Questions: When you drop a 0.39 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s^2 toward the earth's surface. According to Newton's third law, the apple must exert an equal but opposite force on Earth. If the mass of the earth 5.98 x 10^24 kg, what is the magnitude of the earth's acceleration toward the apple? Answer in units of m/s^2.

Solution Steps

According to Newton's third law, the force exerted by the apple on Earth is equal in magnitude and opposite in direction to the force exerted by Earth on the apple. The force exerted by Earth on the apple is given by:

\[ F = m_{\text{apple}} \cdot a_{\text{apple}} \]

where \( m_{\text{apple}} = 0.39 \, \text{kg} \) and \( a_{\text{apple}} = 9.8 \, \text{m/s}^2 \).

Calculating the force:

\[ F = 0.39 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 3.822 \, \text{N} \]

Using Newton's second law, the acceleration of Earth due to the force exerted by the apple can be calculated as:

\[ a_{\text{earth}} = \frac{F}{m_{\text{earth}}} \]

where \( m_{\text{earth}} = 5.98 \times 10^{24} \, \text{kg} \).

Substituting the values:

\[ a_{\text{earth}} = \frac{3.822 \, \text{N}}{5.98 \times 10^{24} \, \text{kg}} \]

Calculating the acceleration:

\[ a_{\text{earth}} = 6.390 \times 10^{-25} \, \text{m/s}^2 \]

Final Answer