Question

When you drop a 0.43 kg apple, Earth exerts

Answer 1

The magnitude of the Earth's acceleration is $7.0\cdot 10^{-25} m/s^2$

Explanation:

First of all, we start by calculating the magnitude of the force exerted by the apple on the Earth. According to Newton's third law, this is equal to the force exerted by the Earth on the apple, which is the weight of the apple, given by:

$F=mg$

where

m = 0.43 kg is the mass of the apple

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$F=(0.43)(9.8)=4.2 N$

Now we can find the Earth's acceleration by applying Newton's second law:

$F=Ma$

where:

F = 4.2 N is the net force exerted by the apple on the Earth

$M=5.98\cdot 10^{24} kg$ is the mass of the Earth

a is the Earth's acceleration

And solving for a, we find:

$a=\frac{F}{M}=\frac{4.2}{5.98\cdot 10^{24}}=7.0\cdot 10^{-25} m/s^2$

Learn more about Newton laws of motion:

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