Question

calculate the acceleration of an apple of mass 200gm,when it falls towards earth.also find the acceleration of the earth towards the apple.(given mass of earth=6*10^24 kg& radius of earth= 6.64*10^6m)​

Answer 1

Answer:
`a_(apple)=9.082m/s^(2)`,
`a_(Earth)=3.027(10)^(-25)m/s^2`

Explanation:

According to Newton’s law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

`F=G(Mm)/(r^2)` (1)

Where:

`F` is the module of the force exerted between the apple and the Earth

`G=6.674x10^(-11)(m^(3))/(kgs^(2))` is the universal gravitation constant.

`M=6(10)^(24)kg` is the mass of the Earth and
`m=200g=0.2kg` is the mass of the apple

`r=6.64(10)^(6)m` is the distance between the apple and the Earth (assuming tha apple is near the surface of the Earth)

On the other hand, according Newton's 2nd Law of Motion the force
`F` is directly proportional to the mass
`m` and to the acceleration
`a` of a body.

So, in the case of the apple:

`F=m.a_(apple)` (2)

`a_(apple)=(F)/(m)` (3)

Substituting
`F` (1) in (3):

`a_(apple)=(F)/(m)=G(M)/(r^2)` (4)

`a_(apple)=6.674x10^(-11)(m^(3))/(kgs^(2))(6(10)^(24)kg)/((6.64(10)^(6)m)^2)` (5)

`a_(apple)=9.082m/s^2` (6)

Now, in the case of the Earth:

`F=M.a_(Earth)` (7)

`a_(Earth)=(F)/(M)` (8)

Substituting
`F` (1) in (8):

`a_(Earth)=(F)/(M)=G(m)/(r^2)` (9)

`a_(Earth)=6.674x10^(-11)(m^(3))/(kgs^(2))(0.2kg)/((6.64(10)^(6)m)^2)` (10)

`a_(Earth)=3.027(10)^(-25)m/s^2` (11)

As we can see, the acceleration of the apple towards the Earth is greater than the acceleration of the Earth towards the apple (although the gravitational force between them is the same), because the mass of the Earth is greater than the mass of the apple.