When you drop a 0.38 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/s2
Don't understand. Please help
To find the magnitude of the Earth's acceleration toward the apple, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F = m * a).
We know that the mass of the Earth (mE) is 5.98 × 10^24 kg and the mass of the apple (ma) is 0.38 kg. The force exerted on the apple by the Earth is equal to the weight of the apple, which is given by the formula W = m * g, where g represents the acceleration due to gravity (9.8 m/s^2 in this case).
According to Newton’s third law, the apple must exert an equal but opposite force on Earth. Since the force exerted on Earth (FE) is equal to the force exerted on the apple, we can write it as FE = ma * aE, where aE is the acceleration of the Earth toward the apple.
Now, we can equate the two forces:
ma * aE = mE * g
We can solve for aE by rearranging the equation:
aE = (mE * g) / ma
Substituting the given values:
aE = (5.98 × 10^24 kg * 9.8 m/s^2) / 0.38 kg
Now we can calculate the expression:
aE ≈ 1.54 × 10^26 m/s^2
Therefore, the magnitude of the Earth's acceleration toward the apple is approximately 1.54 × 10^26 m/s^2.