0
$\begingroup$

I'm really new to science and physics so I apologize in advance if this question is too easy to be asked on this site. I'm retaking high school physics and and I'm using a site that claims covers the whole curriculum of grade 10.

I was always told that gravity = 9.8 m s$^{-2}$ but I recently saw a website where they state that the gravitational acceleration is 9.8 m s$^{-1}$ and I'd just like some clarity on if they're wrong or right or how this works because nowhere on Google can I find anything related to the negative one? And seeing as I just started out I don't want to follow a site for the next two years that's gonna teach me a lot of wrong information. Here's a link to where I found it on the site, it's fairly close to the end of the page, the last row in the table under the heading "Constants in equations".

Any clarification would be appreciated.

$\endgroup$
5
  • $\begingroup$ It's not perfect, but when in doubt at least check Wikipedia. As a basic check consider the units you expect to see (in this case for acceleration) against the ones actually used (in your example a typo). $\endgroup$
    – StephenG
    May 10 '17 at 16:24
  • $\begingroup$ It should be 9.8 m s^-2. And you should make an effort to grasp the concept of acceleration so that you can answer why it is "m s^-2" and not "m s^-1" or "m s^2". $\endgroup$
    – user93237
    May 10 '17 at 16:24
  • $\begingroup$ So the website made a mistake? Or can it be somehow converted to 9.8ms^-1 ? I actually only wanna know if they made a mistake or not coz I'm inclined to believe that if they made a mistake with this then I should probably find another resource for my studies as there might be more mistakes down the line and I'll end up studying the wrong stuff $\endgroup$
    – SeriousLee
    May 10 '17 at 16:28
  • 4
    $\begingroup$ Yes, the website has a typo and, no, "m s^-2" cannot be converted to "m s^-1". $\endgroup$
    – user93237
    May 10 '17 at 16:41
  • $\begingroup$ Bear in .mind that it's not easy to create a work that is 100% free of typos, even the very best textbooks may have some, especially in the early editions, so you need to be on your toes and apply your critical thinking skills while reading. $\endgroup$
    – PM 2Ring
    May 5 '18 at 6:02
2
$\begingroup$

The website made a mistake.

Every object subject to a constant force will accelerate uniformly in time, having a characteristic acceleration which must have SI units $\text m/\text s^2,$ which is to say that "acceleration is a rate of change in velocity per unit time."

Now, generally a force will only impact an object inversely as its mass increases, $a = F/m,$ so if you kick a big massive object and a little object exactly the same, the little object will start to move much faster than the big one. This equation means that the units of force are $\text{kg m}/\text s^2.$

Gravity near the surface of a planet happens to be an approximately constant force for any single object, because the planet's radius is so much larger than the heights that the object is ascending or descending and the size of the object and so forth. However the force of gravity does scale with the mass of that object, $F = m g$ for some constant $g$ that depends on the mass of the planet and its radius, but not on the falling objects.

Since $m$ has units $\text{kg},$ this constant must have the units $\text m/\text s^2,$ and since $a = F/m,$ it is the acceleration that every object on that surface feels due to gravity. That's actually very important: it means that if you fill up a plastic water bottle 1/4 of the way full, and another all of the way full, even though one of them has 4 times the mass and experiences 4 times the force, those two effects perfectly balance out and they both fall exactly the same. If you have the plastic bottles to spare, feel free to do this experiment several times at home, releasing them side-by-side and confirming that they both hit the ground at the same time.

If we were to modify $g$ to have units $\text m/\text s$ then we would probably have to modify Newton's laws to say not $a = F/m$ but rather $v = F/m,$ and this would be disastrous: it would mean, for example, that you could not throw a ball upwards because the moment it left your hand it would have to have negative velocity. We would then have to postulate new "forces" to account for the fact that balls can be thrown in practice, call it an "inertial force" or whatever, that tries to keep it moving the direction it's moving... that's why Newton's laws are so very useful, because they don't require this and you can just say "once that ball leaves your hand, the only forces on it are gravity and wind resistance."

$\endgroup$
2
  • 1
    $\begingroup$ Your first sentence says it all. The rest of your answer is unnecessary. $\endgroup$ May 10 '17 at 16:55
  • 4
    $\begingroup$ Have you seen my answers? I like including unnecessary things. $\endgroup$
    – CR Drost
    May 10 '17 at 16:56
1
$\begingroup$

Do you the Newton law of gravitation Which states that 1)Force of attraction between two masses is directly proportional to the product of masses. 2)This force is inversely proportional to square of distance between them. So,let us say for an object residing on earth the Force applied on it by earth is $$F_{earth}=\frac{GMm}{R^2}$$(where $M$ mass of earth and $m$ mass of the or man on earth)And the $R$ is the radius of earth. So $F=ma$ Again by newtons second law of motion So $F_{earth}=mg$ Putting this $$mg=\frac{GMm}{R^2}$$ $$g=\frac{GM}{R^2}$$ Where the unit of the Gravitational constant can be found out by Unit of force is $N$ Masses are in kg and radius is in "m" So G From $$G=\frac{F_{earth}R^2}{Mm}$$Hence the unit is $\frac{Nm^2}{kg^2}$ now Putting this in the gravitational acceleration we get $$\frac{N}{kg}$$ which is again we know $F=ma$ so $a=N/kg$ $$a=(\frac{kgm}{s^2})/(kg)$$ which resolves out to give

$$m/s^2$$

$\endgroup$
1
  • $\begingroup$ You should show how N/kg cancels down to m/s^2. $\endgroup$ May 10 '17 at 16:51

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.