Verified answer

Free fall acceleration at distance is 17.8 m/s^2

Given:

W = 20 N

r = 4.0E+6 m

m = 0.50 kg

d = 2.0E+6 (altitude)

Solve for gravity at surface.

W = m * g

(20 N) = (0.50 kg) * g

g = (20 N) / (0.50 kg)

g = 40 m/s^2

Use this to solve for the mass of the planet, where G is the Universal Gravitational Constant

M = [g * r^2] / G

M = [ (40 m/s^2) * (4.0E+6 m)^2 ] / (6.67428E-11 m^3/kg-s^2)

M = [ (40 m/s^2) * (1.6E+13 m^2) ] / (6.67428E-11 m^3/kg-s^2)

M = [ 6.4E+14 m^3/s^2 ] / (6.67428E-11 m^3/kg-s^2)

M = 9.589E+24 kg

Find the total radial distance from the center out to the object's raised location

D = r + d

D = (4.0E+6 m) + (2.0E+6 m)

D = 6.0E+6 m

Find free-fall acceleration at new distance

a = GM / D^2

a = [ (6.67428E-11 m^3/kg-s^2) * (9.589E+24 kg) ] / (6.0E+6 m)^2

a = [ 6.4E+14 m^3/s^2 ] / (3.6E+13 m^2)

a = 17.8 m/s^2