# Getting 600 n/kg when calculating the gravity for Earth [closed]

## 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.

• $g=\dfrac{GM}{R^{2}}=9.81\ldots$
– Eli
Commented Feb 3, 2023 at 13:56
• I think one problem is because earth is not a perfect sphere(so it's volume is not $\frac{4}{3}\pi r^3$). Commented Feb 3, 2023 at 14:29
• Well yeah, but I dont think that would skew the output by a factor of 60, but still thnaks for the help! Commented Feb 3, 2023 at 14:30
• What's the factor of $62$ for? Were you calculating the weight of $62\operatorname{kg}$?
– J.G.
Commented Feb 3, 2023 at 14:36
• 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. Commented Feb 3, 2023 at 15:32

$$g=-\frac{GM}{r^2}$$
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).