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 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 × 1024 kg, what
is the magnitude of the earth’s acceleration
toward the apple?
Answer in units of m/s2.
Don't round answer.

Answer 1

$$Ma=mg$$

$$a=\frac{m}{M}g=\frac{0.35}{5.98\cdot 10^{24}}9.8=5.7\cdot 10^{-25}\frac{m}{s^2}$$