Do We Take Gravity = 9.8 M/s² For All Heights When Solving Problems ...

1. 1. Home
   2. Questions
   3. Tags
   4. Users
   5. Unanswered
2. Teams

   Ask questions, find answers and collaborate at work with Stack Overflow for Teams.

   Try Teams for free Explore Teams
3. Teams

4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams

Teams

Q&A for work

Connect and share knowledge within a single location that is structured and easy to search.

Learn more about Teams Do we take gravity = 9.8 m/s² for all heights when solving problems? Why or why not? Ask Question Asked 13 years, 9 months ago Modified 9 years, 3 months ago Viewed 41k times 8 $\begingroup$

Do we take gravity = 9.8 m/s² for all heights when solving problems?

• newtonian-gravity
• physical-constants
• approximations

Share Cite Improve this question Follow edited Oct 12, 2014 at 22:59 Qmechanic♦ 212k4848 gold badges584584 silver badges2.3k2.3k bronze badges asked Feb 28, 2011 at 5:06 user2312user2312 $\endgroup$ 1

• $\begingroup$ I take it as 10 - it makes doing the approximation head-doable $\endgroup$ – user56903 Commented Jun 19, 2015 at 10:06

Add a comment |

7 Answers 7

Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 18 $\begingroup$

No, the value $9.8\frac{\mathrm{m}}{\mathrm{s}^2}$ is an approximation that is only valid at or near the Earth's surface. You can go a few miles up or down and it'll still be good enough, but once you get any significant distance away from the surface of Earth, you would need to use a different value for gravitational acceleration. You can calculate the value from Newton's law of gravitation, $F = Gm_1m_2/r^2$, and you'll get

$$g = \frac{GM}{r^2} = \frac{3.99\times 10^{14}\ \mathrm{m^3/s^2}}{r^2}$$

where $M$ is the mass of the Earth and $r$ is the distance from the Earth's center to the point for which you are doing the calculation.

Share Cite Improve this answer Follow edited Nov 3, 2014 at 4:06 answered Feb 28, 2011 at 5:14 David ZDavid Z 77.2k2727 gold badges185185 silver badges293293 bronze badges $\endgroup$ 6

• 2 $\begingroup$ It's strange that you would measure the height in miles when the value for gravity is in meters per second squared. $\endgroup$ – LDC3 Commented Nov 3, 2014 at 2:53
• $\begingroup$ Not really; miles are a common unit for height. And it's trivial to convert when needed. $\endgroup$ – David Z Commented Nov 3, 2014 at 2:57
• $\begingroup$ Maybe you could cite that $g = \frac{GM}{r^2} \approx \frac{4 \cdot 10^{14}}{r^2}$ (error: 0.3%). r in meters, g in m/s² $\endgroup$ – André Chalella Commented Nov 3, 2014 at 3:00
• 2 $\begingroup$ Maybe if you are a geologist, but for most scientist, we rarely use anything but SI units. $\endgroup$ – LDC3 Commented Nov 3, 2014 at 3:03
• 1 $\begingroup$ @LDC3 some scientists rarely use anything but SI units, I'm sure, but many branches of science have their own conventional unit systems. In particle physics we use natural units ($c$ and $\hbar$ set to 1), in condensed matter they often use some lattice spacing as a length unit, in cosmology they use megaparsecs or the Hubble radius, and so on. The point is, a qualified scientist is capable of understanding the science regardless of what units are used. $\endgroup$ – David Z Commented Nov 3, 2014 at 4:03

| Show 1 more comment 14 $\begingroup$

To expand a little on David's point assume we move from the nominal "surface" where $g$ is $9.8\text{ m}/\text{s}^2$ to another point at radius $r + \Delta r$. How much does the acceleration of gravity change? $$ g = \frac{GM}{(r+\Delta r)^2} = \frac{GM}{r^2(1 + \Delta r/r)^2} $$ and as long as $\Delta r$ is small compared to $r$ we can reasonably approximate this as $$ g \approx \frac{GM}{r^2}\left(1 - 2\frac{\Delta r}{r}\right) . $$ Well, the radius of the Earth is about $6000 \text{ km}$ so the approximation is good at less than 1% error for around $30\text{ km}$ up or down from the nominal surface, which is all the land and sea floor, and a bit up and down from there.

It is also worth noticing that due to variations in the local mass density of the Earth the measured value of $g$ even at the surface can vary by several tenth of a percent.

Share Cite Improve this answer Follow edited Apr 13, 2017 at 12:39 CommunityBot 1 answered Feb 28, 2011 at 5:25 dmckee --- ex-moderator kittendmckee --- ex-moderator kitten 86.7k99 gold badges175175 silver badges346346 bronze badges $\endgroup$ 2

• 3 $\begingroup$ In fact, the measured variations in $g$ are very useful to geophysicists, oil prospectors, etc. $\endgroup$ – Ted Bunn Commented Feb 28, 2011 at 14:56
• 1 $\begingroup$ I was reading an article on the use of optical lattice clocks today which explained how such clocks allow an even more precise measurement of these changes - useful to evaluate height of water table, prospect for oil and gas, etc. $\endgroup$ – Floris Commented Nov 3, 2014 at 4:18

Add a comment | 6 $\begingroup$

$g$ becomes $ g \approx 9.7 \frac{m}{s^2}$ at a height of about 35km, so it would be ok to use the value $9.81$ for "down to earth" problems.

The relevant wikipedia article has lots of useful information, like for example the following approximation formula for different heights: $$ g_h=g_0\left(\frac{r_e}{r_e+h}\right)^2 $$ Where $g_h$, is the gravity measure at height $h$ above sea level; $r_e$, is the Earth's mean radius and $g_0$, is the standard gravity.

Share Cite Improve this answer Follow answered Feb 28, 2011 at 5:25 EelvexEelvex 1,26011 gold badge1414 silver badges2020 bronze badges $\endgroup$ 2

• $\begingroup$ that's not an approximation, it's exact (as long as you assume earth as a point mass...) $\endgroup$ – Tobias Kienzler Commented Feb 28, 2011 at 8:50
• 2 $\begingroup$ @Tobias: It's an approximation in the sense that it treats earth 1) as a point or a perfect sphere 2) not rotating, etc... $\endgroup$ – Eelvex Commented Feb 28, 2011 at 9:51

Add a comment | 2 $\begingroup$

It might also be worth mentioning that $g$ isn't even constant over the earth's surface at sea level. Depending on the mass distribution and the shape (not perfectly spherical!) of the earth, different parts of the world have different $g$.

Share Cite Improve this answer Follow answered Feb 28, 2011 at 16:10 LagerbaerLagerbaer 15k44 gold badges7676 silver badges8383 bronze badges $\endgroup$ Add a comment | 2 $\begingroup$

The approximation of 9.81 m/s^2 is a generalisation. The exact value is most likely different at a specific location, due to the distance from the centre of the earth to the point being evaluated.

The reference to "surface of the earth" is also a relative since the earth is known not to be perfectly round due to centrifugal forces making the radius greater at the equator.

Also, since the earth is spinning the same centrifugal forces have a slight influence on object mass at the evaluation point.

In metrology laboratories, the exact value for g is displayed for that exact location.

Share Cite Improve this answer Follow answered Aug 19, 2015 at 10:03 DaveDave 2111 bronze badge $\endgroup$ Add a comment | 1 $\begingroup$

Acceleration due to gravity, g is not a universal constant like G. Its calculated by formula mentioned in previous answers. So, for a constant mass system, g depends only on r (distance between center of earth & object in problem). As r = R + h (R is radius of earth & h is height of object from surface) & R is constant, g depends mainly on height. The relation: Increase the height, g will become less (as per formula)

The value 9.8 m/s² is valid for the object at the surface of earth (at sea level). When height is small (with respect to radius of earth), the value is slightly less than 9.8 m/s². So, this variation can be neglected for a high school etc problems. When accuracy is important (due to scientific reasons etc), the value of g can't be 9.8 m/s².

Once again, This consideration is valid only for constant mass system. Plus, for relativistic systems, the formula isn't valid with constant space & time scale.

Share Cite Improve this answer Follow answered Feb 28, 2011 at 15:33 Earth is a SpoonEarth is a Spoon 5,11422 gold badges3232 silver badges4646 bronze badges $\endgroup$ Add a comment | 0 $\begingroup$

As we go above or below the surface of earth the value of g decreases since g is inversely proportional to height

Share Cite Improve this answer Follow answered Jun 19, 2015 at 6:18 aishaish 2122 bronze badges $\endgroup$ Add a comment | Highly active question. Earn 10 reputation (not counting the association bonus) in order to answer this question. The reputation requirement helps protect this question from spam and non-answer activity.

• Featured on Meta
• We’re (finally!) going to the cloud!
• More network sites to see advertising test [updated with phase 2]

Linked

27 Does gravity get stronger the higher up you are on a mountain? 12 Does a cooling object lose mass as it radiates? -7 Why does Earth accelerate every object toward itself with acceleration of $9.8\, m/s^2$ 1 Should chemists at different altitudes factor in the specific gravity value when measuring mass via spring scale? 0 Why is the equation for electric potential energy so counter-intuitive? 1 About Earth Gravity and its characteristics in different scenarios 1 Do all planets rotate around the sun with the same acceleration?

Related

0 From escape velocity to gravitational acceleration 0 Gravity Relative to Mass? 0 How much potential energy is stored in a city's elevators? 2 Is $F=mg$ derived from Newton's law of universal gravitation $F=Gm_1m_2/r^2$? 0 Why does the direction of gravity not change in an accelerating elevator?

Hot Network Questions

• Outlet Wiring Gone Wrong
• Why motion of gas never stops?
• How to demystify why my degree took so long on my CV
• Does Russel's Paradox apply to predicate logic also?
• Alexander's dictum
• relative pronouns and their order in instruction manuals
• Line breaks do not fit well in mathmode
• How can I use a terminal emulator as an efficient application launcher?
• Simple successor gate
• To which extent I should let my PI know that I am not feeling very well with my PhD study
• Where can I find an up-to-date map of Stockholm Central Station that shows the platform layout?
• Why does differentiation with respect to time commute with the Fourier Transform?
• How to inflict self damage anywhere in Fallout 2?
• How do I report to Springer a scientific fraud to a cryptographic paper published in Springer?
• Prospective employers tell me my field is obsolete. How can I reinvent myself?
• Difference between English short stories and short English stories
• If LATERAL is optional for table-valued functions, then why does this query error without it?
• A fantasy movie with two races, "Big Ones" (=us) and smaller ones, about saving a newborn baby from a cruel queen
• Who is the "Sea-queen" mentioned in "Oedipus", and why is she referenced?
• Is this basic string encryption in PHP secure?
• Connectedness of complement of intersection of two balls
• Apple falling from boat mast
• Use of pronoun "en" referring to the subject rather than indirect object
• Use of “12 m.” for noon and “12 p.m.” for midnight

more hot questions Question feed Subscribe to RSS Question feed

To subscribe to this RSS feed, copy and paste this URL into your RSS reader.