Question

When you drop a 0.44 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/

Answer 1

The magnitude of the Earth's acceleration toward the apple is approximately 7.20 × 10^-25 m/s².

Step-by-step explanation:

To find the magnitude of the Earth's acceleration toward the apple, we can use Newton's third law of motion. Newton's third law states that for every action, there is an equal and opposite reaction. In this case, the apple exerts a force on Earth, and Earth exerts an equal and opposite force on the apple.

The force that the apple exerts on Earth can be calculated using the formula F = ma, where F is the force, m is the mass of the apple, and a is the acceleration of the apple. According to the problem, the mass of the apple is 0.44 kg and the acceleration is 9.8 m/s². Plugging these values into the formula, we get F = (0.44 kg)(9.8 m/s²) = 4.312 N.

Since the force exerted by the apple on Earth is equal and opposite to the force exerted by Earth on the apple, the magnitude of the Earth's acceleration toward the apple can be calculated using the formula F = ma, where F is the force, m is the mass of the Earth, and a is the acceleration of the Earth. According to the problem, the mass of the Earth is 5.98 × 10^24 kg. Now we can solve for a: a = F / m = 4.312 N / (5.98 × 10^24 kg) ≈ 7.20 × 10^-25 m/s². Therefore, the magnitude of the Earth's acceleration toward the apple is approximately 7.20 × 10^-25 m/s².

Answer 2

7.2×10⁻²⁵ m/s²

Step-by-step explanation:

F = ma

(0.44 kg) (9.8 m/s²) = (5.98×10²⁴ kg) a

a = 7.2×10⁻²⁵ m/s²