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
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.