Question

When you drop a 0.41 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 .

Answer 1

Answer:

$a= 6.7191\times 10^{-25}\ m/s^2$

Explanation:

According to the second law of newton:-

$F=m\times a$

Where, F is the force

m is the mass

a is the acceleration

For Apple:- Given that:

m = 0.41 kg

a = 9.8 m/s²

Thus,

$F=0.41\times 9.8\ N=4.018\ N$

This is the same force which the apple exerts on the Earth.

For Apple:- Given that:

m = $5.98\times 10^{24}\ kg$

$F=4.018\ N$

a = ?

Thus,

$4.018=5.98\times 10^{24}\times a$

$a=\frac{4.018}{5.98\times 10^{24}}\ m/s^2$

$a= 6.7191\times 10^{-25}\ m/s^2$

Earth’s acceleration toward the apple = $6.7191\times 10^{-25}\ m/s^2$

Answer 2

The magnitude of the Earth’s acceleration toward the apple is $6.72$ × $10^{22}$ kg.

Given the following data:

• Mass of apple = 0.41 kg

• Acceleration of apple = $9.8\; m/s^2$

• Mass of Earth = $5.98$ × $10^{24}\; kg$

To find the magnitude of the Earth’s acceleration toward the apple, we would apply Newton's Second Law of Motion:

Mathematically, Newton's Second Law of Motion is given by this formula;

$Force = Mass$ × $acceleration$

$Force = 0.41$ × $9.8$

Force = 4.018 Newton.

For the Earth’s acceleration toward the apple, we would apply Newton's Second Law of Motion and as such use the same force:

$Acceleration = \frac{Force}{Mass} \\\\Acceleration = \frac{4.018}{5.98(10^{22})}$

Acceleration = $6.72$ × $10^{22}$ kg

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