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Chapter 12: Problem 41

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?

Short Answer

Expert verified
Answer: The maximum height the rock will reach above the surface of the moon is approximately 12,460 meters.

Step by step solution

01

Identify the Initial and Final States

The initial state is when the rock is thrown with an initial speed of \(40.0 \mathrm{~m/s}\) from the surface of the moon. At this point, the rock has both kinetic and potential energy. The final state is when the rock reaches its maximum height, at which point it has zero velocity, and its kinetic energy is zero. At this moment, the rock has only potential energy.
02

Calculate the Initial Kinetic Energy

The initial kinetic energy of the rock can be calculated using the formula: \(KE = \frac{1}{2} mv^2\), where m is the mass of the rock and v is the initial velocity. \(KE = \frac{1}{2} (2.00 \mathrm{~kg}) (40.0 \mathrm{~m/s})^2 = 1600 \mathrm{~J}\)
03

Calculate the Initial Gravitational Potential Energy

The initial gravitational potential energy of the rock can be calculated using the formula: \(PE = -G\frac{Mm}{r}\), where G is the gravitational constant (\(6.674 \times 10^{-11} \mathrm{N m^2/kg^2}\)), M is the mass of the moon, m is the mass of the rock, and r is the distance from the center of the moon to the rock (initially equal to the radius of the moon). \(PE = -6.674 \times 10^{-11} \mathrm{N m^2/kg^2} \frac{(8.00 \times 10^{18} \mathrm{~kg})(2.00 \mathrm{~kg})}{6.30 \times 10^4 \mathrm{~m}} = -1.687 \times 10^3 \mathrm{~J}\)
04

Apply the Conservation of Mechanical Energy

According to the conservation of mechanical energy, the initial total mechanical energy (kinetic + potential) is equal to the final potential energy. \(KE_{initial} + PE_{initial} = PE_{final}\) Replacing the values found in steps 2 and 3: \(1600 \mathrm{~J} - 1.687 \times 10^3 \mathrm{~J} = -G\frac{Mm}{r_{final}}\) Solve for \(r_{final}\): \(-6.674 \times 10^{-11} \mathrm{N m^2/kg^2} \frac{(8.00 \times 10^{18} \mathrm{~kg})(2.00 \mathrm{~kg})}{r_{final}} = -87 \mathrm{~J}\) \(r_{final} = 7.546 \times 10^4 \mathrm{~m}\)
05

Calculate the Maximum Height

To find the maximum height, subtract the initial radius of the moon from the final distance: \(height = r_{final} - r_{initial} = 7.546 \times 10^4 \mathrm{~m} - 6.30 \times 10^4 \mathrm{~m} = 1.246 \times 10^4 \mathrm{~m}\) The maximum height the rock will reach above the surface of the moon is approximately 12,460 meters.

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