The gravitational attraction on the surface of the Moon (lunar surface) is about 1/6th of that on the surface of the Earth. An astronaut can throw a ball up to a height of 8 m from the surface of the Earth. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?
48 m
Explanation
According to the work-energy theorem provided in the context, the work done to raise an object to a height h is W = mgh, which equals the gravitational potential energy at that height. When the ball is thrown upwards, its initial kinetic energy (1/2 ) is entirely converted into potential energy (mgh) at the maximum height. Since the astronaut throws the ball with the same upward velocity, the initial kinetic energy is the same on both the Earth and the Moon. Because the gravitational attraction (g) on the Moon is 1/6th of that on the Earth, the height reached will be inversely proportional to the acceleration due to gravity. Thus, the ball will travel 6 times higher on the Moon.
Solution Steps
Step 1: At maximum height, all initial kinetic energy is converted to potential energy. Using the work-energy theorem, mgh = 1/2 mu², which simplifies to h = u² / 2g.
Step 2: Since the ball is thrown with the same upward velocity (u) on both the Earth and the Moon, the height h is inversely proportional to the acceleration due to gravity (g). So, h_m / h_e = g_e / g_m.
Step 3: Given that the gravitational attraction on the Moon is 1/6th of that on the Earth, g_m = 1/6 g_e. Therefore, g_e / g_m = 6.
Step 4: Substituting the values, h_m = 6 × h_e = m = 48 m.