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Chapter 11: Problem 158

A satellite will travel indefinitely in a circular orbit around the earth if the normal component of its acceleration is equal to \(g(R / r)^{2},\) where \(g=9.81 \mathrm{m} / \mathrm{s}^{2}, R=\) radius of the earth \(=6370 \mathrm{km},\) and \(r=\) distance from the center of the earth to the satellite. Assuming that the orbit of the moon is a circle with a radius of \(384 \times 10^{3} \mathrm{km},\) determine the speed of the moon relative to the earth.

Short Answer

Expert verified
The moon's speed relative to Earth is approximately 1017 m/s.

Step by step solution

01

Understand the Problem Context

We need to find the speed of the moon in its circular orbit around the Earth. The problem gives us the formula for the normal component of acceleration that must equal the gravitational acceleration reduced by square of ratio of Earth radius to orbit radius.
02

Recall the Formula for Centripetal Acceleration

For an object in circular motion, the centripetal acceleration is given by the formula: \( a_c = \frac{v^2}{r} \), where \( v \) is the speed and \( r \) is the radius of orbit.
03

Equate Gravitational Acceleration and Centripetal Acceleration

The satellite's centripetal acceleration must equal the gravitational component given by \( g\left(\frac{R}{r}\right)^2 \). Hence, \( \frac{v^2}{r} = g\left(\frac{R}{r}\right)^2 \).
04

Solve for Velocity \(v\)

Rearrange the equation to solve for \(v\):\[ v^2 = g r \left(\frac{R}{r}\right)^2 \]\[ v^2 = g \frac{R^2}{r} \]\[ v = \sqrt{g \frac{R^2}{r}} \]
05

Substitute Known Values

Substitute the given values into the equation:\( g = 9.81 \text{ m/s}^2 \), \( R = 6370 \text{ km} = 6.37 \times 10^6 \text{ m} \), and \( r = 384 \times 10^3 \text{ km} = 3.84 \times 10^8 \text{ m} \).\[ v = \sqrt{9.81 \times \frac{(6.37 \times 10^6)^2}{3.84 \times 10^8}} \]
06

Perform the Calculation

Calculate the velocity:\[ v = \sqrt{9.81 \times \frac{4.0569 \times 10^{13}}{3.84 \times 10^8}} \]\[ v = \sqrt{9.81 \times 1.056 \times 10^5} \]\[ v \approx \sqrt{1.035 \times 10^6} \]\[ v \approx 1017 \text{ m/s} \]
07

Conclusion

The speed of the moon relative to the Earth is approximately 1017 m/s.

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Key Concepts

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

Satellite Motion
Satellites, whether they are artificial or natural like the moon, move in a path called an orbit around a central body due to gravitational forces. Specifically, this kind of movement is termed 'satellite motion'. An orbit can be circular or elliptical, but for simplicity, let's focus on circular orbits in this context.
  • When a satellite orbits the Earth, it is constantly falling towards the planet due to gravity but its horizontal speed ensures that it keeps missing the Earth, staying in a stable path.
  • The altitude of the orbit affects the satellite's speed and period. For a circular orbit, a higher altitude means a slower speed.
  • The balance between gravitational pull and the satellite's inertia keeps it in orbit. This balance requires precise calculations to maintain.
By understanding these dynamics, we can calculate the necessary speed for a satellite to maintain its orbit, such as the moon's speed around the Earth. This helps us comprehend both the mechanics of man-made satellites and natural ones.
Centripetal Acceleration
Centripetal acceleration is crucial for understanding satellite motion. It is the acceleration that keeps a satellite moving in a circular path, pointing towards the center of the circle. Unlike other accelerations, centripetal acceleration doesn't change the satellite's speed; it changes its direction.
  • It is calculated using the formula: \( a_c = \frac{v^2}{r} \), where \( v \) is the velocity of the satellite, and \( r \) is the radius of the orbit.
  • The requirement here is that the centripetal force is supplied by gravity.
  • A greater velocity or a smaller radius results in higher centripetal acceleration since short radii and high speeds make tighter and faster circles.
This concept is key to ensuring that a body remains in its orbit by effectively compensating for gravitational pull.
Gravitational Acceleration
Gravitational acceleration affects all objects under the influence of a massive body like Earth. It is the acceleration caused by Earth's gravity when an object is in free fall towards it. For satellites in orbit, this concept ties directly with maintaining their path.
  • Near Earth's surface, gravitational acceleration is approximately \( g = 9.81 \ \text{m/s}^2 \).
  • As the distance from Earth increases, gravitational acceleration decreases proportionally.
  • For a satellite in orbit, gravity provides the necessary centripetal force to keep it moving along its circular path.
  • The formula \( g\left(\frac{R}{r}\right)^2 \) represents the adjusted gravitational acceleration, considering the radius of Earth \( R \) and the distance \( r \) from the center of Earth to the satellite.
Understanding gravitational acceleration allows us to align our predictions with the forces experienced in an orbit, which is essential for maintaining the correct velocity and path.

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Most popular questions from this chapter

The three-dimensional motion of a particle is defined by the position vector \(r=\left(R t \cos \omega_{n} t\right) \mathbf{i}+c t \mathbf{j}+\left(R t \sin \omega_{n} t\right) \mathbf{k} .\) Determine the magnitudes of the velocity and acceleration of the particle. (The space curve described by the particle is a conic helix.)

Instruments in an airplane which is in level flight indicate that the velocity relative to the air (airspeed) is \(120 \mathrm{km} / \mathrm{h}\) and the direction of the relative velocity vector (heading) is \(70^{\circ}\) east of north. Instruments on the ground indicate that the velocity of the airplane (ground speed) is \(110 \mathrm{km} / \mathrm{h}\) and the direction of flight (course) is \(60^{\circ}\) east of north. Determine the wind speed and direction.

The acceleration of a particle is defined by the relation \(a=k t^{2}\). (a) Knowing that \(v=-8 \mathrm{m} / \mathrm{s}\) when \(t=0\) and that \(v=+8 \mathrm{m} / \mathrm{s}\) when \(t=2 \mathrm{s}\), determine the constant \(k .(b)\) Write the equations of motion, knowing also that \(x=0\) when \(t=2 \mathrm{s}\).

The path of a particle \(P\) is a limafon. The motion of the particle is defined by the relations \(r=b(2+\cos \pi t)\) and \(\theta=\pi t\) where \(t\) and \(\theta\) are expressed in seconds and radians, respectively. Determine (a) the velocity and the acceleration of the particle when \(t=2 \mathrm{s},(b)\) the value of \(\theta\) for which the magnitude of the velocity is maximum.

Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit \(100 \mathrm{mi}\) above the surface of the planet. (See information given in Probs. \(11.153-11.154 .)\) $$ \begin{array}{l}{11.155 \text { Venus: } g=29.20 \mathrm{ft} / \mathrm{s}^{2}, R=3761 \mathrm{mi}} \\ {11.156 \mathrm{Mars}: g=12.17 \mathrm{ft} / \mathrm{s}^{2}, R=2102 \mathrm{mi}} \\ {11.157 \text { Jupiter: } g=75.35 \mathrm{ft} / \mathrm{s}^{2}, R=44,432 \mathrm{mi}}\end{array} $$
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