Question

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 .

Answer 1

$F=ma=0.44\cdot9.8=4.312\ (N)$

$a=F/M=4.312/(5.98\cdot10^{24})=7.2\cdot10^{-25}\ (m/s^2)$ . Answer