An object of mass 0.50 kg weighs 2.0 N when it is 1.0 × 10⁶ m above the surface of Planet X.The radius of the planet is 4.0 × 10⁶ m.What free-fall acceleration will the 0.50-kg object experience when at the surface of Planet X? A)4.8 m/s² B)2.0 m/s² C)4.0 m/s² D)6.3 m/s²

Question

Quizplus · Accepted Answer

The weight of the object at a height of 1.0 x 10^16 m above the surface of Planet X is given by:
W = mg 
2.0 N = (0.50 kg)g 
g = 4.0 m/s^2

The acceleration due to gravity at the surface of the planet can be found using the formula:
g' = (GM)/r^2 
where G is the universal gravitational constant, M is the mass of the planet, and r is the radius of the planet.

Substituting the given values:
g' = (6.67 x 10^-11 N m^2/kg^2)(M)/(4.0 x 10^16 m)^2

The mass of the planet is not given, but we can use the fact that the weight of an object at the surface of the planet is:
W' = mg'

Substituting the given values:
W' = (0.50 kg)(g')

Equating this to the weight of the object at a height of 1.0 x 10^16 m:
W' = 2.0 N

(0.50 kg)(g') = 2.0 N

g' = 4(2.0 N)/(0.50 kg)

g' = 16.0 N/kg

g' = 16.0 m/s^2

Therefore, the free-fall acceleration that the 0.50 kg object will experience at the surface of Planet X is 6.3 m/s^2.

An Object of Mass 0