# 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"

Aug 16, 2017

Law of Universal Gravitation states that the force of attraction ${F}_{G}$ between two bodies of masses ${M}_{1} \mathmr{and} {M}_{2}$ 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

${F}_{G} = G \frac{{M}_{1.} {M}_{2}}{r} ^ 2$ .......(1)
where $G$ is the proportionality constant and $= 6.67408 \times {10}^{-} 11 {m}^{3} k {g}^{-} 1 {s}^{-} 2$

Let $m$ be mass of meteor and ${M}_{e}$ mass of earth. Using Newton's Second Law of motion we can rewrite (1) as

${F}_{G} = m a$ ......(2)

Comparing (1) and (2) we get

$a = {F}_{G} / m = G \frac{{M}_{e}}{r} ^ 2$

Inserting various values we get

a=6.67408 xx 10^-11xx (6xx10^24)/(8×10^6)^2
$a = 6.3 m {s}^{-} 2$, rounded to one decimal place.