# If gravity is independent of mass, why does Jupiter pull an object more than the Earth does?

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Gravity and Mass

## Main Question or Discussion Point

Objects fall on Earth at 9.8 m/sec² independent of mass. If gravity is independent of mass why does Jupiter pull an object more than the Earth does? Is that inconsistency within the law or in my perception?

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fresh_42
Mentor
Summary:: Gravity and Mass

Objects fall on Earth at 9.8 m/sec² independent of mass.
That's not true. The object pulls on earth, too, and the bigger the mass the more it pulls. However, every single mass is so small in comparison to earth, even the big ones, that we cannot measure a difference between a feather and an iron ball. But if the mass is on planetary scale, things change.

And even if we have only one planet and a feather, then the acceleration on Jupiter is no longer $9.81$, it is $24.9$ because Jupiter is heavier. (It would be even more, if Jupiter were smaller.)

Dale
Mentor
Summary:: Gravity and Mass

If gravity is independent of mass why does Jupiter pull an object more than the Earth does?
The force of gravity is not independent of mass. It is proportional to mass.

Ibix
If gravity is independent of mass why does Jupiter pull an object more than the Earth does?
The acceleration due to gravity does not depend on your mass (you can't just say "gravity" - as Dale notes that could mean many different things, not all of which are independent of your mass). However, it does depend on Earth's mass. Or Jupiter's mass if you were on Jupiter. So it's different on different planets.

sophiecentaur
Gold Member
So it's different on different planets.
The gravitational force (between two masses) depends on the separation of the masses (inverse square law etc.). But you have to go a long way above the Earth's surface before the separation between you and the Earth changes significantly compared with the Earth's radius.
It's worth noting that the gravitational force on satellites in low Earth orbit is virtually the same as it is on the ground. Astronauts in the ISS are subject to the same gravity as we are, down here; it's what keeps them in orbit.

I think the OP is getting confused with the speed at which an object is falling vs the force.
Looking at the gravitational force equation
F = GMm/r2
Here G is the Gravitational Constant, M is the mass of the bigger object (say our Earth), r is the distance (for example the radius of the Earth). All three of them, G, M, and r do not change considerably on the Earth.
So, we write
g = GM/r2 that is the acceleration due to gravity. That is nearly 9.8m/s2 on the surface of the Earth.
THen the force equation on Earth could be written as
F = mg

You could see that the force depends upon the mass of the smaller object, say a stone.
Whereas the acceleration and the speed at which an object falls down does not depend upon the mass of the object.
There are three assumptions overall to simply everything.
- M (mass of the Earth) is extremely large compared to m (mass of the object)
- All velocity and acceleration equations ignore the effect of the Air Drag. Then only, a feather or a stone could fall down at the same speed
- Change of the acceleration due to gravity is very small as the distance travelled is very less compared to the radius of the Earth.

rcgldr
Homework Helper
A more generic way to describe Anand Sivaram's posts. If you have two masses, m1, and m2, and use the common center of mass of m1 and m2 as a reference frame. Then the rate of acceleration of m1 towards the common center of mass will be due to the intensity of the gravitational field of m2: $-G \cdot m2 / r^2$, and the acceleration of m2 towards the common center of mass will be due to the intensity of the gravitational field of m1: $-G \cdot m1 / r^2$ , and the rate of acceleration of both masses towards each other is $-G \cdot (m1 + m2) / r^2$ .

vanhees71
Gold Member
2019 Award
The problem is of course most simply described in the center-of-mass frame, where $\vec{R}=(m_1 \vec{x}_1+m_2 \vec{x}_2)/(m_1+m_2)=0$, where $\vec{x}_1$ and $\vec{x}_2$ are the positions of the Sun and the planet in this frame. Then we define the relative coordinates as $\vec{r}=\vec{x}_2-\vec{x}_1$. Then the equation of motion for $\vec{r}$ reads
$$\mu \ddot{\vec{r}}=-\frac{G m_1 m_2}{r^3} \vec{r}, \quad \mu = \frac{m_1 m_2}{m_1+m_2}.$$
Indeed the acceleration of the relative vector pointing from the Sun to the planet thus is
$$\ddot{\vec{r}}=-\frac{G(m_1+m_2)}{r^3} \vec{r}.$$
Solving this equation of motion leads to an ellipse for $\vec{r}$ with one of its foci in the center of mass (the origin of the center-of-mass frame).

The positions of the Sun and the planet are then
$$\vec{x}_1=-\frac{m_2}{m_1+m_2} \vec{r}, \quad \vec{x}_2=\frac{m_1}{m_1+m_2} \vec{r}.$$
For $m_1 \rightarrow \infty$ you get $\vec{x}_1 \simeq 0$ and $\vec{x}_2 \simeq \vec{r}$.

vanhees71
Gold Member
2019 Award
Sure since $g=G M_{\text{Earth}}/R_{\text{Earth}}^2$ the gravitational acceleration $g$ depends on the mass of the Earth.

ZapperZ
Staff Emeritus
Summary:: Gravity and Mass

Objects fall on Earth at 9.8 m/sec² independent of mass. If gravity is independent of mass why does Jupiter pull an object more than the Earth does? Is that inconsistency within the law or in my perception?
As has been stated, this is all due to confusion between "gravitational force" and "gravitational acceleration".

Gravitational force is WEIGHT. It certainly depends on the mass of the object that is falling.

Gravitational acceleration is independent of the mass m of the falling object. But as has been pointed out, it certainly is NOT independent of the mass M of the planet that the object is on.

Still, all of this is in the limit of m << M, ensuring that the center of mass of the m-M system is practically the center of M, so that the planet doesn't move as m falls towards it.

BTW, using the word "gravity" in this case is rather vague and generic. You need to specify exactly what this is. One doesn't know if you mean gravitational force or gravitational acceleration, and there's a good chance that you do not realize that either by your usage and by your question. So if you did a bit more work in figuring out what exactly that you are asking, you might be able to clarify this issue yourself if you look up the definitions.

Zz.

Thanks for the answers. I must say I knew the answer to this however I was interested in seeing the way different people express it differently. In this way one can see the same idea from many angles. This is pleasureful and informative. Thanks again. So there is an inconsistency in saying a feather and a cannon ball fall at the same rate in a vacuum.
Another question is how is, say, the Moon "aware" of the Earth's presence to have an attraction in the first place. What process allows masses to be "conscious" of each other given that 2 way communication requires a process. Sometimes noodling around with simple questions is a useful start to more in depth concepts. Like masses being concentrated loci of energy vortexes. Then for gravity to be "felt" at a distance then these seperate concentrations of energy vortices (m) must communicate by a process. What is that process? Einstein's space compression is I believe the current view of G. Still, how would a mere compression of space account for gm1 and gm2 ? Moreover the G at the center of the earth is 0. Apology if this is a waste of your time.

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So there is an inconsistency in saying a feather and a cannon ball fall at the same rate in a vacuum.
You've been using a figure : 9.8ms-2, for Earth gravitational acceleration accurate to only one decimal place : you don't get to say "inconsistency" when said difference doesn't show up until some 20 decimal places, later.

Dale
Mentor
I must say I knew the answer to this ...
Another question is
And since you probably also know the answer to this question we will close this thread so that people can help the sincere people on the forum. In the future, please be honest in your questions. If you know the answer and just want to hear other explanations then simply say so. The approach you took here is not conducive to building a good community.