Question

What is the acceleration of a meteor at a distance `8xx10^6"m"` from the centre of the Earth? Take the mass of the Earth to be `6xx10^(24)"kg"`

Answer 1

Law of Universal Gravitation states that the force of attraction `F_G` between two bodies of masses `M_1 and M_2` is directly proportional to the product of masses of the two bodies. it is also inversely proportional to the square of the distance `r` between the two.
Mathematically stated

`F_G =G (M_1.M_2)/r^2` .......(1)
where `G` is the proportionality constant and `=6.67408 xx 10^-11 m^3 kg^-1 s^-2`

Let `m` be mass of meteor and `M_e` mass of earth. Using Newton's Second Law of motion we can rewrite (1) as

`F_G=ma` ......(2)

Comparing (1) and (2) we get

`a=F_G/m=G (M_e)/r^2`

Inserting various values we get

`a=6.67408 xx 10^-11xx (6xx10^24)/(8×10^6)^2`
`a=6.3ms^-2`, rounded to one decimal place.