Question

A uniform solid sphere of radius R = 3.6 km produces a gravitational acceleration of ag on its surface. At what distance from the sphere's center are there points (a) inside and (b) outside the sphere where the gravitational acceleration is ag/8?

Answer 1

Part A:

gravitational acceleration is given by:

ag = GM/R^2

Inside the sphere gravitational acceleration is given by:

a = G*m/r^2

m = density*Volume

suppose at distance 'r' from center gravitational acceleration is ag/8, then

Volume = 4*pi*r^3/3

density = Total mass/total Volume = M/(4*pi*R^3/3)

So for r < R,

a = G*(M/(4*pi*R^3/3))*(4*pi*r^3/3)/r^2

a = G*M*r/R^3

Now when a = ag/8

ag/8 = G*M*r/R^3

G*M*/(8*R^2) = G*M*r/R^3

r/R = 1/8

r = R/8 = 3.6/8 = 0.45 km

Part B

for r > R

a = GM/r^2

when a = ag/8

ag/8 = GM/r^2

G*M/(8*R^2) = GM/r^2

(r/R)^2 = 8

r = R*sqrt 8

r = 2.83*R

r = 2.83*3.6 = 10.188 km

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