Question

When you drop a 0.4 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s° 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 x 1024 kg, what is the magnitude of the earth's acceleration toward the apple?
Answer in units of m/s?.

Answer 1

Step-by-step explanation:

Force on the apple :

F = m a

F = .4 kg * 9.81 m/s^2 = 3.924 N

Now for the Earth :

F = ma

3.924 N = 5.98 x 10^24 kg * a

a = 3.924 / (5.98 x 10^24 ) = 6.56 x 10^-25 m/s^2