How would you relate the respective solar gravitational pulls of two planets? I was asked this question:


  How much weaker is the gravitational pull of the Sun on Neptune at 30 AU, than it is on Earth at 1 AU?


How would I even begin to determine this?
For the most part, you can assume Newton's law of universal gravitation, (see link for more information). Basically, this law can be stated as the following.

For the gravitational force between two celestial objects (or really, any two objects with mass), the force is given by the equation,

$$ F_G = G\frac{m_1 m_2}{r^2} $$

Given the mass of the two objects, $m_1$, $m_2$, and the distance between them, $r$. The constant of proportionally given by $G$, the gravitational constant (not to be confused with $g$, the acceleration of an object with respect to the conditions on the surface on the Earth).

For the specific problem(s) that you are given, you can say that, using the information from Wikipedia, for the mass of the Sun, Earth, and Neptune,

$$ m_\text{Sun} = 1.988 \times 10^{30} \text{ kg} $$
$$ m_\text{Earth} = 5.972 \times 10^{24} \text{ kg} $$
$$ m_\text{Neptune} = 1.024 \times 10^{26} \text{ kg} $$

Your question gives us a preliminary distance. The $r$ for the distance between the Sun and the Earth is given as $1 \text{ AU}$, the $r$ for the distance between the Sun and Neptune given as $30 \text{ AU}$. Note that $1 \text{ AU} = 1.496 \times 10^{11} \text{ m}$. 

Therefore, for the gravitational pull (really another way of saying the gravitational force) between the Sun and Neptune, 

$$F_{N \leftrightarrow S} = \left( 6.674×10^{-11} \text{ N m$^2$ kg$^{-2}$} \right) \frac{\left( 1.024 \times 10^{26} \text{ kg} \right) \left( 1.988 \times 10^{30} \text{ kg} \right)}{\left( 5.983 \times 10^{12} \text{ m} \right)^2} $$

Therefore, for the gravitational pull between the Sun and the Earth, 

$$F_{E \leftrightarrow S} = \left( 6.674×10^{-11} \text{ N m$^2$ kg$^{-2}$} \right) \frac{\left( 5.972 \times 10^{24} \text{ kg} \right) \left( 1.988 \times 10^{30} \text{ kg} \right)}{\left(1.496 \times 10^{11} \text{ m}\right)^2} $$

Solving for these quantities (note that the answer that one might get may be numerically different depending on the sources of the planatary constants),

$$F_{N \leftrightarrow S} = 3.796 \times 10^{20} \text{ N}$$

$$F_{E \leftrightarrow S} = 3.541 \times 10^{22} \text{ N}$$

A bit of extra information, the gravitational force felt by the Earth or Neptune from the Sun is exactly the same as the force that the Earth or Neptune acts upon the Sun, as per Newton's third law of motion. But, by now, I suspect that the problem is answered to where taking the ratio of the two is a trivial manner.

Suggested by uhoh:
The force exerted by the Sun on Neptune (and vice versa) is about 93.3 times weaker than that of the force exerted by the Sun on the Earth(and vice versa).