Question

When you drop a 0.44 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

`7.217*10^(-25) }` m/s²

Step-by-step explanation:

We can calculate by the following formulas.

m =0.44 Kg

g =9.81 m/s²

M= 5.98*
`10^(24)` Kg

F=m*g

=0.44*9.81

=4.316 N

F=M*a

a=
`(F)/(M)`

a=
`(4.316)/(5.98^(24) )`

a=
`7.217*10^(-25) }` m/s²