1
$\begingroup$

By Newton's third law of motion there will be equal and opposite forces of gravity exerted on two objects despite their masses. Hence why does a larger objects exert a larger gravitational pull? In other words why does force of gravity increase with mass?

$\endgroup$
8
  • 7
    $\begingroup$ "Why do people say a larger object exerts a larger gravitational pull?" I guess by "larger" they mean "more massive." Anyways, in that case they say it because it is true. Consider a test mass of mass $m$ placed one meter away from a big mass $M$ versus the same test mass of mass $m$ placed one meter away from a small mass $\mu$. In the former case the force felt by the test mass is big: $G M m$. In the latter case the force felt by the test mass is small: $G \mu m$. $\endgroup$
    – hft
    Aug 18, 2022 at 23:23
  • $\begingroup$ Also consider the accelerations. physics.stackexchange.com/q/454332/123208 & physics.stackexchange.com/q/3534/123208 $\endgroup$
    – PM 2Ring
    Aug 19, 2022 at 0:30
  • 2
    $\begingroup$ @Kuhlambo if I were to hear someone say "the gravitational pull of Jupiter is much larger than that of Earth", I'd assume that to mean that the gravitational force due to Jupiter on some third object is higher than that due to Earth, not that the force of gravity exerted on Earth by Jupiter is larger than that exerted on Jupiter by Earth. $\endgroup$
    – muru
    Aug 19, 2022 at 9:46
  • 1
    $\begingroup$ If by "larger" they mean massive, then you are right by Netwon's third law the force is the same. However, what is meant by "pull" here might not be the force but how they perceive the resulting acceleration, the acceleration would be greater on a smaller mass than on a larger mass. $\endgroup$
    – Mauricio
    Aug 19, 2022 at 9:47
  • 1
    $\begingroup$ How is it that this question needs to be "more focused", please? I see many reasons for at least re-wording, if not discarding it but how does "focus" come into it, anyone? $\endgroup$ Aug 19, 2022 at 19:20

6 Answers 6

10
$\begingroup$

It is just a small semantic confusion.

Newton's third law tells that the two forces below are equal regardless the masses:

two bodies attracting each other

Notice that the two forces are acting on different bodies.

When we say "larger objects exert a larger gravitational pull" we compare them acting on an identical third object.

two bodies attracting an identical third body

(opposing forces still there, but not shown)

$\endgroup$
8
$\begingroup$

The gravitational pull is proportional to both masses: $$F =\frac{G m_1 m_2}{ r^2}$$ So if you increase either $m_1$ or $m_2$ while keeping the other constant then the force will increase.

there will be equal and opposite forces of gravity exerted on two objects

That would correspond to swapping $m_1$ and $m_2$. That would effectively increase the mass of one and decrease the other. So you cannot consider a swap to be just an increase in mass.

$\endgroup$
5
$\begingroup$

There are two different ways of describing gravitation involved here. One is the two-body problem, where there is a force being determined.

The other, is a 'gravitational field' picture, which identifies a single body's gravity contribution in the surrounding space, independent of the presence of other masses.

The gravity field around a large body (Earth) is greater in magnitude (because it is proportional to the mass of Earth) than the gravity field around a cat (which is proportional to the lesser mass of the cat). For practical purposes, like standing upright, we need to accomodate the Earth's gravity but not the cat's. So, we identify the Earth's gravity as the larger influence.

As is the way with felines, the cat will also ignore the gravity due to humans.

$\endgroup$
1
  • $\begingroup$ +1 for cats ignoring people. $\endgroup$
    – Timm
    Aug 20, 2022 at 20:12
4
$\begingroup$

By Newton's third law of motion there will be equal and opposite forces of gravity exerted on two objects despite their masses. Hence why does a more massive object exert a larger gravitational pull?

First the first part, you are correct $ {\vec {F}}_{{A\to B}}=-{\vec {F}}_{{B\to A}}$ should always hold, or we a are in trouble.

Let's say you just let an apple ($m$) fall and look at the forces between apple and earth ($M$). Let's say, away from earth ("up") is the positive direction, so force on earth by the apple is: $$F_{{a\to e}}=\frac{G m M}{ r^2}$$ And force on the apple by earth is $$F_ {{e\to a}} =-\frac{G m M}{ r^2}$$ right?

So, you are right opposite and equal forces apply.

So the other part of the question is also answered, a more massive object like earth doesn't pull more on the apple, than the apple does on the earth, it's actually equal and opposite.

But why doesn't earth rush towards the apple, like the apple does towards it, you ask?

Well, you know Newton's laws so apply the 2nd one here. Calculate the acceleration for each individually and see why one would be much much bigger.

(hint: it's really easy to see, you don't need a calculator, just use the fact that the earth is about $10^{25} $ times heavier... but you can of course ask for help...)

And a bonus question: What would happen if the earth was actually pulling the apple more than the other way around?

$\endgroup$
1
  • $\begingroup$ +1 for mentioning equal forces but non equal accelerations, which is what I think OP was the root of the confusion. $\endgroup$
    – Aubreal
    Aug 19, 2022 at 15:37
0
$\begingroup$

Presumably because of approximation, a body with much larger mass may not appear to move, for example in the case of your mass acting on Earth's.

$\endgroup$
-1
$\begingroup$

Consider the force between two cannonballs.

Now add another cannonball beside the first one. That doubles the amount of force on the third cannonball, because both of the first two are acting on it.

Add a third cannonball beside the first one. Three sets of interactions each like what you got when there was only one.

When there's more mass there's more force. There's always that equal and opposite reaction, but it adds up to more.

Incidentally, imagine that gravity doesn't act instantly but only at lightspeed. Then by the time the force from right now arrives at the other weight, the other weight might have moved. The distance and direction could be different. So maybe it isn't an equal and opposite reaction after all?

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.