When you drop a 0.44 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s2 to- ward the earth’s surface. According to New- ton’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/s .
"F=ma=0.44\\cdot9.8=4.312\\ (N)"
"a=F\/M=4.312\/(5.98\\cdot10^{24})=7.2\\cdot10^{-25}\\ (m\/s^2)" . Answer
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