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Two cannonballs that each weigh \(4.00 \mathrm{kN}\) on Earth are floating in space far from any other objects. Determine the mutually attractive gravitational force acting on them when they are separated, center-to-center, by \(10.0 \mathrm{~m}\).

Short Answer

Expert verified
The gravitational force is approximately \(1.11 \times 10^{-7} \mathrm{~N}\).

Step by step solution

01

Understand the Problem

We are tasked with finding the gravitational force between two cannonballs in space, weighing 4.00 kN each, and separated by a distance of 10.0 m. We will use the formula for gravitational force: \[ F = \frac{{G imes m_1 \times m_2}}{{r^2}} \] where \( F \) is the gravitational force, \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{kg}^2 \), \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the separation distance.
02

Convert Weight to Mass

Weight is given in kilonewtons (kN), a force measure, and needs to be converted to mass using Earth's gravitational acceleration \( g = 9.81 \mathrm{~m/s^2} \). Convert weight to mass: \[ m = \frac{W}{g} = \frac{4.00 \times 10^3}{9.81} \approx 407.75 \text{ kg} \] Each cannonball has a mass of approximately 407.75 kg.
03

Substitute Values into Gravitational Force Formula

Now that we have the masses and the distance, substitute into the formula: \[ F = \frac{{(6.674 \times 10^{-11}) \times 407.75 \times 407.75}}{{(10.0)^2}} \]
04

Calculate the Gravitational Force

Perform the calculations: First, calculate the numerator: \[ (6.674 \times 10^{-11}) \times 407.75 \times 407.75 = 1.110 \times 10^{-5} \] Then divide by the square of the distance: \[ \frac{1.110 \times 10^{-5}}{100} = 1.110 \times 10^{-7} \mathrm{~N} \]
05

Round and Report the Result

Round the result to a sensible number of significant figures. The gravitational force at this distance is: \[ F \approx 1.11 \times 10^{-7} \mathrm{~N} \] This is the attractive force between the two cannonballs.

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Key Concepts

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

Newton's law of universal gravitation
Newton's law of universal gravitation is a cornerstone principle in physics, elucidating how every point mass attracts every other point mass in the universe with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This can be mathematically expressed as:\[F = \frac{G \times m_1 \times m_2}{r^2}\]where:
  • \( F \) is the gravitational force between two objects,
  • \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{kg}^2 \),
  • \( m_1 \) and \( m_2 \) are the masses of the two objects in kilograms,
  • \( r \) is the distance between the centers of the two masses in meters.
This law beautifully shows the universality of gravitational force, a principle that applies equally here on Earth and to celestial objects far away in space. Thus, even when objects are floating in space far from other influences, they still exert gravitational forces on each other due to their masses.
mass-weight conversion
In physics, it is crucial to understand the distinction between mass and weight, particularly when converting from one to the other. Weight is a force, measured in newtons (N), that results from the gravitational pull on an object's mass. On Earth, weight can be calculated as \( W = m \times g \), where \( g \approx 9.81 \mathrm{~m/s^2} \) is the gravitational acceleration. To convert from weight to mass, which is a measure of the amount of matter in an object (measured in kilograms), you rearrange the formula:\[m = \frac{W}{g}\]In the original exercise, each cannonball had a weight of \(4.00\) kN, equivalent to \(4.00 \times 10^3\) N. By applying the conversion formula, their mass is found to be approximately \(407.75\) kg per cannonball. This conversion is crucial for calculating gravitational force, which depends directly on the masses of the interacting bodies. Converting weight to mass enables accurate application of Newton's gravitational formula outside the preferred context of Earth's gravitational influence.
significant figures
Significant figures are an important concept in ensuring the precision and accuracy of numerical results in scientific calculations. They represent the digits in a number that carry meaningful information about its precision. For example, in the original solution's calculation of gravitational force, the final result was rounded to three significant figures:\[F \approx 1.11 \times 10^{-7} \mathrm{~N}\]When rounding using significant figures, you usually match the lowest number of significant figures from the data you have. Here, the weights were given as \(4.00\) kN, suggesting three significant figures. Therefore, presenting the gravitational force with three significant figures maintains consistency and reflects the accuracy level of the input values. Proper use of significant figures helps communicate the certainty of measurements and ensures consistency throughout calculations, aiding in clearer interpretation and validation of results.

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Most popular questions from this chapter

A 300 -g mass hangs at the end of a string. A second string hangs from the bottom of that mass and supports a \(900-\mathrm{g}\) mass. ( \(a\) ) Find the tension in each string when the masses are accelerating upward at \(0.700 \mathrm{~m} / \mathrm{s}^{2}\). Don't forget gravity. (b) Find the tension in each string when the acceleration is \(0.700 \mathrm{~m} / \mathrm{s}^{2}\) downward.

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.

Two forces act on a point object as follows: \(100 \mathrm{~N}\) at \(170.0^{\circ}\) and \(100 \mathrm{~N}\) at \(50.0^{\circ} .\) Find their resultant.

A \(5.0\) -kg object is to be given an upward acceleration of \(0.30 \mathrm{~m} / \mathrm{s}^{2}\) by a rope pulling straight upward on it. What must be the tension in the rope? The free-body diagram for the object is shown in Fig. \(3-8(b)\). The tension in the rope is \(F_{T}\), and the weight of the object is \(F_{W}=m g=(5.0 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=49.1 \mathrm{~N}\). Using \(\sum F_{y}=m a_{y}\) with \(u p\) taken as positive, $$ +\uparrow \sum F_{y}=F_{T}-m g=m a_{y} \quad \text { or } \quad F_{T}-49.1 \mathrm{~N}=(5.0 \mathrm{~kg})\left(0.30 \mathrm{~m} / \mathrm{s}^{2}\right) $$ from which \(F_{T}=50.6 \mathrm{~N}=51 \mathrm{~N}\). As a check, we notice that \(F_{T}\) is larger than \(F_{W}\), as it must be if the object is to accelerate upward.

A force of \(100 \mathrm{~N}\) makes an angle of \(\theta\) with the \(x\) -axis and has a scalar \(y\) -component of \(30 \mathrm{~N}\). Find both the scalar \(x\) -component of the force and the angle \(\theta\). (Remember that the number \(100 \mathrm{~N}\) has three significant figures whereas \(30 \mathrm{~N}\) has only two.)Begin your analysis by drawing a diagram. Here the data are sketched roughly in Fig. \(3-4\). We wish to find \(F_{x}\) and \(\theta\). Since $$ \sin \theta=\frac{30 \mathrm{~N}}{100 \mathrm{~N}}=0.30 $$ \(\theta=17.46^{\circ}\), and thus, to two significant figures, \(\theta=17^{\circ} .\) Then, using the \(\cos \theta\) $$ F_{x}=(100 \mathrm{~N}) \cos 17.46^{\circ}=95 \mathrm{~N} $$

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