/*! 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 39 Compute the mass of the Earth, a... [FREE SOLUTION] | 91Ó°ÊÓ

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Compute the mass of the Earth, assuming it to be a sphere of radius \(6370 \mathrm{~km}\). Use the fact that \(\mathrm{g}=9.81 \mathrm{~m} / \mathrm{s}^{2}\) and give your answer to three significant figures. Let \(M_{E}\) be the mass of the Earth, and \(m\) the mass of an object. The weight of the object on the planet's surface is equal to \(m g .\) It is also equal to the gravitational force \(G\left(M_{E} m\right) R_{E}^{2}\), where \(R_{E}\) is the Earth's radius. Hence, $$ m g=G \frac{M_{E} m}{R_{E}^{2}} $$ from which \(\quad M_{E}=\frac{g R_{E}^{2}}{G}=\frac{\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(6.37 \times 10^{6} \mathrm{~m}\right)^{2}}{6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}}=5.97 \times 10^{24} \mathrm{~kg}\)

Short Answer

Expert verified
Earth's mass is approximately \(5.97 \times 10^{24} \ kg\).

Step by step solution

01

Understand the Given Relation

We are given that an object's weight on the Earth's surface is described by two expressions: 1. Newton's law of gravity: \(mg = G \frac{M_E m}{R_E^2}\).2. The gravitational force equation describes the relationship between the gravitational constant \(G\), the Earth's mass \(M_E\), and the Earth's radius \(R_E\).
02

Relate Weight and Gravitational Force

The equation that equates weight with gravitational force is: \[ mg = G \frac{M_E m}{R_E^2} \]By cancelling the mass \(m\) on both sides, we simplify to find Earth's mass: \[ M_E = \frac{g R_E^2}{G} \].
03

Insert Known Values

Substitute the known values for \(g = 9.81\ m/s^2\), \(R_E = 6.37 \times 10^6\ m\), and \(G = 6.67 \times 10^{-11}\ N\cdot m^2/kg^2\) into the equation:\[ M_E = \frac{(9.81)(6.37 \times 10^6)^2}{6.67 \times 10^{-11}} \].
04

Calculate Earth's Mass

Perform the calculation step by step:1. Compute \(R_E^2 = (6.37 \times 10^6)^2 = 4.06169 \times 10^{13} \ m^2\).2. Calculate \(g R_E^2 = 9.81 \times 4.06169 \times 10^{13} = 3.9844 \times 10^{14} \ N\cdot m^2\).3. Finally, \( M_E = \frac{3.9844 \times 10^{14}}{6.67 \times 10^{-11}} = 5.97 \times 10^{24}\ kg \).
05

Conclude with the Result

After performing the calculations, we find the mass of the Earth to be approximately \(5.97 \times 10^{24} \ kg\) to three significant figures.

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Key Concepts

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

Newton's Law of Gravity
Newton's law of gravity is a fundamental principle that explains how two objects are attracted to each other by a force called gravity. This law states that every point mass attracts every other point mass in the universe with a force that is directly proportional to the product of their masses. It is also inversely proportional to the square of the distance between their centers.

In the formula, it is expressed as:
  • F = G \(\frac{m_1 m_2}{r^2}\)
Here, \(F\) is the gravitational force between the two masses, \(m_1\) and \(m_2\) are the masses, \(r\) is the distance between the centers of the two masses, and \(G\) is the gravitational constant. It is fascinating because it not only helps us understand our planet's interaction with objects but also deals with the behavior of celestial bodies, explaining orbital motions and much more.
Gravitational Constant
The gravitational constant, denoted by the symbol \(G\), is a key quantity in Newton's law of gravity. It describes the inherent strength of gravitational interaction in the universe. Its value is approximately \(6.674 \times 10^{-11}\) \(\mathrm{N}\cdot \mathrm{m}^2/\mathrm{kg}^2\), signifying that gravity is a relatively weak force—especially noticeable in space studies and measurements.

This constant is used to calculate the gravitational attraction between any two masses. To put this into perspective, it is the factor that transmits the influence of gravity. Without the value of \(G\), we would not be able to compute the gravitational forces and hence could not calculate important measurements like the Earth's mass. Its usage is ubiquitous in physics, appearing in formulas that explain phenomenons from the arcs of celestial orbits to even events like black holes.
Radius of the Earth
The radius of the Earth is a critical input for calculations involving the planet's physical properties, such as volume, surface area, and gravitational interactions. In this context, the radius is especially important to calculate the mass of Earth using gravitational principles.

The Earth is approximated as a sphere with an average radius of about 6,370 kilometers. Having this measurement helps determine how much the gravitational force influences objects on its surface and also provides a basis for comparison within celestial measurements. It is important to note that the Earth is not a perfect sphere; it is slightly flattened at the poles and bulging at the equator. However, for many calculations like estimating the mass of Earth, treating it as a sphere is sufficiently accurate.
Significant Figures
Significant figures are a way of expressing precision in numerical calculations. When you are dealing with scientific computations, it is crucial to represent a number to a specific amount of significant figures to convey the accuracy and reliability of the measurement.
  • Significant figures reflect the number of meaningful digits in a measured or calculated quantity.
  • In scientific notation, significant figures consist of all the non-zero digits and any zeroes that are between them or are final zeros in a decimal number.

For example, in the mass of the Earth calculation, the value is expressed to three significant figures as \(5.97 \times 10^{24}\) kg. This ensures that the reported number accurately represents the calculated value without overstating the precision, which is particularly important in scientific contexts where slight discrepancies can have meaningful impacts. Therefore, understanding and applying significant figures correctly provides clarity and consistency in the communication of scientific data.

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Most popular questions from this chapter

A 200-N wagon is to be pulled up a \(30^{\circ}\) incline at constant speed. How large a force parallel to the incline is needed if friction effects are negligible? The situation is shown in Fig. \(3-16(a)\). Because the wagon moves at a constant speed along a straight line, its velocity vector is constant. Therefore, the wagon is in translational equilibrium, and the first condition for equilibrium applies to it. We isolate the wagon as the object. Three non-negligible forces act on it: (1) the pull of gravity \(F_{W}\) (its weight), directed straight down; (2) the applied force \(F\) exerted on the wagon parallel to the incline to pull it up the incline; (3) the push \(F_{N}\) of the incline that supports the wagon. These three forces are shown in the free-body diagram in Fig. \(3-16\). For situations involving inclines, it is convenient to take the \(x\) -axis parallel to the incline and the \(y\) -axis perpendicular to it. After taking components along these axes, we can write the first condition for equilibrium: $$ \begin{array}{lll} +\sum F_{x}=0 & \text { becomes } & F-0.50 F_{W}=0 \\ +\sum F_{y}=0 & \text { becomes } & F_{N}-0.866 F_{W}=0 \end{array} $$ Solving the first equation and recalling that \(F_{W}=200 \mathrm{~N}\), we find that \(F=0.50 F_{W}\). The required pulling force to two significant figures is \(0.10 \mathrm{kN}\).

Two blocks, of masses \(m_{1}\) and \(m_{2}\), moving in the \(x\) -direction are pushed by a force \(F\) as shown in Fig. 3-19. The coefficient of friction between each block and the table is \(0.40 .(a)\) What must be the value of \(F\) if the blocks are to have an acceleration of \(200 \mathrm{~cm} / \mathrm{s}^{2} ?\) How large a force does \(m_{1}\) then exert on \(m_{2}\) ? Use \(m_{1}=300 \mathrm{~g}\) and \(m_{2}=500 \mathrm{~g}\). Remember to work in SI units. The friction forces on the blocks are \(F_{f 1}=0.40 m_{1} g\) and \(F_{f 2}=0.40 m_{2} g\). Take the two blocks in combination as the object for discussion; the horizontal forces on the object from outside (i.e., the external forces on it) are \(F\), \(F_{f 1}\), and \(F_{f 2}\). Although the two blocks do push on each other, those pushes are internal forces; they are not part of the unbalanced external force on the two-mass object. For that object, $$ \pm \sum F_{x}=m a_{x} \text { becomes } F-F_{f 1}-F_{\mathrm{f} 2}=\left(m_{1}+m_{2}\right) a_{x} $$ (a) Solving for \(F\) and substituting known values $$ F=0.40 g\left(m_{1}+m_{2}\right)+\left(m_{1}+m_{2}\right) a_{x}=3.14 \mathrm{~N}+1.60 \mathrm{~N}=4.7 \mathrm{~N} $$ (b) Now consider block \(m_{2}\) alone. The forces acting on it in the \(x\) -direction are the push of block \(m_{1}\) on it (which we represent by \(F_{b}\) ) and the retarding friction force \(F_{12}=0.40 m_{2} g\). Then, for it, $$ \pm \sum F_{x}=m a_{x} \quad \text { becomes } \quad F_{b}-F_{f 2}=m_{2} a_{x} $$ We know that \(a_{x}=2.0 \mathrm{~m} / \mathrm{s}^{2}\) and so $$ F_{b}=F_{\mathrm{f} 2}+m_{2} a_{x}=1.96 \mathrm{~N}+1.00 \mathrm{~N}=2.96 \mathrm{~N}=3.0 \mathrm{~N} $$

Having hauled it to the top of a tilted driveway, a child is holding a wagon from rolling back down. The driveway is inclined at \(20^{\circ}\) to the horizontal. If the wagon weighs \(150 \mathrm{~N}\), with what force must the child pull on the handle if the handle is parallel to and pointing up the incline?

The Earth's radius is about \(6370 \mathrm{~km}\). An object that has a mass of \(20 \mathrm{~kg}\) is taken to a height of \(160 \mathrm{~km}\) above the Earth's surface. ( \(a\) ) What is the object's mass at this height? ( \(b\) ) How much does the object weigh (i.e. how large a gravitational force does it experience) at this height?

Imagine a planet having a mass twice that of Earth and a radius equal to \(1.414\) times that of Earth. Determine the acceleration due to gravity at its surface.

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