We know that the acceleration due to gravity at the surface of the Earth is 9.8m

Question

We know that the acceleration due to gravity at the surface of the Earth is 9.8m/s2. We also know that the acceleration due to gravity falls off as 1/r2 above the surface of the Earth, with r being measured from the center of the planet.

Knowing that the M term in gravity for the Earth can be set equal to the amount of mass contained inside a sphere of radius r centered upon the middle of the Earth in cases where we are investigating rEarth, find the value of g when we are 1/3 of the way between the center and surface of the Earth.

____  m/s2.

Explanation / Answer

The acceleration due to gravity at a distance 'r' from the earth's center is given by:

g = GM/r^2

where G = gravitational constant = 6.67*10^-11

M = amount of mass contained inside a sphere of radius r centered upon the middle of the Earth

So, for r = R = radius of earth = 6371 km = 6.37*10^6 m,

M = mass of earth = 6*10^24 kg

we get g = 6.67*10^-11*6*10^24/(6.37*10^6)^2 = 9.81 m/s2

We know, the volume of sphere = (4/3)*pi*r^3

and mass of portion of earth contained inside the radius r is directly proportional to volume

So, the mass is directly proportional to r^3

So, for r = (1/3) of R(radius of earth from center to surface) = R/3,

mass is directly proportional to (R/3)^3 = R^3/27

So, mass is (1/27) times the total mass of earth

So, m = (1/27)*6*10^24 kg

r = R/3 = 6.37*10^6/3

So, g = Gm/r^2

= 6.67*10^-11*((1/27)*(6*10^24))/(6.37*10^6/3)^2

= 3.29 m/s2 <---------------answer

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