# On Masses Related to a Greater Gravitational Force

Small Introduction

Today I was speaking with my Physics teacher and we ended up discussing the Law of Universal Gravitation, essentially going from the three postulates it has (I am not sure of this affirmation), and the widely known proportionality: $$F\propto \frac{m_1 m_2}{r^2}$$ To the known Newton's equation: $$F=G\frac{m_1m_2}{r^2}$$

Question

Now from that equation I deduced that since the gravitational force is proportional to the product of both masses, then it follows that the greater the mass of a body is, the greater the constant g, but my professor told me that wasn't the case, that it was dependent of the planet. That sounds just wrong, is that correct? Because as far as I can see, the sun has a greater g value than, for example, Jupiter (which has already a greater value than the Earth), and their masses are congruent to their gravitational force.

• It's better to use \propto for $F\propto \cdots$. I read "$F\alpha m$" as "F times alpha times m". – user12029 May 25 '17 at 3:33
• $g=F/m=GM/r^2$. – user126422 May 25 '17 at 3:39
• Willy, precisely because of that equation is my statement, where the Force is proportional to the mass of the planet. – Pablo Ivan May 25 '17 at 22:49
• may be you have a typo? you said your professor said it was dependent on the planet, same as you do? – user126422 May 26 '17 at 1:05
• Ah, sorry, I got confused. I meant "the force is proportional to the product (M)(m)". – Pablo Ivan May 26 '17 at 2:20

It depends on your phrasing. The statement under question is, "the greater the mass of a body is, the greater the constant g". But the real question is which body we're talking about.

The whole idea of having a little "g" is for the common circumstance in our daily lives, where one body is fixed in space and the other object is free to accelerate. Say that $g=G m_1/{R^2}$, $R$ being the radius of the planet and $m_1$ being the mass of the planet (the fixed object). Does "the mass of a body" refer to the fixed object $m_1$ or the moving object $m_2$? If it refers to the stationary object $m_1$ then the original statement is true. If it refers to the moving object then the original statement is false.

So it is possible you are both correct and it was just a miscommunication.

In the case where both objects are free to move, things are much more complicated; you get things like the reduced mass $\frac{1}{\mu}=\frac{1}{m_1}+\frac{1}{m_2}$ and you have to talk in more precise/general terms, so it's good to avoid!

• Yes, by saying body I referred to the planet, yes. – Pablo Ivan May 25 '17 at 22:47

If you want to calculate the acceleration due to gravity standing on the surface of an object, use $F = G\frac{mM}{r^2}$, assuming that $M$ is the mass of the planet and $m$ is my mass. Because $F=ma$ too, the acceleration due to gravity near the surface of the planet ($g$) is due to both the total mass of the planet and the radius of the planet too.

• And precisely because it is proportional to the mass of the planet, I assume the greater mass it has the greater force it will have. – Pablo Ivan May 25 '17 at 22:46

The gravitational acceleration $g$ is dependent on both the mass of the planet and the distance from the center of the planet. A rocky planet and a gas planet of the same mass will have different surface gravities because the "surface" (such as it is) of a gas planet is much farther from the center than the surface of the rocky planet. This is of course reflected in the fact that the acceleration is reciprocal to the center-to-center distance. This is a special case ignoring the mass of one body.