Question

An object's weight is W0 on the earth.

a. What would be its weight (in terms of W0) if the earth had twice its present mass, but was the same size?

b.What would be its weight (in terms of W0) if the earth had half its present radius, but the same mass?

c. What would be its weight (in terms of W0) if the earth had half its present radius and half its present mass?

d.What would be its weight (in terms of W0) if the earth had twice its present radius and twice its present mass?

Answer 1

F = G*m1*m2/d²

m1 = mass of first body, m2 = mass of second body, d = distance between their centers.

F = G*m_earth/d_earth²

Weight is equal to mass * force due to gravity. so W₀ = m*F

a.

The force due to gravity would be doubled because the value for M_earth would be doubled. , Thus the weight would also double.2*W₀

b.

Assuming the object lies exactly on the edge of the earth's radius, the force due to gravity would be quadrupled (4x). This is because of the algebra that goes on in the denominator(d/2)². 4*W₀

c.

Hals the radius gives 4W₀, but halving the mass would halve the force due to gravity. Putting the two together, you can see that the final force would be equal to 2W₀.

d.

Twice the radius would mean the opposite of what happens whenyou halve it, so the force due to gravity would be cut to onefourth its current value. Twice the mass means that youdouble the force due to gravity. Putting those together meansthat the overall value for the force of gravity is halved. Thus, the answer would be W₀/2