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 2 .

Question

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 × 1024 kg, what is the magnitude of the earth’s acceleration toward the apple? Answer in units of m/s 2 .

Answer 1

Let

m = 0.39 kg be the mass of the apple,

g = 9.8 m/s² be the acceleration of the apple towards the Earth,

M = 5.98×10²⁴ kg be the mass of the Earth,

a (to be computed) be the acceleration of the Earth towards the apple.

By Newton's Second Law the force acting on the apple has magnitude

F_a = mg

And the force acting on the Earth has magnitude

F_e = Ma

By Newton's Third Law

F_e = F_a

Ma = mg

a = mg/M = 0.39×9.81/5.98×10²⁴ = 6.4×10^-23m/s²