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Chapter 8: Problem 22

The gravitational force of the earth on a freely falling ball of mass one kilogram is \(9.8 \mathrm{~N}\). The acceleration of the earth towards the ball is (1) \(9.8 \mathrm{~m} \mathrm{~s}^{-2}\) (2) negligible. (3) slightly less than \(9.8 \mathrm{~m} \mathrm{~s}^{-2}\) (4) more than \(9.8 \mathrm{~m} \mathrm{~s}^{-2}\)

Short Answer

Expert verified
Answer: The acceleration of the Earth towards the ball is negligible, approximately 1.64 * 10^{-24} m/s^2.

Step by step solution

01

Write down the given information

We have a ball of mass 1 kg and the gravitational force of the Earth on the ball is 9.8 N.
02

Write the Newton's second law of motion

Newton's second law of motion states that the force acting on an object is equal to the mass of the object times its acceleration, i.e., F = ma.
03

Write down the gravitational force formula

The gravitational force formula is given by F = G * (m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.
04

Find the acceleration of the ball

We know that F = ma and F = 9.8 N, and m (mass of the ball) = 1 kg. Therefore, the acceleration of the ball is a = F/m = 9.8 N / 1 kg = 9.8 m/s^2.
05

Find the mass of the Earth

We know that G = 6.674 * 10^{-11} N*(m/kg)^2, and the radius of the Earth is approximately 6.371 * 10^6 meters. From the gravitational force formula, we can calculate the mass of the Earth (m_e): F = G * (m * m_e) / r^2 9.8 N = (6.674 * 10^{-11} N*(m/kg)^2) * (1 kg * m_e) / (6.371 * 10^6 m)^2 After solving for m_e, we get: m_e ≈ 5.972 * 10^{24} kg
06

Find the acceleration of the Earth towards the ball

Now, we will use Newton's second law of motion again, but this time for the Earth, with the same gravitational force: F = m_e * a_e 9.8 N = 5.972 * 10^{24} kg * a_e a_e = 9.8 N / (5.972 * 10^{24} kg) a_e ≈ 1.64 * 10^{-24} m/s^2 The acceleration of the Earth towards the ball is extremely small, thus the answer is: (2) negligible.

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