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The lowest point in the Western Hemisphere is found at Badwater in Death Valley, California where the elevation is \(86 \mathrm{~m}\) below sea level. Using the shell theorem, determine the value of \(g\) at that location.

Short Answer

Expert verified
The value of g at Badwater in Death Valley, approximately 86 m below the sea level is \( g\approx 9.819 \) m/s².

Step by step solution

01

Understand Shell Theorem

Shell theorem states that the gravitational force acting on a body situated inside a homogenous shell is zero. According to this theorem, any mass situated at a depth below the Earth's surface does not contribute to the net gravitational force experienced by bodies on or above the surface. Thus, when calculating the value of \(g\) in Death Valley, which is 86 m below sea level, we only consider the mass of the Earth above this point.
02

Create the Equation for Gravitational Force

The gravity acceleration formula is \( g = G \frac{M}{r^2}\), where \( G \) is the gravitational constant (\(6.67430(15) x 10^{-11} m^3 kg^{-1} s^{-2}\)), \(\ M \) is the Earth's mass (\(5.972 × 10^{24}\) kg) and \(r\) is the distance from the center of the earth to the point under consideration.
03

Adjust Sphere's Radius

Since Badwater is 86 m below sea level, we have to subtract this value from Earth's radius (\(6.371 × 10^{6}\) m). The adjusted radius becomes \(6.371 × 10^{6} - 86 = 6.370914 × 10^{6}\) m.
04

Substitute The Values in The Formula

Replace \( M \), \( G \), and \( r \) in the formula from Step 2. This yields \[ g = 6.67430(15) x 10^{-11} m^3 kg^{-1} s^{-2} \frac{5.972 × 10^{24}}{ (6.370914 × 10^{6})^2} \] Teaser out the math and this yields \( g\approx 9.819 \) m/s²
05

Interpret The Result

The calculated value of \( g \approx 9.819 \) m/s² implies that the gravitational acceleration at Badwater in Death Valley, which is 86 m below the sea level, is ever so slightly more than the average value at sea level (which is about 9.8 m/s²).

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Key Concepts

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

Shell Theorem
The Shell Theorem is a fundamental concept in physics proposed by Isaac Newton. It essentially helps us understand how gravity works within the confines of a spherical shell of mass. According to the Shell Theorem:
  • A spherically symmetric body affects external objects as though all of its mass were concentrated at a point at its center.
  • Inside a hollow spherical shell, the net gravitational force on a particle is zero.
This theorem is particularly useful when dealing with objects below the Earth’s surface, such as in the case of Death Valley. By applying the Shell Theorem, we consider only the mass of the Earth that is above the location in question. This simplifies the calculations for gravitational acceleration at depths below the surface, like the 86 meters below sea level at Badwater.
Gravity Formula
The Gravity Formula provides a way to calculate the gravitational force exerted by a mass. It is given by the equation: \[ g = G \frac{M}{r^2} \]where:
  • \( G \) is the gravitational constant, a fixed value in physics.
  • \( M \) is the mass of the object exerting the force, such as the Earth, in kilograms.
  • \( r \) is the distance from the center of the mass to the point where the force is being measured, in meters.
This formula equips us with the method to determine the acceleration due to gravity at any distance from the center of a gravitating body. By substituting the known mass of the Earth and its radius, we can calculate the gravitational acceleration experienced at various elevations and depths.
Earth's Radius
Earth's Radius is a critical factor when working with gravitational calculations. The average radius of Earth is approximately 6.371 million meters (aka 6.371 × 10^6 m).
When an object is located at a position below sea level, as in our original exercise, this radius needs to be adjusted. If Death Valley is 86 meters below sea level, then this value must be subtracted from the Earth's average radius to get the correct distance from the center to the point of interest. This provides the adjusted radius necessary for precise calculations in the gravity formula. Understanding the Earth's radius and making the necessary adjustments ensures that our gravitational acceleration computations are accurate.
Gravitational Constant
The Gravitational Constant, denoted as \( G \), is a pivotal element in gravitational calculations. It is a fixed number in physics representing the strength of gravity and appears in Newton's law of universal gravitation. The value of \( G \) is approximately:
6.67430(15) × 10^{-11} m^3 kg^{-1} s^{-2}.
This constant allows us to quantify the force of attraction between two masses. By incorporating \( G \) into the gravity formula, we can determine the specific gravitational force at any given location, integrating the mass of the Earth and the distance from its center. It is essential for achieving accurate results in calculating gravitational acceleration. Without understanding \( G \), predicting how objects are drawn towards each other within Earth’s gravitational field would be incomplete.

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Most popular questions from this chapter

Spherical storage tanks are used to hold natural gas. A storage facility contains a row of five such tanks, each with a volume of \(5000 \mathrm{~m}^{3}\). Determine the net gravitational force on the sphere at the end of the row of tanks due to the other tanks if they are filled with natural gas (natural gas' density is \(0.8 \mathrm{~kg} / \mathrm{m}^{3}\) ) and constructed from steel (steel's density is \(8000 \mathrm{~kg} / \mathrm{m}^{3}\) ) that is \(10 \mathrm{~cm}\) thick. The tanks are \(75 \mathrm{~m}\) apart (this distance is the closest one from the right edge of one tank to the left edge of the adjacent one).

On Earth, froghoppers can jump upward with a takeoff speed of \(2.8 \mathrm{~m} / \mathrm{s}\). Suppose you took some of the insects to an asteroid. If it is small enough, they can jump free of it and escape into space. (a) What is the diameter (in kilometers) of the largest spherical asteroid from which they could jump free? Assume a typical asteroid density of \(2.0 \mathrm{~g} / \mathrm{cm}^{3}\). (b) Suppose that one of the froghoppers jumped horizontally from a small hill on an asteroid. What would the diameter (in \(\mathrm{km}\) ) of the asteroid need to be so that the insect could go into a circular orbit just above the surface?

One consequence of the shell theorem is that if you are inside a uniform spherical object (which is a reasonable approximation for Earth or a star), the gravitational force you experience actually increases as you move radially away from the center until you reach the outer edge of the object. The gravitational force on a small object of mass \(m\) inside a spherically symmetric object can be written as $$ \vec{F}(r)=-\Gamma m \vec{r} $$ where \(\Gamma\) is a constant with units of \(\mathrm{s}^{-2}\). Formulate an expression for the gravitational potential energy function when an object of mass \(m\) is inside such a spherical object.

A huge, uniform sphere that has an inner radius of \(9500 \mathrm{~m}\) and an outer radius of \(10,000 \mathrm{~m}\) has a total mass of \(100,000 \mathrm{~kg}\) (Figure 10-18). (a) What is the net force on a 1-kg pointlike object that is \(50,000 \mathrm{~m}\) from the surface of the sphere? (b) What is the mass per unit volume of the sphere?

According to Newton's universal law of gravitation, \(\overrightarrow{\boldsymbol{F}}=-\frac{G m_{1} m_{2}}{r^{2}} \hat{r}\), if the distance \(r\) is doubled, the force is A. four times as much as the original value. B. twice as much as the original value. C. the same as the original value. D. one-half of the original value. E. one-fourth of the original value.

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