2
$\begingroup$

Objects intrinsically resist to be accelerated due to their masses. A clear example would be kicking a soccer ball vs kicking a bowling ball. The latter ball will resist much more to be accelerated than the first one due to its greater mass (intrinsic property).

What if we position them in a inertial frame of reference in space? If we push both previous balls with the same force we will obtain different accelerations due to the balls' different masses, isn't it?

$\endgroup$
2
  • $\begingroup$ The answer is simply yes. I can't get the real question. $\endgroup$
    – Yashas
    Aug 6 '17 at 3:36
  • $\begingroup$ I was curious about the differences between acceleration due resulting force (i.e. someone pushing a stroller; acceleration depends on stroller's mass) and acceleration due gravity (acceleration does not depend on mass object) $\endgroup$
    – JD_PM
    Aug 6 '17 at 15:28
2
$\begingroup$

Because measure of inertia (mass $m$) and the gravitational charge (mass $m_g$) happen to be the same.

That's a good question, actually, even if not that clearly formulated.

From Newton's 2nd law, $F=dp/dt$, with constant net force and mass, one gets

$$a=F/m$$

And Newton' gravitation law says that $F_g= GMm_g/R^2$, so for $F=F_g$, then

$$a=\frac{GM}{R^2}\frac{m_g}{m}.$$

The "gravity" you have in mind probably the one we feel in day to day, close to the surface of Earth, which means $R\approx \mathrm{const}$ and

$$a\propto\frac{m_g}{m},$$

from where you get that, if $m_g=m$, then $a$ is a constant: which we commonly denote by $g$.

$\endgroup$
4
  • $\begingroup$ Okey, a doubt came to my mind. I want to compare an elevator uniformly accelerated in space and an elevator subjected to g. As a = g, theoretically we could not distinguish between both frame of references. But thinking, you realize there is gravitational differencial in earth's elevator. The top of the elevator is slightly farther from earth's center than the elevator's floor. Consequently, the floor will experience a slightly major gravitational force comparing with the top of the elevator (tidal force). So what can we do with tidal force factor? Discard it when it is not significant? $\endgroup$
    – JD_PM
    Jul 30 '17 at 18:30
  • $\begingroup$ And I will improve the way I formulate my questions, thank you for the advertisement $\endgroup$
    – JD_PM
    Jul 30 '17 at 18:31
  • $\begingroup$ @JD_PM, without putting too much thought into it, it seems to me you can get this tidal effect by rotating the elevator (not spinning) with a big radius, with its top pointing to the center of rotation. But that's a good question, maybe you should actually post it separately. By the way, do you know that your mental experiment is a usual way to introduce General relativity? That's the theory that gives a deeper explanation to the otherwise pretty strange coincidence of $m_g=m$. $\endgroup$
    – stafusa
    Jul 30 '17 at 18:38
  • $\begingroup$ Yes you are right! In fact, this experiment is used to explain equivalence principle. My doubts are focused on how tidal force affects equivalence principle, but this is a GR topic. Thank you for editing my question $\endgroup$
    – JD_PM
    Jul 30 '17 at 20:04
1
$\begingroup$

Although your question isn't clear, I'll try answering what I've understood from your question. In your question, you have assumed force is constant. In case of constant force, yes acceleration will vary inversely with mass, i.e., as you say, kicking a bowling ball will produce lesser acceleration than a soccer ball. However, in the case of acceleration due to gravity, the force is not the same for both the balls. Instead, it's greater for the ball with greater mass and lesser for the ball with lesser mass. (The acceleration is given by GM$_e$/R$^2$ and as you can see it doesn't depend on the mass of the ball.) The net effect is that the acceleration is the same for both balls. Why this is so can easily be derived from Newton's laws of Gravitation - which I'll leave to you. Hope this answers your question.

$\endgroup$
1
  • $\begingroup$ Your answer is useful! My doubts were more on why acceleration due gravity does not depend on body's mass. With your answer and some research now the issue is solved. It is about understanding how gravity acts in a body: F = mg. As we can consider earth as an inertial frame of reference ( although it is true earth surface is accelerated, but just on the order of 0.01), we can use Newton's second law F = ma. So mg=ma being g = GMe/R^2. So now I have clear why acceleration due gravity does not depend on object's mass $\endgroup$
    – JD_PM
    Jul 30 '17 at 17:58
0
$\begingroup$

Well, the thing is that the gravitational force the earth or any other body exerts on another is given by gM, where g is the acceleration due to gravity for that body, a constant for any given body and M is the mass of the other body. In case of the earth g* equals g and hence the Force of gravity= Mg and hence, the acceleration of any body of mass M is Mg/M= g. You see, this does obey Newton's second law. Hope this helps you.

$\endgroup$
0
$\begingroup$

If we push both previous balls with the same force we will obtain different accelerations due balls' masses isn't it?

Yes. In any inertial frame we can count on Newtons 2nd law:

$$\sum F=ma$$

The masses resist acceleration, so the larger mass resist acceleration more when the forces are equal.

Why acceleration due resulting force depend on mass and acceleration due gravity not?

You can turn it around and ask yourself: why would it? Remember that the gravitational force is different on different masses. Half the mass only has half the weight, so it is also only pulled in half as much by gravity.

View it like this: What gravity does is to pull in every single "particle" equally. If there are double as many "particles", then the pull in each is still the same and each accelerates the same amount.

$\endgroup$

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