Basic Problem:

So, I'm trying to figure out how to calculate the gravitational force of the earth. I am using Desmos to graph the equation so that will explain where $x$ and $y$ come from. Whenever I put in the information for Earth I got around 600 for the gravitational force I got 600 n/kg(which should be 9.8 n/kg).

Equation Details:

I'm using the following equation: $$y=0.000000000066743\frac{\left(\left(\frac{4}{3}\pi x^{3}\right)\cdot5520\right)62}{\left(x+0.8\right)^{2}}$$ Im basing everything on the following equation for gravity: $$F=G{\frac{m_1m_2}{r^2}}$$ The part that says $(43πx3)⋅5520$ is for calculating the mass of Earth based on x, which is the radius. The $(43πx3)$ part gets the volume in $m^{3}$ and then multiples it by $5520$, which is the number of kilograms 1 cubic meter of Earth is. I then put in 62 for $m_2$ since that is the average weight in kg of a human. I then did some research and figured out that G, the gravitational constant, is $0.000000000066743$. Now, to get $r^{2}$, I did $(x+0.8)^2$ since x is the radius, thus the distance from the center of the Earth to the crust, and then added 0.8 since that is half the average height of a human.

What Have I Tried:

I have tried checking if my density is correct by multiplying it by the volume of the Earth and it was correct. I can't find the gravitational constant from another source so that could be a possibly incorrect thing. I double-checked that my volume equation was correct and I also checked that I'm using the right units of measurement. Thanks for any help and feel free to ask questions about any equations/anything in general.

  • 1
    $\begingroup$ $g=\dfrac{GM}{R^{2}}=9.81\ldots $ $\endgroup$
    – Eli
    Commented Feb 3, 2023 at 13:56
  • $\begingroup$ I think one problem is because earth is not a perfect sphere(so it's volume is not $\frac{4}{3}\pi r^3$). $\endgroup$
    – MpH81679
    Commented Feb 3, 2023 at 14:29
  • $\begingroup$ Well yeah, but I dont think that would skew the output by a factor of 60, but still thnaks for the help! $\endgroup$ Commented Feb 3, 2023 at 14:30
  • $\begingroup$ What's the factor of $62$ for? Were you calculating the weight of $62\operatorname{kg}$? $\endgroup$
    – J.G.
    Commented Feb 3, 2023 at 14:36
  • $\begingroup$ You've calculated the F os gravity on a 62kg object on the earth's surface. Remember that F = ma, so a = F/m. To get 9.8, which is the acceleration, you need to divide the force by 62g to get the acceleration of the 62kg object. This is 60/62 which is close to 9.8. Simplifying the procedure. you get the equation in @Eli's comment above. $\endgroup$ Commented Feb 3, 2023 at 15:32

1 Answer 1


The gravitational acceleration can be calculated from the formula below:


Where G is the gravitational constant, $6.6743 × 10^{11} m^3 kg^{-1} s^{-2}$ (it is better to express really small or large quantities in scientific notation), $M$ is the mass of the planet, and $r$ is the radius of the planet. Thus,

$$g=-\frac{(6.6743 × 10^{-11} m^3 kg^{-1} s^{-2})(5.97219 × 10^{24} kg)}{(6.3781 × 10^6m)^2}≈-9.8 m/s^2$$

Thus, the gravitational acceleration on Earth is about $9.8 m/s^2$. To find the gravitational force on an object on the surface of the earth, multiply this number by the mass of the object (in kilograms). For instance, and object that weighs $2 kg$ has a graviation pull of about 9.6 Newtons (of force).


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