When you drop a 0.43 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

The acceleration of the earth is 7.05 * 10^-25 m/s²

Explanation:

Step 1: Data given

mass of the apple = 0.43 kg

acceleration = 9.8 m/s²

mass of earth = 5.98 * 10 ^24 kg

Step 2: Calculate the acceleration of the earth

Following the third law of Newton F = m*a

F(apple) = F(earth) = m(apple)*a(apple)

F(apple) = 0.43 kg * 9.8 m/s² = 4.214 N

a(earth) = F(apple/earth)/m(earth)

a(earth) = 4.214N /5.98 * 10 ^24 kg

a(earth) = 7.05 * 10^-25 m/s²

The acceleration of the earth is 7.05 * 10^-25 m/s²

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