Question

When you drop a 0.35 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s^2 toward the earths surface. According to Newton’s third law, the apple must exert an equal but opposite force on Earth. If the mass of the earth is 5.98x10^24 kg, what is the magnitude of the earth’s acceleration toward the apple? Answer in units of m/s^2.

Answer 1

The force on Apple by the earth is

$F=mg= 0.35kg\times 9.8ms^{-2} = 3.43N$

By Newton's third law, the apple also apply the same magnitude force on earth.

$F_{Earth}=M a = 3.43N$

$a = \frac{3.43N}{M} = \frac{3.43N}{5.98\times10^{24}kg} = 5.74\times10^{-25}ms^{-2}$