when you drop a 0.36 kg apple, earth exerts a force on it that accelerates it at 9.8 m/s² toward the earths…

when you drop a 0.36 kg apple, earth exerts a force on it that accelerates it at 9.8 m/s² toward the earths surface. according to newtons 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 earths acceleration toward the apple? answer in units of m/s².

when you drop a 0.36 kg apple, earth exerts a force on it that accelerates it at 9.8 m/s² toward the earths surface. according to newtons 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 earths acceleration toward the apple? answer in units of m/s².

Answer

Explanation:

Step1: Calculate the force on the apple

According to Newton's second - law $F = ma$. Here, $m = 0.36\ kg$ and $a=9.8\ m/s^{2}$. So $F = 0.36\times9.8\ N$. $F=0.36\times9.8 = 3.528\ N$

Step2: Calculate the Earth's acceleration

According to Newton's third - law, the force the apple exerts on the Earth is $F'=F = 3.528\ N$. And from Newton's second - law $F'=M_{E}a_{E}$, where $M_{E}=5.98\times 10^{24}\ kg$. Then $a_{E}=\frac{F'}{M_{E}}$. $a_{E}=\frac{3.528}{5.98\times 10^{24}}\ m/s^{2}\approx5.9\times 10^{-25}\ m/s^{2}$

Answer:

$5.9\times 10^{-25}\ m/s^{2}$