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Chapter 3: Problem 81

An object that weighs \(2700 \mathrm{~N}\) on the surface of the Earth is raised to a height (i.e., altitude) of two Earth radii above the surface. What will it weigh up there?

Short Answer

Expert verified
The object weighs 300 N at the altitude of two Earth radii.

Step by step solution

01

Understand the Gravitational Force Formula

The weight of an object is the gravitational force exerted on the object by the Earth. The formula for gravitational force is \( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \), where \( F \) is the force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two objects (in this case, the Earth and the object), and \( r \) is the distance between the centers of the two masses.
02

Calculate the Initial Weight

The initial weight given is \(2700 \mathrm{~N}\), which is the force when the object is on the Earth's surface. The distance here is equal to the Earth's radius \( R \).
03

Determine the New Distance

When the object is raised to an altitude of two Earth radii above the surface, the new distance from the center of the Earth becomes \(3R\) (one Earth radius \(R\) plus the two additional Earth radii).
04

Apply the Gravitational Force Formula at New Distance

Using the formula \( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \), the new weight \( F' \) at the distance of \(3R\) is calculated as follows: \[ F' = \frac{G \cdot m_1 \cdot m_2}{(3R)^2} = \frac{F}{3^2} = \frac{2700}{9} \].
05

Calculate the New Weight

Calculate the new weight: \( F' = \frac{2700}{9} = 300 \mathrm{~N} \).

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Key Concepts

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

Weight Calculation
Weight is a force that results from the gravitational pull between two masses. On Earth, it is the force that pulls an object towards the center of the planet. The formula used to calculate weight is derived from Newton's law of universal gravitation. This force is directly proportional to the mass of the object and inversely proportional to the square of the distance between the object and the Earth's center.

When calculating weight using the formula, we use the gravitational force equation: \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]- **F** is the gravitational force or weight.- **G** is the gravitational constant. It's a fixed value used to calculate gravitational forces.- **m_1** and **m_2** represent the mass of the Earth and the mass of the object, respectively.- **r** is the distance from the center of the Earth to the object. By plugging values into this formula, we can determine the weight of the object at various distances from the Earth's center.
Newton's Law of Universal Gravitation
Newton's law of universal gravitation is a fundamental principle that describes how objects attract each other with a force perfectly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

- This law can be expressed as: \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]Where this equation is at the core of calculations involving gravitational forces.

It tells us that:- All objects with mass exert a gravitational pull on others.- Gravitational force decreases rapidly as the distance between objects grows, exemplified by the square of the distance (**r虏**) in the denominator.Thus, understanding this relationship is key to comprehending weight differences at varying altitudes.
Distance and Weight Relationship
The relationship between distance and weight is critical in understanding gravitational forces. As an object moves away from the Earth, its weight decreases due to the increased distance from the Earth's center.

This decrease in weight can be calculated by recalling that gravitational force is inversely proportional to the square of the distance. So, when the distance triples, as in our exercise, the force (or weight) is reduced by \(3^2 = 9\).
When the object is moved from the Earth's surface to a height of two Earth radii away:- The total distance from the Earth's center is now three times the Earth's radius (3R).- This increase in distance results in a weight that is a fraction of the original weight, calculated as the initial weight divided by 9.
This illustrates how even small increases in height can dramatically reduce weight due to the gravitational force's dependence on distance.

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Most popular questions from this chapter

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.

The mythical planet Mongo has twice the mass and twice the radius of Earth. Compute the acceleration due to gravity at its surface. We know from Problem \(3.40\) that in general $$ g_{R}=\frac{G M}{R^{2}} $$ Then for the Earth at its surface $$ g_{E}=\frac{G M_{E}}{R_{E}^{2}}=g=9.81 \mathrm{~m} / \mathrm{s}^{2} $$ where \(R_{E}\) is the Earth's radius and \(M_{E}\) is its mass. For Mongo $$ g_{M}=\frac{G M_{M}}{R_{M}^{2}}=\frac{G\left(2 M_{E}\right)}{\left(2 R_{E}\right)^{2}} $$ and or $$ \begin{array}{c} g_{M}=\frac{1}{2} \frac{G M_{E}}{R_{E}^{2}}=\frac{1}{2}\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right) \\ g_{M}=4.91 \mathrm{~m} / \mathrm{s}^{2} \end{array} $$

A cord passing over a light frictionless pulley has a \(7.0\) -kg mass hanging from one end and a \(9.0-\mathrm{kg}\) mass hanging from the other, as seen in Fig. 3-20. (This arrangement is called Atwood's machine.) Find the acceleration of the masses and the tension in the cord. Because the pulley is easily turned, the tension in the cord will be the same on each side. The forces acting on each of the two masses are drawn in Fig. 3-20. Recall that the weight of an object is \(m g\). It is convenient in situations involving objects connected by cords to take the overall direction of motion of the system as the positive direction. That's often indicated by the direction of motion of the pulley when the system is let free to move. In the present case, the pulley would turn clockwise, and so we take \(u p\) positive for the \(7.0\) -kg mass, and down positive for the \(9.0\) -kg mass. (If we do this, the acceleration will be positive for each mass. Because the cord doesn't stretch, the accelerations are numerically equal.) Writing \(\sum F_{y}=m a_{y}\) for each mass in turn, $$ +\uparrow \sum F_{y A}=F_{T}-(7.0)(9.81) \mathrm{N}=(7.0 \mathrm{~kg})(a) \text { and }+\downarrow \sum F_{y B}=(9.0)(9.81) \mathrm{N}-F_{T}=(9.0 \mathrm{~kg})(a) $$ Add these two equations and the unknown \(F_{T}\) drops out, giving $$ (9.0-7.0)(9.81) \mathrm{N}=(16 \mathrm{~kg})(a) $$ for which \(a=1.23 \mathrm{~m} / \mathrm{s}^{2}\) or \(1.2 \mathrm{~m} / \mathrm{s}^{2}\). Now substitute \(1.23 \mathrm{~m} / \mathrm{s}^{2}\) for \(a\) in either equation and obtain \(F_{T}=77 \mathrm{~N}\).

A constant force acts on a \(5.0 \mathrm{~kg}\) object and reduces its velocity from \(7.0 \mathrm{~m} / \mathrm{s}\) to \(3.0 \mathrm{~m} / \mathrm{s}\) in a time of \(3.0 \mathrm{~s}\). Determine the force. We must first find the acceleration of the object, which is constant because the force is constant. Taking the direction of motion as positive, from Chapter 2 $$ a=\frac{v_{f}-v_{i}}{t}=\frac{-4.0 \mathrm{~m} / \mathrm{s}}{3.0 \mathrm{~s}}=-1.33 \mathrm{~m} / \mathrm{s}^{2} $$ Use \(F=m a\) with \(m=5.0 \mathrm{~kg}\) : $$ F=(5.0 \mathrm{~kg})\left(-1.33 \mathrm{~m} / \mathrm{s}^{2}\right)=-6.7 \mathrm{~N} $$ The minus sign indicates that the force is a retarding force, directed opposite to the motion.

A horizontal cable pulls a \(200-\mathrm{kg}\) cart along a horizontal track. The tension in the cable is \(500 \mathrm{~N}\). Starting from rest, (a) How long will it take the cart to reach a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) ? (b) How far will it have gone?
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