Question

23. Two spheres, with radii of R, are in contact with each other and attract each other with a force of F. If the radii of both of the spheres are cut to half while the density remains the same, what is the new gravitational force between them?

Answer 1

If the radii of both spheres are cut in half while the density remains the same, the new gravitational force between the spheres will be one-fourth of the original attraction.

Step-by-step explanation:

The gravitational force between two objects is given by Newton's law of gravitation: Fgravity = (G x M₁ x M₂) / R², where G is the gravitational constant, M₁ and M₂ are the masses of the two objects, and R is their separation.

If the radii of both spheres are cut in half while the density remains the same, the new masses of the two spheres will also be halved.

Therefore, the new gravitational force between the spheres will be F/4, which is one-fourth of the original attraction.

Answer 2

Drop by a factor of 64

Step-by-step explanation:

By Newtons law of gravity,

`F=(Gm₁m₂)/(r^(2) )` where;

G = Universal gravitaional constant

r = distance between center of gravities of two objects

m₁,m₂ = masses of thee objects.

F = Gravitational force.

m =
`(4)/(3)π r³ρ` (mass = volume into density.

So when radius is halved, mass drops by a factor of 8,

m' =
`(4)/(3)π (r/2)³ρ`

=
`(1)/(6)π r³ρ`

So in substitution to the equation,

`F'=(G(m₁/8)(m₂/8))/(r^(2) )`

`F' = F/64`