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"

1 Answer
A08
Aug 16, 2017

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.

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