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

Question

Gunnar Roth

Physics

31 December, 19:34

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

+1

Answers (1)

Know the Answer?

Anna Arias · Answer 1

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