Question

When you drop a 0.42 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:

6.9×10⁻²⁵ m/s²

Explanation:

m = Mass

a = Acceleration

F = ma

Force exerted on apple

$F=0.42\times 9.8\\\Rightarrow F=4.116\ N$

From Newton's third law the opposite force would be the same but the mass is different which means the acceleration would be different

$a=\frac{F}{M}\\\Rightarrow a=\frac{4.116}{5.98\times 10^{24}}\\\Rightarrow a=6.9\times 10^{-25}\ m/s^2$

The magnitude of the earth’s acceleration toward the apple is 6.9×10⁻²⁵ m/s²

Answer 2

Answer:

a = 6.97 × 10⁻²⁵ m/s²

Explanation:

given,

mass = 0.42 kg

g = 9.8 m/s²

mass of the earth = 5.98 × 10²⁴ kg

magnitude of earth acceleration = ?

F = m g

F = 0.42 × 9.8

F = 4.17 N

according to newtons third law Force exerted by the earth will be equal to force exerted by apple

m a = 4.17 N

$a = \dfrac{4.17}{5.98 \times 10^{24}}$

a = 0.697 × 10⁻²⁴ m/s²

a = 6.97 × 10⁻²⁵ m/s²

Hence, acceleration of earth toward apple is equal to  a = 6.97 × 10⁻²⁵ m/s²