Two objects that may be considered point masses are initially separated by a distance d. the separation distance is then decreased to d/3. how does the gravitational force between these two objects change as a result of the decrease? a. the force will not change since it is only dependent on the masses of the objects. b. the force will be nine times larger than the initial value. c. the force will be three times larger than the initial value. d. the force will be one third of the initial value. e. the force will be one ninth of the initial value. if an object at the surface of the earth has a weight w, what would be the weight of the object if it was transported to the surface of a planet that is one-sixth the mass of earth and has a radius one third that of earth? a. 3wb. 4w/3c. wd. 3w/2e. w/3

a) B. The force will be nine times larger than the initial value.

b) D. 3W/2

Explanation:

a) If we can take both masses as point masses, the attractive force between them, is given by Newton's Universal Law of gravitation:

Fg = G*m₁*m₂ / r₁₂² (1)

where r₁₂ is the distance between the center of the masses.

If we reduce the distance in such a way that r₁₂f = r₁₂₀/3, replacing in (1), we get:

Fgf = G*m₁*m₂  / (r₁₂₀/3)² = 9* (G*m₁*m₂/ r₁₂₀²) = 9* Fg₀

The result is reasonable, as the gravitational force is inversely proportional to the square of the distance between masses, which is consequence of our 3-D space.

b) We call weight, to the force that earth exerts on any mass on the planet, based on the application of Newton's 2nd Law to the gravitational force between Earth and the mass, as follows:

Fg = G*m*me / re² = m*a

⇒ a = ge = G*me/re² (2)

If we apply the same considerations to the same mass on the surface of a planet with a mass equal to one-sixth the mass of Earth, and a radius one third that of Earth, we can apply the same equation as above:

Fgp = G*m*(me/6) / (re/3)² = m*ap

⇒ap = gp = (G*me/re²)*(9/6) = 3/2* ge

As the mass is the same, we conclude that the gravitational force exerted by the unknown planet on the mass (which we call weight) is 3/2 times the one experimented on Earth's  surface.

So, the right answer is D. 3/2W.

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explanation:

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