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Chapter 12: Problem 34

What is the magnitude of the free-fall acceleration of a ball (mass \(m\) ) due to the Earth's gravity at an altitude of \(2 R\), where \(R\) is the radius of the Earth?

Short Answer

Expert verified
Answer: The magnitude of the free-fall acceleration is given by the expression \(a = G\frac{M}{(3R)^2}\), where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the radius of the Earth.

Step by step solution

01

Gravitational force formula

We start by recalling the formula for the force between two objects due to gravity: \(F = G\frac{Mm}{r^2}\) where \(F\) is the gravitational force, \(G\) is the gravitational constant, \(M\) is the mass of the Earth, \(m\) is the mass of the ball, and \(r\) is the distance between the centers of the two objects. In this case, the distance \(r\) is equal to the sum of the Earth's radius and the altitude of the ball, so \(r = R + 2R = 3R\). We can now rewrite the formula as: \(F = G\frac{Mm}{(3R)^2}\)
02

Equate gravitational force to mass times acceleration

The gravitational force acting on the ball is also equal to its mass times its free-fall acceleration, which we'll denote as \(a\). So we can write: \(F = ma\)
03

Solve for acceleration

Now we can equate the expressions from steps 1 and 2, and solve for the acceleration: \(ma = G\frac{Mm}{(3R)^2}\) Divide both sides by the mass of the ball, \(m\): \(a = G\frac{M}{(3R)^2}\) That's the expression for the magnitude of the free-fall acceleration of the ball at an altitude of \(2R\).

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Most popular questions from this chapter

Two planets have the same mass, \(M .\) Each planet has a constant density, but the density of planet 2 is twice as high as that of planet \(1 .\) Identical objects of mass \(m\) are placed on the surfaces of the planets. What is the relationship of the gravitational potential energy, \(U_{1},\) on planet 1 to \(U_{2}\) on planet \(2 ?\) a) \(U_{1}=U_{2}\) b) \(U_{1}=\frac{1}{2} U_{2}\) c) \(U_{1}=2 U_{2}\) d) \(U_{1}=8 U_{2}\) e) \(U_{1}=0.794 U_{2}\)

Two 30.0 -kg masses are held at opposite corners of a square of sides \(20.0 \mathrm{~cm} .\) If one of the masses is released and allowed to fall toward the other mass, what is the acceleration of the first mass just as it is released? Assume that the only force acting on the mass is the gravitational force of the other mass. a) \(1.5 \cdot 10^{-8} \mathrm{~m} / \mathrm{s}^{2}\) b) \(2.5 \cdot 10^{-8} \mathrm{~m} / \mathrm{s}^{2}\) c) \(7.5 \cdot 10^{-8} \mathrm{~m} / \mathrm{s}^{2}\) d) \(3.7 \cdot 10^{-8} \mathrm{~m} / \mathrm{s}^{2}\)

Standing on the surface of a small spherical moon whose radius is \(6.30 \cdot 10^{4} \mathrm{~m}\) and whose mass is \(8.00 \cdot 10^{18} \mathrm{~kg}\) an astronaut throws a rock of mass 2.00 kg straight upward with an initial speed \(40.0 \mathrm{~m} / \mathrm{s}\). (This moon is too small to have an atmosphere.) What maximum height above the surface of the moon will the rock reach?

Express algebraically the ratio of the gravitational force on the Moon due to the Earth to the gravitational force on the Moon due to the Sun. Why, since the ratio is so small, doesn't the Sun pull the Moon away from the Earth?

Compare the magnitudes of the gravitational force that the Earth exerts on the Moon and the gravitational force that the Moon exerts on the Earth. Which is larger?
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