An object of mass 0.50 kg is transported to the surface of Planet X where the object’s weight is measured to?
be 20 N. The radius of the planet is 4.0E6 m. What free fall acceleration will the 0.50-kg object experience when transported to a distance of 2.0E6 m from the surface of this planet?
Answer by Earth Man
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
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