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Chapter 13: Problem 9

When a falling meteoroid is at a distance above the Earth's surface of 3.00 times the Earth's radius, what is its acceleration due to the Earth's gravitation?

Short Answer

Expert verified
The acceleration of the meteoroid due to the Earth's gravitation when it is at a distance three times the Earth's radius away from the Earth's surface is approximately \(0.867 \, m/s^2.\)

Step by step solution

01

Data extraction

There are two data given: \n\n1) The distance of meteoroid from the surface of the earth, which is three times the Earth's radius\n\n 2) The Earth's radius \( r = 6.37 \times 10^6 \, m \)\n\nThe universal gravitational constant \( G = 6.674 \times 10^{-11} \, m^3 kg^{-1} s^{-2} \) and the Earth's mass \( M = 5.98 \times 10^{24} \, kg \) are known.
02

Calculate the total distance

The total distance (r_total) from the center of the Earth to the meteoroid is the sum of the Earth's radius and the distance of the meteoroid above the Earth's surface. This gives us a result of \( r_{total} = 3r = 3(6.37 \times 10^6\, m) = 1.911 \times 10^{7}\, m \)
03

Apply the Formula for Gravitational Acceleration

The formula for gravitational acceleration is: \( a = \frac{GM}{r^2} \) \n\nSubstitute the given values into the formula to find the acceleration: \( a = \frac{ \left(6.674 \times 10^{-11} \, m^3 kg^{-1} s^{-2}\right) (5.98 \times 10^{24} \, kg) }{(1.911 \times 10^7 \, m )^2} \)
04

Calculate the Acceleration

Calculate the value for acceleration: \( a \approx 0.867 \, m/s^2 \)

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Key Concepts

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

Universal Gravitational Constant
The universal gravitational constant, often denoted by the symbol \( G \), is a key part of understanding the nature of gravitational force. It appears in Newton's law of universal gravitation, which describes how the force between two masses is related to their separation and their masses. The value of \( G \) is approximately \( 6.674 \times 10^{-11} \, m^3 kg^{-1} s^{-2} \). This constant is essential because it allows us to calculate the gravitational force between objects on a universal scale.
  • Newton's law of universal gravitation: \( F = \frac{GMm}{r^2} \)
  • \( G \) is a very small number, highlighting that gravitational force is comparatively weak compared to other fundamental forces.
For everyday calculations like those involving the Earth and objects near it, \( G \) is combined with the mass of Earth to calculate gravitational acceleration. Despite its weak nature, gravity is significant because it affects objects over large distances and masses.
Earth's Radius
The Earth's radius is a crucial measurement for calculations involving the planet's gravitational effects. The average radius is about \( 6.37 \times 10^6 \, m \). This value is used as a reference point in calculating distances of objects above the Earth's surface, such as satellites and meteoroids.
  • Distance from the center of Earth affects gravitational pull.
  • Greater distance from the Earth's center means less gravity's impact.
In the context of gravitational calculations, the radius helps determine how far an object is from the Earth's core, not just its surface, which is critical for accurately calculating forces acting on it.
Mass of Earth
The mass of Earth, denoted by \( M \), is approximately \( 5.98 \times 10^{24} \, kg \). This astounding number plays a major role in the calculation of gravitational forces that objects experience on or near the Earth.
  • Earth's mass is a determinant of the gravitational pull experienced by objects.
  • It affects the calculation of gravitational acceleration, which is \( a = \frac{GM}{r^2} \).
Along with the universal gravitational constant \( G \) and the distance from Earth's center, the mass allows us to calculate how much force an object will experience due to gravity. The larger the mass of a planetary body, the stronger its gravitational pull on nearby objects.

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Most popular questions from this chapter

(a) What is the minimum speed, relative to the Sun, necessary for a spacecraft to escape the solar system if it starts at the Earth's orbit? (b) Voyager 1 achieved a maximum speed of \(125000 \mathrm{km} / \mathrm{h}\) on its way to photograph Jupiter. Beyond what distance from the Sun is this speed sufficient to escape the solar system?

Studies of the relationship of the Sun to its galaxy-the Milky Way- -have revealed that the Sun is located near the outer edge of the galactic disk, about 30000 lightyears from the center. The Sun has an orbital speed of approximately \(250 \mathrm{km} / \mathrm{s}\) around the galactic center. (a) What is the period of the Sun's galactic motion? (b) What is the order of magnitude of the mass of the Milky Way galaxy? Suppose that the galaxy is made mostly of stars of which the Sun is typical. What is the order of magnitude of the number of stars in the Milky Way?

Derive an expression for the work required to move an Earth satellite of mass \(m\) from a circular orbit of radius \(2 R_{E}\) to one of radius \(3 R_{E}\).

Two spheres having masses \(M\) and \(2 M\) and radii \(R\) and \(3 R\) respectively, are released from rest when the distance between their centers is \(12 R\). How fast will each sphere be moving when they collide? Assume that the two spheres interact only with each other.

A \(200-\mathrm{kg}\) object and a \(500-\mathrm{kg}\) object are separated by \(0.400 \mathrm{m} .\) (a) Find the net gravitational force exerted by these objects on a 50.0 -kg object placed midway between them. (b) At what position (other than an infinitely remote one) can the \(50.0-\mathrm{kg}\) object be placed so as to experience a net force of zero?
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