Question

When you drop a 0.37kg apple earth exerts a force on it that accelerates it at 9.8m/s^2 toward the earths surface. According to newtons third law, the apple must exert an equal but opposite force on earth.

Answer 1

force applied due to earth on the apple is given as

$F = ma$

given that

m = 0.37 kg

a = 9.8 m/s^2

now we have

$F = m*a$

$F = 0.37* 9.8 = 3.63 N$

now as per Newton's III law equal and opposite force is applied on earth

so now we can find acceleration of Earth using same equation as we used above

$F = ma$

here we know that

m = mass of earth

F = 3.63 N

$3.63 = 5.98 * 10^{24} * a$

$a = \frac{3.63}{5.98 * 10^{24}$

$a = 6.1 * 10^{-25} m/s^2$

so acceleration of earth is very small and given by above value