# Is force of gravity the same in the water?

We know that $$F_n = m*a$$ where $$F$$ is a net force.

From this equation and gravity law, we can get $$a = g=9.8$$ for which case we mention $$F = m*g$$.

But This is only true when an object is falling and there's no upward force exerting on it. The reason I'm saying this is we derive it from the following: $$\large m * a = \frac{G * M * m}{r^2}$$

Note that if there was any other force exerting on an object upward, the above equation wouldn't be true, as the left side of the equation is the net force, while the right side is only downard force. So we conclude that F = mg is only true when there's only gravity force exerting on an object downward

Now, imagine a can of beans inside a water. We conclude that downward force is gravity and equal to m*g. This seems to me wrong as in order to conclude that downward force is 9.8*g, we had to assume that net force is only downard, but we know there's boyant force there as well. So isn't it wrong we say the downward force of the can is 9.8 * m ?

• Re, "we conclude that $F=mg$ is only true when there's only gravity force exerting on an object" That depends entirely on what you think $F$ means. Physical laws only make sense if you interpret them in the way that they were meant to be interpreted. $F=mg$ is always true if you think that $F$ is the force due to gravitation between some object and the Earth and, that $m$ is the mass of the object and, that the object is close to the surface of the Earth. If, on the other hand, you think that $F$ represents the sum of all forces acting on the object, then $F=mg$ will not always be true. Apr 28, 2023 at 19:26
• Well, I meant that to derive a, people do ma = GMm/r^2 and from this, they get a = 9.8. Note that they do it as if ma is treated as downward force while there's no upward force. In that case, it's correct. but we know ma is the net force. in a water, where you got object(and Imagine, i want to calculate a here), I can't presume upward force to be 0. That was my confusion. I guess, we just presume that upward is 0 and we calculate a for it and it becomes the same even in a water
– Matt
Apr 28, 2023 at 20:18
• There is no need to use $*$ to indicate multiplication. Surely your physics textbook doesn’t do that. Apr 28, 2023 at 21:25
• That's why $g$ is defined as : Near the surface of the Earth, an object in free fall in a vacuum will accelerate at approximately 9.8 m/s^2. Notice "free fall" and "vacuum". And yes, force of gravity acts same in every medium, i.e. is invariant to medium change. However, other forces may not, such as drag force, buoyancy and etc. May 2, 2023 at 21:02

The point is that $$g=9.8$$ m/s$$^2$$ is a parameter that tells us the strength of the gravitational force on an object of mass $$m$$ near the surface of the Earth, via $$mg$$. At this point, despite having units of acceleration, $$g$$ is not an acceleration: it's just a number that tells us how to compute the strength of the gravitational force.

If gravity is the only force acting, then the acceleration of the object is $$g$$, because $$a = \frac{F_{\textrm{net}}}{m} = \frac{F_G}{m} = \frac{mg}{m}=g\,.$$ However, if there are other forces acting, then the acceleration will be different than $$g$$, because $$\vec{a} = \frac{\vec{F}_{\textrm{net}}}{m} = \frac{\vec{F}_G + \vec{F}_{\textrm{other}}}{m}\,,$$ which won't in general have a magnitude of $$g$$.

So, then, where does $$g$$ come from? It comes from an approximation of the gravitational force that we make when we're near the surface of the Earth.

Here's the whole story. As long as we're near the surface of the Earth, we can approximate the gravitational force in the following way. Suppose $$y$$ is the height of an object above the Earth's surface, and $$y$$ is much smaller than the radius $$R$$ of the Earth. Then the gravitational force exerted by the Earth on an object of mass $$m$$ is $$F_G = G\frac{m_1M}{(R+y)^2}\,,$$ where $$M$$ is the mass of the Earth. We can expand this function in a Taylor series about $$y=0$$ (or, equivalently, we can use the binomial theorem on $$(R+y)^{-2}$$) to get $$F_G = G\frac{m_1M}{(R+y)^2} \approx G\frac{M}{R^2} m - 2G\frac{M}{R^2}\left(\frac{y}{R}\right)m + 3G\frac{M}{R^2}\left(\frac{y}{R}\right)^2m+\cdots\,.$$ Since $$y$$ is much smaller than $$R$$, $$y/R$$ is much smaller than 1, as is any power of $$y/R$$, so we can safely neglect all but the first term as an approximation. We are left with $$F_G\approx \frac{GM}{R^2}m\,,$$ and if you compute $$GM/R^2$$ for the Earth, you get about $$9.8$$ m/s$$^2$$. Thus, as a model for how the gravitational force works, we can write $$F_G \approx mg\,.$$

As you can see in this derivation, we never said anything about an acceleration (equivalently, we didn't say anything about the net force on an object). The parameter $$g$$ just isn't the acceleration of an object when we write down $$F_G=mg$$; that is not the role of $$g$$ in this equation. Instead, the role of $$g$$ is to determine the strength of the gravitational force on an object near the surface of the Earth. Then, we can use this to compute the acceleration of an object, and, as above, in the special case where $$F_G$$ is the only force acting, it turns out that $$a=g$$.

• Excellent answer, as it directly addresses what seems to be the key confusion: A body's acceleration is not generally $g$, even though that parameter has units of acceleration. The derivation of $g$ is also a handy review. Apr 28, 2023 at 21:24
• You mention the following: "So, then, where does 𝑔 come from? It comes from an approximation of the gravitational force that we make when we're near the surface of the Earth." - That means you make another assumption that there is no force acting upwards on an object, otherwise, you wouldn't get g = 9.8. That was my confusion
– Matt
Apr 28, 2023 at 21:41
• @Matt. No, I don't agree. In fact, that's sort of the whole point of my post. I am not calculating the net force on an object. All I am doing is "rewriting" the gravitational force in a different way. Other forces can still act. That doesn't change the value of $GM/R^2$, because how could it? Re-read my post and see if you can see what I mean here. This idea about thinking about what other forces are acting is a red herring. The question is, "How do I write down the gravitational force?" and the answer is $mg$ as long as we're near the surface of the Earth. Apr 28, 2023 at 21:49
• You say: "If gravity is the only force acting, then the acceleration of the object is 𝑔" and "However, if there are other forces acting, then the acceleration will be different than 𝑔". So thats the point. in your comment, you say: "Other forces can still act. That doesn't change the value of 𝐺𝑀/𝑅2". True but if other forces act, g won't be correct.
– Matt
Apr 29, 2023 at 5:18
• @Matt. Everything you've said is correct. So I'm not sure what your issue is. Unless: Your last sentence is "g won't be correct". If you mean, "g isn't the acceleration of the object", than you are correct, but if you mean "we can't use g to calculate the gravitational force", then you are wrong. Which did you mean? Apr 30, 2023 at 15:51

You have two forces acting on the can: gravity point downwards and buoyancy pointing upward. The net force on the can is the sum of the two. If the can floats the net force is zero. The can sinks to a point where the buoyancy force is equal and opposite to the gravity force.

So isn't it wrong we say the downward force of the can is 9.8 * m ?

No, that's a correct way of saying it. The downward force is 9.8m and the upward force is -9.8m and hence the net force is zero.

• Thanks for the answer. Though, I'm curious how you derive 9.8 while you're in a water. I showed you how I derived it while object starts to fall down. if you do the same while you're in a water, ma = G M m / r^2 won't be true as I said in my question, hence how would you get a to be 9.8 then ?
– Matt
Apr 28, 2023 at 19:24
• a = 9.8 m/s^2 only applies to the gravitational force. They buoyancy force depends the immersed volume, the density of the can and the density of the immersing liquid. However, if it floats (and ONLY if it floats), you know that buoyancy force must be equal and opposite of the gravitational, so you can just use this instead. Apr 29, 2023 at 19:35

You say net force $$F_{net}=ma$$ which is how it's often introduced in very early physics. A much better way to state Newton's 2nd Law is the "sum of the forces" equals $$ma,$$ or

$$F_1+F_2+F_3+...=\sum F = ma$$

You add all forces in a given direction, and the sum of those equals the mass times acceleration in that direction. I think this may help with your confusion.

The force of gravity

$$F_g = m \left(\frac{GM_{Earth}}{R_{Earth}^2}\right) = mg$$

is only one of the possible forces that would be included in that sum.

• Well, if an object is in a water, how would you calculate a ? you would write F(downward) + F(upward) = ma. Then no way you can calculate a. It seems even if object is in a water and you want to calculate a, you treat F(upward) to be 0 . This was my question. why treat upward force 0 ? otherwise you wouldn't get a=9.8.
– Matt
Apr 28, 2023 at 20:15
• No, $F_{upward}$ is a buoyancy force, which is equal to the weight of the water displaced. If the object is moving, there is an additional force of fluid drag. You must find expressions or values for all of the summed forces if you want to solve for the unknown $a$. Apr 28, 2023 at 21:54

There is are already great answer above , but when you write F= GMM/ R^2 what you are calculating is not the net force, you are only calculating force due to gravity (only anf force due to gravity).

and it is F=ma

so m*a( where a is acceleration due to gravity)= GMm/r^2

You cannot calculate the net force via the equation GMm/R^2. It is only the force due to gravity that you are calculating. Hence ma=Gmm/r^2 where a=g=acceleration due to gravity.