Question

Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron and are spherical, what diameter in feet must they have for their mutual gravitational attraction to be 1 lb?

Answer 1

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

1. $F=\frac{GMm}{R^{2} }$

Where:

• F => Gravitational force or force of attraction between two masses
• M => Mass of asteroid 1
• m => Mass of asteroid 2
• R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

• M=m

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

• R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

• m=ρV

This is going to be helpful because the volume of a sphere is:

• $\frac{4}{3}\pi r^{3}$

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

• $F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }$
• $F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }$
• $F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2} }{36}$

Now let´s plug in the values we know:

1. $F = 1 lb$     mutual gravitational attraction force
2. $G = 6.67(10)^{-11}$     gravitational constant
3. $ρ_{iron} =491.5 \frac{lb}{ft^{3} }$

• $1= \frac{6.67(10)^{-11} \pi ^{2} r^{4} (491.5)^{2}}{36}$

Solve for r and multiply by 2 because 2r = diameter

• $d=2\sqrt[4]{\frac{1}{7.07(10)^{-5} } }$

Result is d = 21.81 Feet