 \times 10^7} \]. Calculate \( U_{\text{initial}} \).
04

Calculate Final Potential Energy

Substitute the new radius, \(7.0 \times 10^6 \, \text{m}\), into the formula: \[ U_{\text{final}} = -\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24} \times 6200}{7.0 \times 10^6} \]. Calculate \( U_{\text{final}} \).
05

Calculate Work Done by the External Force

The work done by the external force to change the satellite's orbit is the change in gravitational potential energy. Calculate \( W = U_{\text{final}} - U_{\text{initial}} \). Substitute the values from Steps 3 and 4 and solve for \( W \).
06

Solve for Work

Perform the calculations from the previous step to find the value of \( W \), which represents the work done by the external force.

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Key Concepts

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

Satellite Orbit
Satellites orbit the Earth because they travel at high speeds, constantly falling toward the earth without ever hitting it. This balancing act between forward momentum and gravitational pull keeps a satellite in orbit. A satellite moves in a circular path due to the gravitational force exerted by the Earth, which acts as the centripetal force. The speed at which a satellite travels depends on the mass of Earth and the distance from its center to the satellite. The orbital speed ensures that the satellite remains in motion around Earth without falling back.
The type of orbit can affect the satellite's function and the energy required for its maintenance. A geostationary orbit, for example, is circular and revolving at a rate that keeps the satellite over the same geographical point, requiring very precise control. Meanwhile, a low Earth orbit involves the satellite moving across the Earth’s surface, often faster than the rotation of the Earth itself.
Understanding these dynamics is crucial for calculating the energy or work needed to change the satellite's orbit.
Work Done by External Force
When considering the work needed to change a satellite's orbit, we think about how much energy must be added or taken away. The work done by an external force is essentially the energy required to move the satellite from one orbit to another. This involves overcoming the difference in gravitational potential energy between the two orbits.
Gravitational potential energy influences the amount of work required. The formula for work done, when moving a satellite from a higher orbit to a lower one, is given by the change in gravitational potential energy:
  • Calculate the initial potential energy for the original orbit radius.
  • Find the final potential energy for the new orbit radius.
  • Determine the work done:
    \[ W = U_{\text{final}} - U_{\text{initial}} \ \]

In simple terms, this calculation tells us how much energy is needed to push or pull the satellite to its new path, aligning it with the desired orbit's length and shape.
Circular Orbit Radius
The radius of a circular orbit refers to the average distance from the center of Earth to the satellite. It plays a critical role in determining both the speed of the satellite and its gravitational potential energy. The larger the radius, the more potential energy the satellite has, and the slower it travels if just held by gravity. Conversely, a smaller radius generally means a faster orbit due to the stronger gravitational pull at play.
To calculate the gravitational potential energy for any given radius, we utilize the formula:
  • \[ U = -\frac{G \, M \, m}{r} \ \]
  • Where \(G\) is the gravitational constant, \(M\) is the mass of Earth, \(m\) is the mass of the satellite, and \(r\) is the radius of the orbit.
Understanding the circular orbit radius is crucial for satellite trajectory planning and energy calculations. It helps visualize the path and speed a satellite must maintain to accomplish its mission. Accurate computation of the orbit radius ensures efficient energy use, keeping satellites operational and fulfilling their roles correctly.

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Most popular questions from this chapter

A cable lifts a \(1200-\mathrm{kg}\) elevator at a constant velocity for a distance of \(35 \mathrm{~m}\). What is the work done by (a) the tension in the cable and (b) the elevator's weight?

The motor of a ski boat generates an average power of \(7.50 \times 10^{4} \mathrm{~W}\) when the boat is moving at a constant speed of \(12 \mathrm{~m} / \mathrm{s}\). When the boat is pulling a skier at the same speed, the engine must generate an average power of \(8.30 \times 10^{4} \mathrm{~W}\). What is the tension in the tow rope that is pulling the skier?

A 63 -kg skier coasts up a snow-covered hill that makes an angle of \(25^{\circ}\) with the horizontal. The initial speed of the skier is \(6.6 \mathrm{~m} / \mathrm{s}\). After coasting a distance of \(1.9 \mathrm{~m}\) up the slope, the speed of the skier is \(4.4 \mathrm{~m} / \mathrm{s}\). (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?

A motorcycle (mass of cycle plus rider \(=2.50 \times 10^{2} \mathrm{~kg}\) ) is traveling at a steady speed of \(20.0 \mathrm{~m} / \mathrm{s} .\) The force of air resistance acting on the cycle and rider \(2.00 \times 10^{2} \mathrm{~N}\). Find the power necessary to sustain this speed if (a) the road is level and (b) the road is sloped upward at \(37.0^{\circ}\) with respect to the horizontal.

Refer to Interactive Solution \(6.51\) for a review of the approach taken in problems such as this one. A \(67.0\) -kg person jumps from rest off a \(3.00\) -m-high tower straight down into the water. Neglect air resistance during the descent. She comes to rest \(1.10 \mathrm{~m}\) under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is nonconservative.

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