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

Short Answer

Expert verified
The distance \( h \) is approximately equal to 637 km above the Earth's surface.

Step by step solution

01

Use the formula for gravitational acceleration

The formula used to calculate the acceleration due to gravity at a certain distance from the center of the Earth is given by \( g = \frac{GM}{r^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the distance from the center of the Earth.
02

Relate the two accelerations

Since we know both the acceleration at the surface \( g_0 = 9.80 \text{ m/s}^2 \) and the acceleration at the distance \( g = 0.980 \text{ m/s}^2 \), we can write \( \frac{g}{g_0} = \left(\frac{R}{R + h}\right)^2 \), where \( R \) is the Earth's radius and \( h \) is the height above the surface.
03

Solve for \( h \)

Rearrange the equation to solve for \( \frac{R}{R + h} \) and then \( h \):1. \( \frac{0.980}{9.80} = \left(\frac{R}{R + h}\right)^2 \)2. \( \frac{1}{10} = \left(\frac{R}{R + h}\right)^2 \)3. \( \sqrt{\frac{1}{10}} = \frac{R}{R + h} \)4. \( \frac{1}{\sqrt{10}} = \frac{R}{R + h} \)5. Solve for \( h \), rearrange to find \( h = \frac{R}{\sqrt{10} - 1} - R \)

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Key Concepts

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

Gravitational Constant
The gravitational constant, represented by the symbol \( G \), is a key factor in the calculation of gravitational forces between two masses. The value of \( G \) is approximately \( 6.674 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2 \). This constant is universal, meaning it does not change regardless of the masses involved or where in the universe these masses are located.

Understanding \( G \) is essential because it helps us calculate the gravitational force exerted by two bodies, such as a planet and an object. In the equation \( F = \frac{GMm}{r^2} \), \( G \) works as a proportionality factor. It ensures that the force is directly proportional to the product of the masses \( M \) and \( m \), and inversely proportional to the square of the distance \( r \) between their centers.

The gravitational constant helps maintain consistency in gravitational calculations, ensuring concepts like weight and orbit are predictable and stable. In problems like the one given, \( G \) remains constant, allowing us to solve for other variables such as distance without changing its value.
Mass of the Earth
The Earth's mass is a fundamental component in calculating the gravitational acceleration at any given point near or on its surface. This mass is symbolized by \( M \) in formulas like \( g = \frac{GM}{r^2} \). The approximate mass of Earth is \( 5.972 \times 10^{24} \, \text{kg} \).

The Earth's mass is crucial in determining the gravitational pull it has on objects, providing the base for gravity-related calculations. When considering the exercise problem, Earth's mass remains constant and is a vital determinant of the gravitational acceleration, affecting everything from the fall of an apple to the orbiting of satellites.

Because \( g \) is directly proportional to Earth's mass in the formula, understanding changes or variations in \( g \) at different distances can help one grasp how gravity diminishes with distance from the surface. Although the mass itself doesn't change, the effect of its gravity does.
Earth's Radius
Earth's radius, denoted as \( R \), is essential when determining the distance at which gravitational effects take place. The average radius of Earth is about \( 6,371 \, \text{km} \).

This measurement is critical when studying gravitational forces because it provides a baseline for calculating how gravity decreases with distance. In the formula \( g = \frac{GM}{r^2} \), \( r \) is the distance from the Earth's center, meaning \( R + h \) is used when objects are above the surface.

In our problem, the radius is used to relate surface gravity to gravity at a certain height. It acts as a baseline distance from the center, helping to calculate \( h \), the height above Earth's surface where a reduced gravity force is felt. This understanding is key to solving for \( h \) in various gravity-related queries.
Distance from the Center of the Earth
The distance from the center of the Earth, denoted as \( r \), plays a pivotal role in calculating gravitational force and acceleration. In gravitational equations, \( r \) is the complete radial distance from the Earth's core to the point of interest.

In practice, \( r \) is often expressed as the sum of the Earth's radius \( R \) and an additional height \( h \), where \( r = R + h \). This is particularly useful for determining gravitational force above the earth's surface. The formula \( g = \frac{GM}{r^2} \) illustrates how gravitational acceleration is inversely proportional to the square of \( r \).

In our exercise, understanding \( r \) is crucial to finding out how high an object must be from the Earth's surface to experience a specific gravitational pull, like 0.980 m/s\(^2\). Solving for \( h \) involves working backwards from the known changes in \( g \) to determine how \( r \) must change, illustrating the principle that gravity weakens with distance.

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Most popular questions from this chapter

Planet X rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs 943.0 N on the earth weighs 915.0 N at the north pole of Planet X and only 850.0 N at its equator. The distance from the north pole to the equator is 18,850 km, measured along the surface of Planet X. (a) How long is the day on Planet X? (b) If a 45,000-kg satellite is placed in a circular orbit 2000 km above the surface of Planet X, what will be its orbital period?

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030-kg ears is 6.0 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.

Astronomers have observed a small, massive object at the center of our Milky Way galaxy (see Section 13.8). A ring of material orbits this massive object; the ring has a diameter of about 15 light-years and an orbital speed of about 200 km/s. (a) Determine the mass of the object at the center of the Milky Way galaxy. Give your answer both in kilograms and in solar masses (one solar mass is the mass of the sun). (b) Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star? (c) Many astronomers believe that the massive object at the center of the Milky Way galaxy is a black hole. If so, what must the Schwarzschild radius of this black hole be? Would a black hole of this size fit inside the earth's orbit around the sun?

Two uniform spheres, each with mass \(M\) and radius \(R\), touch each other. What is the magnitude of their gravitational force of attraction?

A planet orbiting a distant star has radius 3.24 \(\times\) 10\(^6\) m. The escape speed for an object launched from this planet's surface is 7.65 \(\times\) 10\(^3\) m/s. What is the acceleration due to gravity at the surface of the planet?

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