-2
$\begingroup$

I have to understand the classic formula for finding acceleration due to gravity's force.

$$ \frac{GmM}{d^2} = m\times9.8\,\mathrm m /\mathrm s^2 $$

dividing the mass of the object out . . .

$$ \frac{GM}{d^2} = 9.8\,\mathrm m /\mathrm s^2 $$

Finding acceleration from the mass x the universal constant and the inverse square of the distance to the center is so incredible and useful but still perplexing. My main question is:

How are the units of Meters per second every second; m/s/s obtained when you are dividing Kilograms of mass? maybe the units of big "G" drives the conversion to m/s/s through the units m3 kg-1 s2 ?

This classical formula is very interesting but very challenging to visualize. The numbers all work but an explanation of the resulting units are difficult to find. Please if anybody is able to visualize the logic behind 9.8 meters a second squared and can share, it would be GREATLY appreciated.

$\endgroup$
  • 2
    $\begingroup$ Its not clear what you are asking. There's nothing magical about 9.8m/s^2... its simply the empirically derived acceleration due to gravity at the surface of the earth. That being said, do you know what the units of G are? I think that if you look up what the units are, the answer to this becomes very trivial. And remember, in the 9.8m/s^2 case, big M is always the mass of the earth, because the "acceleration of gravity" is due to the gravity of the earth (the acceleration due to gravity on other planets is different because their masses are different) $\endgroup$ – Cort Ammon Apr 29 '17 at 3:00
  • $\begingroup$ I've wondered who can be accredited to the empirical work to find 9.8m/s^2 ? If Newton wouldn't have been stumped by G he might have been able to explain this with diagrams. Interesting there still isn't a diagram of this. But, YES, the Units of G do explain this system well. Thanks for your help. $\endgroup$ – Matt Apr 29 '17 at 5:01
  • $\begingroup$ 9.8m/s^2 was found long before Newton. The idea that the distance an object falls is proportional to the square of the time it spends falling dates back to the 14th century (a little bit before Galileo). It's connection to the mass of the earth did not get made until Newton's Law of Universal Gravity, which is the one with G in it. His law explained the motion of the planets better than anything before it. $\endgroup$ – Cort Ammon Apr 29 '17 at 5:34
  • $\begingroup$ It is interesting that the dimensional analysis does not work for the proportion of Force due to gravity with the earths mass and inversely with the distance to the center squared. The dimensionless big G has to cover a lot of bases to balance the equation. $\endgroup$ – Matt Apr 30 '17 at 4:25
  • $\begingroup$ G is not a dimensionless value. Its units are m^3/kg-s^2 $\endgroup$ – Cort Ammon Apr 30 '17 at 4:53
0
$\begingroup$

I think a good way of understanding this is through something commonly referred to as "dimensional analysis."

Replace each variable with its corresponding units:

$ (G)(M)(m)\frac{1}{r^2}=ma$

becomes

$(\frac{m^3}{(Kg) (s^2)})(Kg)(Kg)\frac{1}{m^2}=Kg\frac{m}{s^2}$

Simplification of the left side gives

$Kg\frac{m}{s^2}=Kg\frac{m}{s^2}$

which is clearly true.

Here it becomes important to note that universal constants are often ASSIGNED units in such a way that equations have unit equality, but I understand your desire to have an intuitive understanding of the units.

Take the units of G, and separate them into quantities you can convert into other units.

$\frac{m^3}{(Kg) (s^2)} = (\frac{m}{s^2})\frac{m^2}{Kg}=\frac{N m^2}{(Kg)^2}$

If you think of a very thin bar, it is a line with linear density $d=\frac{Kg}{m}$.

So, $\frac{N m^2}{(Kg)^2}= \frac{N}{d^2}$

So the units of the gravitational constant can be turned into units of force per squared linear density. This may or may not be useful with regards to physical insight, but it is interesting to think about. I've been thinking through this as I've gone along typing, so I'm not sure exactly what "line" d is the linear density of. My intuition would guess that it could be a sort of "density" of energy or something of the like on the line between each body.

The take home here is that the gravitational constant relates the amount of force that is experienced by two objects to how much mass they have and how far apart they are.

As far as 9.8m/s/s goes, because the mass of the earth is SO large compared to most things we might drop, the falling object's mass doesn't play a significant role in determining how quickly objects accelerate. The same logic can be applied to the distance from the Earth.

$\endgroup$
  • $\begingroup$ Thanks Ben! I appreciate you insight in dimensional analysis; perfectly balanced (Hats off to Cavendish for figuring out big "G" and its units!). Although you answered perfectly well, I'll leave the question open to see what additional insight might be shared on the subject. Thanks for helping me out $\endgroup$ – Matt Apr 29 '17 at 4:50
  • $\begingroup$ If it helped you, I'm glad. This is often how I reason out the "meaning" of physical constant. I would encourage this sort of analysis for any constants considering the units are often hard to comprehend in the standard way they are written. In my experience, there is a lot to be learned from dimensional analysis. $\endgroup$ – Ben Apr 29 '17 at 5:22
0
$\begingroup$

What you said about the units of big G driving the units is absolutely correct. Let's step through it using dimensional analysis.

I think it'll be good to look first at how we can simplify the law of universal gravitation.

$$F = m_1a$$

Where $m_1$ is the mass of the object experiencing the acceleration. Let's say $m_2$ is the mass of the body that's causing the acceleration. $$m_1a = \frac{Gm_1m_2}{r^2}$$

Therefore, cancelling the $m_1$ on each side:

$$a = \frac{Gm_2}{r^2}$$

So let's look at the units on the things we know:

$m_1$ is in units kg.

$r^2$ is in units m^2.

So I'm going to write out just the units to see if everything cancels out using the units of $G$, which are $m^3*kg^{-1}*s^{-2}$

Units of $\frac{Gm_2}{r^2}$ are $\frac{(m^3*kg^{-1}*s^{-2}) * kg}{m^2}$

You will then be left with: $\frac{m}{s^2}$ which is indeed the units for acceleration. If you're wondering what the significance of this unit is (i.e. the squared unit term), it's saying how much the velocity (measured in meters per second) is changing per second. So it's (meters per second) per second which we write as $\frac{m}{s^2}$.

$\endgroup$
  • $\begingroup$ Thanks for your clear and concise answer! Really appreciate the help. $\endgroup$ – Matt Apr 29 '17 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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