Question

A solid lead sphere of radius10m (about 66 feet across) has a mass of about 57 million kg. If two to these spheres are floating right next to each other (centers 20m apart) in deep space, the gravitational attraction between the spheres is only 540N (about 100 pounds). How large would this gravitational force be if the distance between the centers of the two spheres were tripled?

Answer 1

Answer 2

Calculation:

1. Original Force (F₁):
   Given:

• Distance (r₁) = 20 m

• Force (F₁) = 540 N

2. New Distance (r₂):
   Tripled distance: r₂ = 3 × r₁ = 3 × 20 m = 60 m

3. Relationship Between Force and Distance:
   Gravitational force follows the inverse-square law:

   $$F\propto \frac{1}{r^2}$$

   So, the new force (F₂) is:

   $$F_2=F_1\times {\left(\frac{r_1}{r_2}\right)}^2$$

4. Plug in the Values:

   $$F_2=540\text{N}\times {\left(\frac{20}{60}\right)}^2=540\text{N}\times {\left(\frac{1}{3}\right)}^2=540\text{N}\times \frac{1}{9}=60\text{N}$$

Thus, tripling the distance reduces the force to 60 N

Answer:
The gravitational force would be 60 N (about 11 pounds) if the distance between the centers of the two spheres were tripled.