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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 distance is approximately 13800 km above Earth's surface.

Step by step solution

01

Understanding the Gravitational Formula

The acceleration due to gravity at a distance, say \(d\), from the center of the Earth is given by \(g_d = \frac{G M}{d^2}\), where \(G\) is the gravitational constant and \(M\) is the mass of the Earth. At the surface, this acceleration is also given by \(g = \frac{G M}{R^2}\), where \(R\) is the radius of the Earth.
02

Relate Surface Gravity to Desired Gravity

We need to find the distance \(h\) above the Earth's surface where the gravity \(g_d\) is 0.980 \(\mathrm{m}/\mathrm{s}^{2}\). Using \(g_d = 0.980\) and \(g = 9.80\), we set up the equation \(\frac{G M}{(R+h)^2} = 0.980\), and \(\frac{G M}{R^2} = 9.80\). Divide these two equations to eliminate \(G\) and \(M\).
03

Solve for \(R+h\)

By dividing, we have \(\frac{9.80}{0.980} = \frac{(R+h)^2}{R^2}\). Simplifying this gives \(10 = \frac{(R+h)^2}{R^2}\). Solving for \(R+h\), this equation becomes \(R+h = R\sqrt{10}\).
04

Calculate Height \(h\)

To find \(h\), rearrange \(R+h = R\sqrt{10}\) to \(h = R(\sqrt{10} - 1)\). Assume \(R = 6.371 \times 10^6\) m (mean radius of Earth). Substitute \(R\) to find \(h\): \(h = 6.371 \times 10^6 (\sqrt{10} - 1)\). Calculate \(\sqrt{10} \approx 3.162\), so \(h \approx 6.371 \times 10^6 \times 2.162\).
05

Compute the Numerical Solution

Perform the final calculation: \(h \approx 6.371 \times 10^6 \times 2.162\). This yields \(h \approx 1.38 \times 10^7\) m, or approximately 13800 km.

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Key Concepts

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

Acceleration Due to Gravity
Acceleration due to gravity is a crucial concept in gravitational physics, especially when dealing with objects close to Earth. It is the rate at which an object accelerates downwards due to Earth's gravitational pull. This acceleration is often denoted as \(g\) and is approximately \(9.8 \ \mathrm{m}/\mathrm{s}^{2}\) at the Earth's surface. This means any object, in the absence of other forces, will accelerate downwards at this rate.
For different altitudes or distances from Earth's surface, the acceleration can change. This is why \(g\) is sometimes replaced with \(g_d\) to indicate the acceleration at a certain distance \(d\) from the Earth's center. The further you move from the Earth, the weaker the gravitational pull, hence the smaller the acceleration due to gravity.
In the textbook problem, you examine how \(g\) changes when moving away from the surface. At a height where gravity is \(0.980 \ \mathrm{m}/\mathrm{s}^{2}\), mathematical steps helped determine that distance, showing gravity's dependence on altitude.
Gravitational Constant
The gravitational constant, denoted as \(G\), plays an integral role in calculations involving gravitational force. Newton's law of universal gravitation uses this constant, which quantifies the strength of gravity. The value of \(G\) is approximately \(6.674 \times 10^{-11} \ \mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}\).
In the context of the provided exercise, \(G\) helps relate gravitational forces between two masses. This constant allows you to understand how force works through distances, such as from Earth's surface upwards. In formulating gravitational potential or calculating weight scales with distance, \(G\) provides the stability and reliability across various scenarios of altitude or distance.
With Earth's mass \(M\) and radius \(R\), \(G\) featured prominently in deriving forms like \(g = \frac{G M}{R^2}\), illustrating how fundamental constants tie deeply into how we calculate or predict gravitational behaviors.
Radius of the Earth
The radius of the Earth is a pivotal measure used in gravitational calculations. Known to be about \(6.371 \times 10^6\ \mathrm{m}\), it serves as a baseline for many calculations regarding gravity and motion near the Earth.
In the exercise, the radius is used to find how gravity changes with distance. The Earth's radius provides a point of reference for understanding how far above the surface you are, translating into varying gravitational forces. When determining the height \(h\) at which the gravitational force is a certain value, you use the radius as part of the formula \(g_d = \frac{G M}{(R+h)^2}\).
This radius aids in solidity and accuracy of predictions or analyses about moving away from Earth. For students or curious minds exploring gravitational physics, Earth's radius offers a grounded way to visualize or calculate different aspects of gravity's effects over distances.

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Most popular questions from this chapter

A thin, uniform rod has length \(L\) and mass \(M .\) A small uniform sphere of mass \(m\) is placed a distance \(x\) from one end of the rod, along the axis of the rod (Fig. 12.34\()\) . (a) Calculate the gravita- tional potential energy of the rod-sphere system. Take the potential energy to be zero when the rod and sphere are infinitely far apart. Show that your answer reduces to the expected result when \(x\) is much larger than \(L\) . (Hint: Use the power series expansion for \(\ln (1+x)\) given in Appendix B. (b) Use \(F_{x}=-d U / d x\) to find the magnitude and direction of the gravitational force exerted on the sphere by the rod (see Section 7.4 ). Show that your answer reduces to the expected result when \(x\) is much larger than \(L .\)

One of the brightest comets of the 20 th century was Comet Hyakutake, which passed close to the sun in early 1996 . The orbital period of this comet is estimated to be about \(30,000\) years. Find the semi-major axis of this comet's orbit. Compare it to the average sun-Pluto distance and to the distance to Alpha Centauri, the nearest star to the sun, which is 4.3 light-years distant.

The planet Uranus has a radius of \(25,560 \mathrm{km}\) and a surface acceleration due to gravity of 11.1 \(\mathrm{m} / \mathrm{s}^{2}\) at its poles. Its moon Miranda (discovered by Kniper in 1948 ) is in a circular orbit about Uranus at an altitude of \(104,000 \mathrm{km}\) above the planet's surface. Miranda has a mass of \(6.6 \times 10^{19} \mathrm{kg}\) and a radius of 235 \(\mathrm{km}\) . (a) Calculate the mass of Uranus from the given data, (b) Calculate the magnitude of Miranda's acceleration due to its orbital motion about Uranus. (c) Calculate the acceleration due to Miranda's gravity at the surface of Miranda. (d) Do the answers to parts (b) and (c) mean that an object released 1 \(\mathrm{m}\) above Miranda's surface on the side toward Uranus will fall up relative to Miranda? Explain.

The mass of Venus is 81.5\(\%\) that of the earth, and its radius is 94.9\(\%\) that of the earth. (a) Compute the acceleration due to gravity on the surface of Venus from these data. (b) If a rock weighs 75.0 \(\mathrm{N}\) on earth, what would it weigh at the surface of Venus?

At a certain instant, the earth, the moon, and a stationary \(1250-\mathrm{kg}\) spacecraft lie at the vertices of an equilateral triangle whose sides are \(3.84 \times 10^{5} \mathrm{km}\) in length. (a) Find the magnitude and direction of the net gravitational force exerted on the spacecraft by the earth and moon. State the direction as an angle measured from a line connecting the earth and the spacecraft. In a sketch, show the earth, the moon, the spacecraft, and the force vector. (b) What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? You can ignore any gravitational effects due to the other planets or the sun.

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