Question

The distance between two objects is increased by 3 times the original distance. How will this change the force of attraction between the two objects?

The new force will be 1

3 of the original.



The new force will be 3 times more than the original.




The new force will be 1

9 of the original.



The new force will be 9 times more than the original.

Answer 1

Answering the question is not really possible unless you are told what type on interaction there is between the two objects.

If there was a spring connected the where Hooke's law is obeyed, \[F=kx\] then the force is proportional to the separation and increasing the separation by a factor of 3 also increases the force threefold.

However if it was a gravitational attraction between the objects then the force obeys an inverse square law \[F \propto \frac {1}{\text{separation}^2}\]. In this case if the separation increased by a factor of three the force would change by a factor \[\frac{1}{3^2}\].

Answer 2

this type of problem is almost exclusively gravitational potential energy, since it doesn't say the two objects are connected at all

Answer 3

All objects have an inherent gravitational attraction to one another, albeit very very small. Mathematically, two objects have a force of attraction of\[\huge \text{F}=\text{G}\frac{m_1m_2}{d^2}\]From this equation, we can build a proportion that gives us the relationship between the two parameters we're asked for: Force and distance.
\[\huge \text{F} \propto \frac{1}{d^2}\]Which is the proportion that Grovr stated earlier.
Now we're told that the distance is increased by a factor of 3, thus we have:\[\huge \text{F} \propto \frac{1}{(3d)^2}\propto \frac{1}{9d^2}\]In order to make this proportion match the one we created directly from the equation, then we need a 1/9 in front of F. Therefore, the force is 1/9 what it was before if we increase the separation between the two masses by a factor of 3. To check our work, we can continue to simplify\[\huge \frac{1}{9}\text{F} \propto \frac{1}{9d^2}\]\[\huge \frac{1}{9}\text{F} \propto \frac{1}{9}\frac{1}{d^2}\]\[\huge \cancel{\frac{1}{9}}\text{F} \propto \cancel{\frac{1}{9}}\frac{1}{d^2}\]Leaving us with our original proportion as\[\huge \text{F} \propto \frac{1}{d^2}\]

Answer 4

Miscellaneous note: all objects share this relationship with another. We see that if we take the limit as the separation approaches infinity that the attractive force is nearly 0, but still existent. Gravity is universal X)