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Chapter 3: Problem 83

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?

Short Answer

Expert verified
(a) 20 kg, (b) 187.0 N

Step by step solution

01

Understanding Object's Mass at Height

The mass of an object does not change based on its location. Whether the object is on the surface of the Earth or at a height, its mass remains constant. Therefore, the mass of the object at a height of \(160 \text{ km}\) above the Earth's surface is still \(20 \text{ kg}\).
02

Calculating Total Distance from Earth's Center

The Earth鈥檚 radius is \(6370 \text{ km}\) and the object is \(160 \text{ km}\) above the Earth's surface. We need to calculate the total distance from the Earth's center to the object. This distance is the sum of the Earth's radius and the height above the Earth's surface: \[R = 6370 + 160 = 6530 \text{ km}\].
03

Remembering the Gravitational Force Formula

The gravitational force \(F\) exerted on an object is given by the formula:\[F = \frac{G \cdot m_1 \cdot m_2}{r^2}\],where \(G\) is the gravitational constant \((6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2)\), \(m_1\) and \(m_2\) are the masses of the two objects (here the Earth and the object), and \(r\) is the distance between their centers of mass.
04

Inputting Earth's Mass and Object Variables

We know the mass of the Earth \(m_1 = 5.972 \times 10^{24} \text{ kg}\), and the mass of the object is \(m_2 = 20 \text{ kg}\). The total distance we calculated is \(r = 6530 \text{ km} = 6.53 \times 10^{6} \text{ m}\) (converted to meters from kilometers).
05

Calculating Gravitational Force

Substitute the values into the formula for gravitational force: \[F = \frac{(6.674 \times 10^{-11}) \cdot (5.972 \times 10^{24}) \cdot (20)}{(6.53 \times 10^{6})^2}\].Carry out the calculations to find the gravitational force. \[F \approx 187.0 \text{ N}\].

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Key Concepts

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

Mass and Weight
In physics, it's crucial to understand the difference between mass and weight. Mass is a measure of the amount of matter in an object, and it does not change regardless of its location. It's measured in kilograms (kg). For example, in our exercise, an object has a mass of \(20\, \text{kg}\) on Earth, and this remains the same even when it is taken to a height of \(160\, \text{km}\) above the Earth's surface. This is because mass is an inherent property of the object.

Weight, on the other hand, depends on gravitational force; it is how much gravity pulls on the object. The formula to find weight is the gravitational force, given by:
\[ F = m imes g \],
where \(F\) is the force (weight), \(m\) is the mass, and \(g\) is the acceleration due to gravity. Weight changes with location because gravity varies across the universe. Thus, the weight of the same object on Earth and on the Moon differs.

  • Mass remains constant irrespective of location.
  • Weight varies because it's influenced by the gravitational pull.
Gravitational Constant
The gravitational constant, denoted by \(G\), is a vital component in the equation for calculating gravitational force. It is a universal constant, which means it has the same value everywhere in the universe. Its value is approximately \(6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2\). This constant helps us understand the strength of gravitational attraction between two masses.

When calculating the gravitational force between the object and the Earth, \(G\) is used along with the mass of Earth and the object, and the distance between their centers of mass. The formula to find the gravitational force between two masses \(m_1\) and \(m_2\) is:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \],
where \(r\) is the distance between the centers of the two masses. This equation shows how the gravitational force depends on both the masses involved and the distance separating them.

  • The gravitational constant \(G\) is a fixed value, crucial for calculations involving gravity.
  • It features in Newton's law of universal gravitation, explaining how masses attract each other.
Distance from Earth's Center
When calculating gravitational force at a height above the Earth's surface, the total distance (\(r\)) from the Earth's center to the object must be determined. This total distance is composed of the Earth's radius plus the object's height above the Earth's surface.

In our exercise, the Earth's radius is \(6370 \text{ km}\). The object is positioned \(160 \text{ km}\) above Earth's surface. Thus, the distance from Earth's center to the object becomes:
\[ R = 6370 + 160 = 6530 \text{ km} \].
Ensure to convert this distance into meters for calculations, as the metric system is typically used in physics. Therefore, \(6530 \text{ km} = 6.53 \times 10^{6} \text{ m}\).

This distance is essential for calculating the gravitational force because it's inversely proportional to the square of the distance. This inverse square law means that as the distance increases, the gravitational force decreases, demonstrating how gravitational force weakens with increasing distance.

  • Total distance calculation: sum of Earth's radius and height above the surface.
  • Metric conversion is necessary for precision in equations.

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Most popular questions from this chapter

A car whose weight is \(F_{W}\) is on a ramp which makes an angle \(\theta\) to the horizontal. How large a perpendicular force must the ramp withstand if it is not to break under the car's weight? As rendered in Fig. \(3-6\), the car's weight is a force \(\overrightarrow{\mathbf{F}}_{W}\) that pulls straight down on the car. We take components of \(\overrightarrow{\mathbf{F}}\) along the incline and perpendicular to it. The ramp must balance the force component \(F_{W} \cos \theta\) if the car is not to crash through the ramp. In other words, the force exerted on the car by the ramp, upwardly perpendicular to the ramp, is \(F_{N}\) and \(F_{N}=F_{W} \cos \theta\).

An astronaut weighs \(480 \mathrm{~N}\) on Earth. She visits the planet Krypton, which has a mass and diameter each ten times that of Earth. Determine her weight at a distance of two Kryptonian radii above that fictional planet.

An \(8000-\mathrm{kg}\) engine pulls a 40000 -kg train along a level track and gives it an acceleration \(a_{1}=1.20 \mathrm{~m} / \mathrm{s}^{2}\). What acceleration \(\left(a_{2}\right)\) would the engine give to a \(16000-\mathrm{kg}\) train? For a given engine force, the acceleration is inversely proportional to the total mass. Thus, $$ a_{2}=\frac{m_{1}}{m_{2}} a_{1}=\frac{8000 \mathrm{~kg}+40000 \mathrm{~kg}}{8000 \mathrm{~kg}+16000 \mathrm{~kg}}\left(1.20 \mathrm{~m} / \mathrm{s}^{2}\right)=2.40 \mathrm{~m} / \mathrm{s}^{2} $$

Consider an essentially spherical homogeneous celestial body of mass \(M\). The acceleration due to gravity in its vicinity beyond its surface at a distance \(R\) from its center is \(g_{R}\). Show that $$ g_{R}=\frac{G M}{R^{2}} $$ Notice that the acceleration drops off as \(1 / R^{2}\). Imagine an object of mass \(m\) at a distance \(R\) from the center of our celestial body. Its weight is \(F_{W}=m g_{R}\), but that's also the gravitation force on it due to the mass \(M\), that is, \(F_{W}=F_{G}\). Hence, $$ m g_{R}=G \frac{m M}{R^{2}} $$

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.
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