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Chapter 5: Problem 38

\(\mathrm{A}\) satellite has a mass of \(5850 \mathrm{kg}\) and is in a circular orbit \(4.1 \times\) \(10^{5} \mathrm{m}\) above the surface of a planet. The period of the orbit is \(2.00 \mathrm{hours}\). The radius of the planet is \(4.15 \times 10^{6} \mathrm{m} .\) What would be the true weight of the satellite if it were at rest on the planet's surface?

Short Answer

Expert verified
The true weight of the satellite on the planet's surface is approximately 51670 N.

Step by step solution

01

Calculate the distance from the center of the planet

The satellite orbits 4.1 \times 10^5 m above the surface of the planet. Given that the radius of the planet is 4.15 \times 10^6 m, the total distance from the center of the planet is calculated as follows: \[ d = 4.1 \times 10^5 \, m + 4.15 \times 10^6 \, m = 4.56 \times 10^6 \, m \] where \(d\) is the distance from the planet's center.
02

Calculate the gravitational force exerted by the planet on the satellite in orbit

The gravitational force \( F \) is found using Newton's law of universal gravitation: \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \] However, instead of calculating this directly, we use the formula for the centripetal force since the satellite is in orbit, where \(r\) is the distance from the center of the planet: \[ F = \frac{4 \pi^2 \cdot m}{T^2} \cdot d \] Substituting the values, \( m = 5850 \, \text{kg}, \) \( T = 2 \times 3600 \, \text{s}, \) and \( d = 4.56 \times 10^6 \, \text{m} \), calculate \( F \): \[ F = \frac{4 \pi^2 \times 5850}{2^2 \times 3600^2} \times 4.56 \times 10^6 = 2.863 \times 10^4 \text{ N} \]
03

Calculate the mass of the planet

Using the relation between centripetal force and gravitational force, we have: \[ F = \frac{G \cdot m_s \cdot M_p}{r^2} = \frac{4 \pi^2 \cdot m_s \cdot r}{T^2} \] Solving for \( M_p \) gives: \[ M_p = \frac{4 \pi^2 \cdot r^3}{G \cdot T^2} \] Substituting \( r = 4.56 \times 10^6 \, \text{m} \) and \( T = 7200 \, \text{s} \), calculate \( M_p \). Using \( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \): \[ M_p = \frac{4 \pi^2 \times (4.56 \times 10^6)^3}{6.674 \times 10^{-11} \cdot (7200)^2} \approx 8.93 \times 10^{24} \text{ kg} \]
04

Calculate the gravitational force on the satellite at planet's surface

To find the weight of the satellite on the planet's surface, we use: \[ F_s = \frac{G \cdot m_s \cdot M_p}{R^2} \] where \( R = 4.15 \times 10^6 \, \text{m} \). Substituting the values for \( m_s \), \( M_p \), and \( R \), we find the true weight: \[ F_s = \frac{6.674 \times 10^{-11} \cdot 5850 \cdot 8.93 \times 10^{24}}{(4.15 \times 10^6)^2} \approx 5.167 \times 10^4 \text{ N} \]

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Key Concepts

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

Satellite Mass
The satellite mass is the amount of matter contained within the satellite, typically measured in kilograms. In most physics problems related to orbital mechanics, knowing the mass of the satellite is crucial for calculating forces and energies. In this particular exercise, the satellite has a mass of \( 5850 \text{ kg} \). This mass affects the gravitational force that the satellite experiences from the planet. It's important to remember that while mass is constant and does not change with location or conditions, weight does change depending on the location due to the different gravitational forces acting on the mass.
Universal Gravitation
Universal gravitation is a fundamental principle in physics that describes the attractive force between two masses. According to this principle, every point mass attracts every other point mass with a force that is proportional to the product of their masses and inversely proportional to the square of their separation distance. Mathematically, it is expressed with the formula:
  • \[F = \frac{G \cdot m_1 \cdot m_2}{r^2}\]
where:
  • \( F \) is the gravitational force,
  • \( G \) is the universal gravitational constant (\( 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \)),
  • \( m_1 \) and \( m_2 \) are the masses of the objects,
  • \( r \) is the distance between the centers of the two masses.
For the satellite and planet in this exercise, universal gravitation determines the gravitational force at any given distance from the planet.
Centripetal Force
When an object moves in a circular path, it experiences a centripetal force, which acts towards the center of the circle to keep the object in motion. In the context of a satellite orbiting a planet, centripetal force is necessary for maintaining the circular orbit. The force required is given by:
  • \[F = \frac{4 \pi^2 \cdot m \cdot d}{T^2} \]
where:
  • \( F \) is the centripetal force,
  • \( m \) is the mass of the satellite,
  • \( d \) is the distance from the center of the circle,
  • \( T \) is the period of orbit.
In this example, the weight of the satellite orthogonally balances with the centripetal force, allowing it to stay in stable circular motion. Understanding centripetal force is essential in determining the forces that keep satellites in orbit.
Orbital Mechanics
Orbital mechanics is the study of the motions of artificial satellites and natural celestial bodies under the influence of forces like gravity. In this exercise, we delve into aspects of orbital mechanics to understand the behavior of a satellite with a specific mass in orbit around a planet. Several important principles can be gleaned:
  • Satellites follow elliptical or circular paths determined by gravitational forces.
  • The distance from the planet, combined with gravitational forces, dictates the velocity needed for stable orbits.
  • The concept of angular velocity, period of orbit, and the relation to centripetal force are integral to these calculations.
In practice, orbital mechanics uses these principles to determine how satellites can be launched, maintained in orbit, and used for various applications, from communication to scientific observation.

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Most popular questions from this chapter

A special electronic sensor is embedded in the seat of a car that takes riders around a circular loop-the-loop ride at an amusement park. The sensor measures the magnitude of the normal force that the seat exerts on a rider. The loop- the-loop ride is in the vertical plane and its radius is \(21 \mathrm{m}\). Sitting on the seat before the ride starts, a rider is level and stationary, and the electronic sensor reads \(770 \mathrm{N}\). At the top of the loop, the rider is upside down and moving, and the sensor reads \(350 \mathrm{N}\). What is the speed of the rider at the top of the loop?

A space laboratory is rotating to create artificial gravity, as the figure indicates. Its period of rotation is chosen so that the outer ring \(\left(r_{0}=\right.\) \(2150 \mathrm{m})\) simulates the acceleration due to gravity on earth \(\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right) .\) What should be the radius \(r_{1}\) of the inner ring, so it simulates the acceleration due to gravity on the surface of Mars \(\left(3.72 \mathrm{m} / \mathrm{s}^{2}\right) ?\)

The second hand and the minute hand on one type of clock are the same length. Find the ratio \(\left(a_{\text {c, second }} / a_{\text {c, minute }}\right)\) of the centripetal accelerations of the tips of the second hand and the minute hand.

A satellite circles the earth in an orbit whose radius is twice the earth's radius. The earth's mass is \(5.98 \times 10^{24} \mathrm{kg}\), and its radius is \(6.38 \times 10^{6} \mathrm{m}\). What is the period of the satellite?

The hammer throw is a track-and-field event in which a \(7.3-\mathrm{kg}\) ball (the "hammer") is whirled around in a circle several times and released. It then moves upward on the familiar curving path of projectile motion and eventually returns to earth some distance away. The world record for this distance is \(86.75 \mathrm{m},\) achieved in 1986 by Yuriy Sedykh. Ignore air resistance and the fact that the ball is released above the ground rather than at ground level. Furthermore, assume that the ball is whirled on a circle that has a radius of \(1.8 \mathrm{m}\) and that its velocity at the instant of release is directed \(41^{\circ}\) above the horizontal. Find the magnitude of the centripetal force acting on the ball just prior to the moment of release.
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