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At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.980 \(\mathrm{m} / \mathrm{s}^{2}\) if the acceleration due to gravity at the surface has magnitude 9.80 \(\mathrm{m} / \mathrm{s}^{2}\) ?

Short Answer

Expert verified
The height is approximately 13,781 kilometers above the Earth's surface.

Step by step solution

01

Identify the Given Values

We are given two values of acceleration due to gravity: at the Earth's surface it is \( 9.80 \, \mathrm{m/s^2} \), and at a certain height it is \( 0.980 \, \mathrm{m/s^2} \). We want to find the height above the Earth's surface where gravity is \( 0.980 \, \mathrm{m/s^2} \).
02

Understand the Formula for Gravitational Acceleration

The acceleration due to gravity at a distance \( r \) from the center of the Earth is given by \( g = \frac{GM}{r^2} \), where \( G \) is the gravitational constant and \( M \) is the mass of the Earth. At the Earth's surface, \( r \) is the Earth's radius \( R_e \).
03

Set Up the Ratio of Gravitational Accelerations

According to the inverse square law, \( \frac{g_1}{g_2} = \left(\frac{r_2}{r_1}\right)^2 \). Let \( g_1 = 9.80 \, \mathrm{m/s^2} \) and \( g_2 = 0.980 \, \mathrm{m/s^2} \), so \( \frac{9.80}{0.980} = \left(\frac{r_2}{R_e}\right)^2 \), where \( R_e \) is the Earth's radius.
04

Solve the Ratio Equation

From \( \frac{9.80}{0.980} = \left(\frac{r_2}{R_e}\right)^2 \), we get \( 10 = \left(\frac{r_2}{R_e}\right)^2 \). Taking the square root, \( \sqrt{10} = \frac{r_2}{R_e} \), so \( r_2 = R_e \sqrt{10} \).
05

Calculate the Height Above the Earth's Surface

The height \( h \) above the Earth's surface is \( r_2 - R_e \). Using \( r_2 = R_e \sqrt{10} \), we have \( h = R_e (\sqrt{10} - 1) \).
06

Substitute the Earth's Radius

Substitute the approximate value for Earth's radius \( R_e \approx 6,371,000 \, \mathrm{m} \) to find \( h = 6,371,000 \times (\sqrt{10} - 1) \). Calculate \( \sqrt{10} \approx 3.162 \).
07

Compute the Final Height

Perform the calculation: \( h = 6,371,000 \times (3.162 - 1) \approx 6,371,000 \times 2.162 \approx 13,780,902 \, \mathrm{m} \). This converts to approximately 13,781 kilometers.

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Key Concepts

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

Inverse Square Law
The inverse square law is a fundamental principle in physics that describes how a force, in this case, gravitational force, is affected by distance. According to this law, the intensity of gravitational acceleration decreases proportionally to the square of the distance from the source of the gravitational field, which is the center of the Earth in this scenario.
This means that if you are twice as far from the center of the Earth, the gravitational acceleration is not just half, but a quarter of what it is at the first distance. This relationship is crucial in understanding how gravitational force diminishes with distance, and it's why you need to calculate ratios as shown in the exercise to find changes in gravity based on distance.
Therefore, using the inverse square law, you set up a ratio between gravitational accelerations at different distances to solve for the unknown distance above Earth's surface.
Earth's Radius
Earth's radius, often represented as \( R_e \), is the distance from the Earth's center to its surface. The average value used in calculations is roughly 6,371 kilometers. This value is important because it factors into many calculations involving Earth's gravitational field.
For example, in the exercise, when calculating how the gravitational force changes at different altitudes, knowing the Earth's radius helps you determine how far from the Earth's surface you are in terms of the total distance from the Earth's center. When solving the problem, you converted the change in gravitational acceleration to a change in distance using the ratio of Earth's radius and the unknown distance \( r_2 \).
Earth’s radius is a standard reference for many geological and astronomical calculations involving the Earth and forms the baseline for approximating distances related to the Earth's gravitational influence.
Gravitational Constant
The gravitational constant, often denoted as \( G \), is a crucial part of the formula for determining gravity's force. Its value is approximately \( 6.674 \times 10^{-11} \, \mathrm{Nm^2/kg^2} \). This constant reflects the strength of gravity and is used in the universal law of gravitation to calculate gravitational forces between two masses anywhere in the universe.
Within the context of Earth's gravity, the gravitational constant helps determine the gravitational pull that the Earth exerts on objects at or near its surface. Although \( G \) itself is not directly used in every step of the calculation in basic exercises, it underlies the concept that gravity decreases as we move away from Earth’s center, as shown by the inverse square law.
The gravitational constant is what allows us to link mass and distance into a meaningful force, giving mathematic expression to the gravitational pull between masses at various distances.
Mass of the Earth
The mass of the Earth, represented as \( M \), is a critical factor when calculating gravitational acceleration. It is approximately \( 5.972 \times 10^{24} \, \mathrm{kg} \), providing the bulk of the gravitational pull experienced at the Earth's surface.
In the formula \( g = \frac{GM}{r^2} \), where \( G \) is the gravitational constant and \( r \) is the distance from Earth's center, Earth's mass \( M \) combines with \( G \) to show how gravity's strength decreases with distance. This is key in exercises like the one above, as it helps us understand why gravity weakens as altitude increases.
The large mass of the Earth contributes to its ability to hold the atmosphere and exert a significant gravitational pull, maintaining equilibrium for objects that exist within its gravitational field. This concept helps in understanding how gravitational force varies with distance, work that is crucial when addressing problems involving changes in gravity over different altitudinal distances.

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Most popular questions from this chapter

The acceleration due to gravity at the north pole of Neptune is approximately 10.7 \(\mathrm{m} / \mathrm{s}^{2} .\) Neptune has mass \(1.0 \times 10^{26} \mathrm{kg}\) and radius \(2.5 \times 10^{4} \mathrm{km}\) and rotates once around its axis in about 16 \(\mathrm{h}\) . (a) What is the gravitational force on a 5.0 -kg object at the north pole of Neptune? (b) What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.)

Rhea, one of Saturn's moons, has a radius of 765 \(\mathrm{km}\) and an acceleration due to gravity of 0.278 \(\mathrm{m} / \mathrm{s}^{2}\) at its surface. Calculate its mass and average density.

Four identical masses of 800 kg each are placed at the corners of a square whose side length is 10.0 \(\mathrm{cm} .\) What is the net gravitational force (magnitude and direction) on one of the masses, due to the other three?

A uniform, spherical, \(1000.0-\mathrm{kg}\) shell has a radius of 5.00 \(\mathrm{m} .\) (a) Find the gravitational force this shell exerts on a \(2.00-\mathrm{kg}\) point mass placed at the following distances from the center of the shell: (i) \(5.01 \mathrm{m},\) (ii) \(4.99 \mathrm{m},\) (iii) 2.72 \(\mathrm{m}\) . (b) Sketch a qualitative graph of the magnitude of the gravitational force this sphere exerts on a point mass \(m\) as a function of the distance \(r\) of \(m\) from the center of the sphere. Include the region from \(r=0\) to \(r \rightarrow \infty.\)

Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 \(\mathrm{km}\) from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 \(\mathrm{km} .\) The astronaut is positioned inside the spaceship such that one of her 0.030 -kg ears is 6.0 \(\mathrm{cm}\) farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer. (a) What is the tension between her ears? Would the astronaut find it difficult to keep from being torn apart by the gravitational forces? (Since her whole body orbits with the same angular velocity, one ear is moving too slowly for the radius of its orbit and the other is moving too fast. Hence her head must exert forces on her ears to keep them in their orbits. \((\) b) Is the center of gravity of her head at the same point as the center of mass? Explain.

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