Two solid spheres of radius R made of the same type of steel are placed in contact, as shown in the figures above. The magnitude of the gravitational force that they exert on each other is F₁. When two other solid spheres of radius 3R made of this steel are placed in contact, what is the magnitude of the gravitational force that they exert on each other?
(A) F₁
(B) 3F₁
(C) 9F₁
(D) 81F₁
(D) 81F₁
In order to solve this physics problem we need 3 formulas. The first formula is
$$F=\frac{GM_{1}M_{2}}{R^2}$$
This formula is used to find F (the gravitational force between two objects), given the masses of the two objects (M₁ and M₂), and the distance between them (R). The second formula that you need to solve this problem is
$$V = \frac{4}{3} \pi R^3$$
This is the formula for the volume of a sphere given its Radius (R). Remember that the two spheres in this problem are solid and are made of steel. The third formula you need for this problem is the density formula. Right now it might not make a whole lot of sense why we need this to solve the problem, but it will make sense as we continue with the solution. The formula for density is
$$\rho = \frac{M}{V}$$
In this formula Ρ is the density, M is the mass, and V is the volume.
The problem gives you two spheres, made out of steel, each with a radius R as shown in this image:
They tell you that the gravitational force between these two spheres is F₁.
Then they tell you to imagine the same scenario, but with two spheres that have a radius 3R, or 3 times the initial radius.
They then ask you to find the new gravitational force in terms of the old gravitational force. In other words, how many times bigger is the gravitational force between these two larger spheres (radius 3R) compared to the gravitational force between the two smaller spheres (radius R)?
Here's the simple explanation to this physics problem. The two small spheres each have a radius R and a mass M. The two large spheres each have a radius 3R and a mass 27M. By tripling the radius of the small spheres, you get two larger spheres which are heavier and have more mass. Specifically, tripling the radius causes the mass of a sphere to increase 27 times the initial mass. Why is it that multiplying the radius of a sphere causes its mass to be 27 times its original mass? The reason is because both small sphere (radius R) and large sphere (radius 3R) are solid and made of steel. Steel has the same density (Ρ) no matter what size the sphere is. If a sphere with radius R has a volume V, then a sphere with radius 3R has a volume 27V. In order to understand why take a look at the following:
$$\begin{aligned} V &=\ \frac{4}{3} \pi R^3 \\ \\ V_{3R} &=\ \frac{4}{3} \pi (3R)^3 \\ \\ &=\ \frac{4}{3} \pi (27R^3) \\ \\ &=\ 27V \end{aligned}$$
This explains why the larger spheres (with radius 3R) have a volume 27 times (27V) the smaller spheres (with radius R). But why does multiplying the volume of something by 27 cause its mass to get multiplied by 27 also? The answer lies in the density formula:
$$\begin{aligned} \rho &=\ \frac{M}{V} \\ \\ &=\ \frac{27M}{27V} \end{aligned}$$
The equations above show that if you multiply the volume by 27, you need to multiply the mass by 27 because the density (Ρ) of steel doesn't change whether you have a small sphere (radius R) or a large sphere (radius 3R). In other words the density of steel or any object remains constant no matter how much of it you have.
The final part of the problem requires the gravitational force formula:
$$\begin{aligned} F&=\ \frac{GM_{1}M_{2}}{R^2} \\ \\ F_{R} &=\ \frac{G(M)(M)}{R^2} \\ \\ F_{3R} &=\ \frac{G(27M)(27M)}{(3R)^2} \\ \\ &=\ \frac{G(27M)(27M)}{9R^2} \\ \\ &=\ \frac{81G(M)(M)}{R^2} \\ \\ &=\ 81F_{R} \end{aligned}$$
As you can see, the new gravitational force between the larger spheres (with radius 3R) is 81 times the gravitational force between the smaller spheres (with radius R). Therefore the answer is (D) 81F₁