Finding density for identical spheres

[Tiven white]

Homework Statement

Two identical spheres of diameter 8.55m are on the surface of the earth. The spheres are touching each other. What is the minimum density the spheres must have such that the gravitational force between them is at least equal to the weight of one of the spheres.

Homework Equations

F= m1*a = (G*m1*m2)/(r^2)
Newtons law of universal gravitation.

The Attempt at a Solution

I used the value for the acceleration of the Earth and multiplied it by the square of the radius between the spheres then divided the result by G to obtain m the mass m of one of the spheres was then divided by the volume of the sphere calculated from ((4/3)*Π*r^3) though this gave a value too high for the density of the sphere the value was like 3*10^10 kg/(m^3) higher than even the density of the planet is there any other insight to an attempt or solution to this problem it would be greatly appreciated.

 

[Dick]

Homework Statement

Two identical spheres of diameter 8.55m are on the surface of the earth. The spheres are touching each other. What is the minimum density the spheres must have such that the gravitational force between them is at least equal to the weight of one of the spheres.

Homework Equations

F= m1*a = (G*m1*m2)/(r^2)
Newtons law of universal gravitation.

The Attempt at a Solution

I used the value for the acceleration of the Earth and multiplied it by the square of the radius between the spheres then divided the result by G to obtain m the mass m of one of the spheres was then divided by the volume of the sphere calculated from ((4/3)*Π*r^3) though this gave a value too high for the density of the sphere the value was like 3*10^10 kg/(m^3) higher than even the density of the planet is there any other insight to an attempt or solution to this problem it would be greatly appreciated.

Your answer is in the right ballpark and I think you are doing the right thing, but I don't think it's quite right. Can you show your work. The density will have to be huge? Gravity is a very weak force.

 

[Tiven white]

OK so where a= (G*m)/(r^2)
9.81 = ((6.67*10^-11)*(m)) / (8.55^2)
So (9.81)*( 8.55^2)/(6.67*10^-11) = m
So m = 1.08 * 10^13 kg


V =( 4/3)*(3.14)*(8.55/2)^3
V = 3.27 * 10^2 (m^3)
So density = (1.08* 10^13) ÷(3.27*10^2)
This implies the density =( 3.3 * 10^10 kg/m^3)

 

[Tiven white]

Could u verify the working as u had asked.for.a sample of the calculations

 

[haruspex]

OK so where a= (G*m)/(r^2)
9.81 = ((6.67*10^-11)*(m)) / (8.55^2)
So (9.81)*( 8.55^2)/(6.67*10^-11) = m
So m = 1.08 * 10^13 kg


V =( 4/3)*(3.14)*(8.55/2)^3
V = 3.27 * 10^2 (m^3)
So density = (1.08* 10^13) ÷(3.27*10^2)
This implies the density =( 3.3 * 10^10 kg/m^3)

Looks right.

 

[Dick]

Looks right.

I agree. I was forgetting to divide the diameter by 2 to get the radius.

 

[Tiven white]

So I was skeptical about why the spheres had to have such a huge density could I hear an explanation why

 

[Dick]

So I was skeptical about why the spheres had to have such a huge density could I hear an explanation why

It's what I said in the first response. It's because G is so small. It's so small it's difficult to even measure using laboratory scale objects, http://en.wikipedia.org/wiki/Gravitational_constant#History_of_measurement It shouldn't be surprising that to get an amount of acceleration comparable to the acceleration produced by the whole Earth you'd need a lot of mass. Hence huge density.