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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 above the surface of the moon is 260 meters.

Step by step solution

01

Write down the necessary formulas.

We have the Universal Law of Gravitation: \[F = G \frac{m_1 m_2}{r^2}\] And the potential energy is given by: \[U = -G \frac{m_1 m_2}{r}\] Here, \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the moon and the rock, respectively, and \(r\) is the distance between their centers. The initial kinetic energy (at the moon's surface) can be calculated as: \[K_i = \frac{1}{2} m v^2\] where \(m\) is the mass of the rock, and \(v\) is its initial speed. Now, we will apply the conservation of mechanical energy, i.e., \[K_i + U_i = U_f\] where \(K_i\) is the initial kinetic energy, \(U_i\) is the initial gravitational potential energy, and \(U_f\) is the final gravitational potential energy at the maximum height.
02

Calculate the initial gravitational potential energy.

When the astronaut throws the rock from the surface of the moon, the distance between the centers is equal to the moon's radius, \(R\). So, the initial gravitational potential energy is: \[U_i = -G \frac{M m}{R}\] Substitute the given values: \[U_i = - (\text{6.674}\times 10^{-11} \text{ Nm/kg}^2) \frac{(8.00 \times 10^{18} \text{ kg})(2.00 \text{ kg})}{(6.30 \times 10^4 \text{ m})}\] Calculate \(U_i\): \[U_i = -1.681 \times 10^7 \text{ J}\]
03

Calculate the initial kinetic energy.

Using the initial speed given, we can calculate the initial kinetic energy: \[K_i = \frac{1}{2}m v^2\] Substituting the values: \[K_i = \frac{1}{2}(2.00 \text{ kg})(40.0 \text{ m/s})^2\] Calculate \(K_i\): \[K_i = 1.600 \times 10^3 \text{ J}\]
04

Apply conservation of energy and solve for the final radius.

We know that \(K_i + U_i = U_f\), hence: \[1.600 \times 10^3 \text{ J} - 1.681 \times 10^7 \text{ J} = -G \frac{(8.00 \times 10^{18}\text{ kg})(2.00\text{ kg})}{R_f}\] Solving for \(R_f\): \[R_f = \frac{(8.00 \times 10^{18}\text{ kg})(2.00\text{ kg})(\text{6.674}\times 10^{-11} \text{ Nm/kg}^2)}{1.681 \times 10^7 \text{ J} - 1.600 \times 10^3 \text{ J}}\] Calculate \(R_f\): \[R_f = 6.326 \times 10^4 \text{ m}\]
05

Calculate the maximum height above the moon's surface

Now that we have the final radius, we can calculate the maximum height the rock reaches above the moon's surface by subtracting the moon's radius from the final radius: \[H = R_f - R \] Substitute the values: \[H = (6.326 \times 10^4 \text{ m}) - (6.30 \times 10^4 \text{ m})\] Calculate \(H\): \[H = 260 \text{ m}\] The maximum height above the surface of the moon is 260 meters.

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Most popular questions from this chapter

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?

Newton's Law of Gravity specifies the magnitude of the interaction force between two point masses, \(m_{1}\) and \(m_{2}\), separated by a distance \(r\) as \(F(r)=G m_{1} m_{2} / r^{2} .\) The gravitational constant \(G\) can be determined by directly measuring the interaction force (gravitational attraction) between two sets of spheres by using the apparatus constructed in the late 18th century by the English scientist Henry Cavendish. This apparatus was a torsion balance consisting of a 6.00 -ft wooden rod suspended from a torsion wire, with a lead sphere having a diameter of 2.00 in and a weight of 1.61 lb attached to each end. Two 12.0 -in, 348 -lb lead balls were located near the smaller balls, about 9.00 in away, and held in place with a separate suspension system. Today's accepted value for \(G\) is \(6.674 \cdot 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\) Determine the force of attraction between the larger and smaller balls that had to be measured by this balance. Compare this force to the weight of the small balls.

Where the International Space Station orbits, the gravitational acceleration is just \(11.4 \%\) less than its value on the surface of the Earth. Nevertheless, astronauts in the space station float. Why is this so?

For two identical satellites in circular motion around the Earth, which statement is true? a) The one in the lower orbit has less total energy. b) The one in the higher orbit has more kinetic energy. c) The one in the lower orbit has more total energy. d) Both have the same total energy.

Satellites in low orbit around the Earth lose energy from colliding with the gases of the upper atmosphere, causing them to slowly spiral inward. What happens to their kinetic energy as they fall inward?

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