Does an apple have its own gravity?

[jeff einstein]

TL;DR Summary

Can I assume that an apple has its own gravity even though it may be very very small?

Hey there,
I just want to know if objects with small mass have their own gravity. To think of it they must have gravity as all matter with mass have gravity. When such a small mass is dropped one meter above the surface of earth would the earths and the small mass gravity combine? If so larger masses will have larger gravity and when held at same height and dropped the gravity must combine resulting the bigger mass to accelerate more.

 

[Ibix]

I just want to know if objects with small mass have their own gravity.

Yes. Look up the Cavendish experiment, which measures the gravitational attraction of spheres with fairly small masses.

When such a small mass is dropped one meter above the surface of earth would the earths and the small mass gravity combine? If so larger masses will have larger gravity and when held at same height and dropped the gravity must combine resulting the bigger mass to accelerate more.

Not exactly, and you need to be careful with what you claim here. In the inertial center of mass frame, the acceleration of each body depends only on the mass of the other. That does give you an overall closing rate that depends on the mass of both, but the objects aren't accelerating more. In a non-inertial frame attached to one object the acceleration of the other object does depend on the mass of both, but there are fictitious forces to take into account.

Generally the mass of the Earth is so much larger than anything in the neighbourhood that it might as well be perfectly stationary, and "the acceleration due to gravity is independent of the object's mass" is fine.

 

[haushofer]

Yes. A lot of people confuse force with acceleration. The 3d law of Newton states that the gravitational force of Earth on the apple is the same (!) as the gravitational force of the apple on the Earth. So if an apple weighs 100 gram, Earth pulls with 1 Newton on the apple, but also the other way around.

But the resulting accelerations are different: that's where Newton's second law kicks in. You can show that the resulting acceleration of the Earth relative to that of the apple is given by the ratio of the mass of the apple and the Earth. That's why the apple falls to the Earth, and (fapp) not the other way around.

 

[PeroK]

TL;DR Summary: Can I assume that an apple has its own gravity even though it may be very very small?

Hey there,
I just want to know if objects with small mass have their own gravity. To think of it they must have gravity as all matter with mass have gravity. When such a small mass is dropped one meter above the surface of earth would the earths and the small mass gravity combine? If so larger masses will have larger gravity and when held at same height and dropped the gravity must combine resulting the bigger mass to accelerate more.

This is a common mistake. The force is greater for a more massive object, but the acceleration is the same. Take two objects of mass $m_1$ and $m_2$ dropped from the same height. The gravitational force on each object depends on the mass, but the acceleration is the same:
$$F_1 = m_1g, \ a_1 = \frac{F_1}{m_1} = g$$$$F_2 = m_2g, \ a_2 = \frac{F_2}{m_2} = g$$

 

[jeff einstein]

just want to clarify does the gravity of the earth and the apple combine?

 

[PeroK]

just want to clarify does the gravity of the earth and the apple combine?

Newton's law of gravitation is that the magnitude of the force between two point masses is:
$$F = \frac{Gm_1m_2}{r^2}$$Where $G$ is the universal gravitational constant and $r$ is the distance between the masses. That equation includes both masses - as it must.

 

[jeff einstein]

so we can conclude that the gravity of the smaller mass has no effect on the rate of fall toward earth. but I want to know why we ignore the gravity of the smaller mass is it because its very insignificant?

 

[PeroK]

so we can conclude that the gravity of the smaller mass has no effect on the rate of fall toward earth. but I want to know why we ignore the gravity of the smaller mass is it because its very insignificant?

No. It's because we use Newton's laws of motion and gravitation. You need to think in terms of the equations. The mass of the falling object cancels out (as I showed above). It's not ignored. It cancels out from the equations - which is a very different thing.

 

[jeff einstein]

this may be a silly question but why is F=m2*g ang not m2*g

 

[jeff einstein]

this may be a silly question but why is F=m2*g ang not m2*g

sorry mistake i meant: this may be a silly question but why is F=m2*g ang not m1*g

 

[PeroK]

sorry mistake i meant: this may be a silly question but why is F=m2*g ang not m1*g

In what context?

 

[jeff einstein]

why cant we say the earth falling towards the smaller object

 

[jeff einstein]

and then we find the gravity acting on earth instead

 

[jeff einstein]

where mass of earth cancels out

 

[PeroK]

why cant we say the earth falling towards the smaller object

That's a different question. The effect of the apple on the Earth is neglected, because the effect is too small to measure.

 

[jeff einstein]

what if we consider the "effect is too small" and say that different ojects dropped have diiferent gravity even though it is very very small of a diiference and so they fall towards earth at different rates

 

[PeroK]

what if we consider the "effect is too small" and say that different ojects dropped have diiferent gravity even though it is very very small of a diiference and so they fall towards earth at different rates

In general, "falling" means one object is small compared to the other. That, I thought, was the context of your original question. You were talking about an apple, after all. Now, you are talking about somehow considering the motion of the Earth in response to the apple falling?

There was a long thread about this recently. My answer is that the effect is too small to measure. And further discussion is unproductive in terms of actually learning physics.

If other people want to indulge you, that's up to them.

 

[jeff einstein]

ok, i have got it now this simple answer is way easier to understand compared to the others on the older thread. so the effect I am talking about is actually present but too small to realize

 

[haushofer]

The dominant gravitational force of Earth makes us sloppy in stating which object exercises which force, and which object is influenced by it. But in this case it's important :)

By the way, just to throw a different situation at you: if you place two marbles of ten grams at a distance of 1 meter of each other in interstellar space, their mutual gravitational interaction will make them collide in ca. 10 days. In that case both marbles accelerate the same (from an inertial point of view).

 

[Dale]

There was a long thread about this recently. My answer is that the effect is too small to measure. And further discussion is unproductive in terms of actually learning physics.

The previous long thread is this one https://www.physicsforums.com/threads/a-contradiction-of-the-equivalence-principle.1055598/

@jeff einstein what is different from this thread and the previous one? Make sure that you do not wander back to the previous thread which was closed for cause.

 

[Vanadium 50]

Since we're covering the same material, here's a quote from that last thread:

Jeff, are you even considering the messages you are getting? You respond within seconds and we are covering the same ground over and over and over.

 

[jeff einstein]

Very much thank you for this i sure couldn't read that

 

[jbriggs444]

so we can conclude that the gravity of the smaller mass has no effect on the rate of fall toward earth. but I want to know why we ignore the gravity of the smaller mass is it because its very insignificant?

When you speak of the "rate of fall toward earth", which of these do you mean?

1) The apple's acceleration relative to an inertial frame tied to the combined earth+apple center of mass.
2) The apples acceleration relative to an accelerating frame tied to the earth center of mass.

The gravitational effect of the apple matters for 2) but not for 1).

 

[jack action]

I just want to know if objects with small mass have their own gravity.

A mass by itself doesn't have a "gravity". Gravity is defined as "a fundamental force which causes mutual attraction between all things that have mass." (source)

So for gravity to exist, you need at least two masses. The resulting force $F$ between two masses $m_1$ and $m_2$, separated by a distance $r$ is:
$$F = \frac{Gm_1m_2}{r^2}$$
Where $G$ is the gravitational constant and is equal to 6.674×10⁻¹¹ N⋅m²/kg². Since the force is applied to both ("for every action, there is an equal and opposite reaction"), we get this:

When such a small mass is dropped one meter above the surface of earth would the earths and the small mass gravity combine?

They don't combine, as each mass doesn't have a "gravity" by itself, but the gravity force shared by them is influenced by both masses.

If so larger masses will have larger gravity

You should now understand that they don't.

the gravity must combine resulting the bigger mass to accelerate more.

Let's find the acceleration of both masses with what we know.

If we have a 100 g (0.1 kg) apple, with a radius of 5 cm (0.05 m) at a 1m distance from the Earth whose mass is 5.9722 X 10²⁴ kg and radius is 6371 km (6371000 m), the acceleration $a_2$ of the apple $m_2$ can be found with:
$$a_2 = \frac{F_2}{m_2} = \frac{G\frac{m_1m_2}{r^2}}{m_2}= \frac{G}{r^2}m_1$$
Similarly, the acceleration $a_1$ of the Earth $m_1$ will be:
$$a_1 = \frac{F_1}{m_1} = \frac{G\frac{m_1m_2}{r^2}}{m_1}= \frac{G}{r^2}m_2$$
The distance $r$ is the distance center-to-center between the two masses:
$$r = 6371000 + 1 + 0.05 = 6371001.05\text{ m}$$
Thus:
$$a_2 = \frac{6.674 \times 10^{−11}}{6371001.05^2} \times 5.9722 X 10^{24} = 9.8\text{ m/s²}$$
Which is close to the standard acceleration due to gravity on Earth, $g$ (more on this here). And the Earth's acceleration is:
$$a_2 = \frac{6.674 \times 10^{−11}}{6371001.05^2} \times 0.1 = 1.6 × 10^{-25}\text{ m/s²} \approx 0\text{ m/s²}$$
You can try with other objects which are heavier and/or at larger distances and you will see that those numbers will not vary much unless the masses are comparable to the Earth's mass or the distance between them is comparable to the Earth's radius. (For example, other planets.)

Also, if the distance between the two is 0, then the acceleration of both is also 0 since they are just stuck together, pressing against one another (because the gravitational force is still present).

 

[vanhees71]

When you speak of the "rate of fall toward earth", which of these do you mean?

1) The apple's acceleration relative to an inertial frame tied to the combined earth+apple center of mass.
2) The apples acceleration relative to an accelerating frame tied to the earth center of mass.

The gravitational effect of the apple matters for 2) but not for 1).

That's not so difficult to explain. All you need is Newton's 2nd Law for the motion of two particles interacting through the gravitational interaction. Let $m_1$ the mass and $\vec{x}_1$ the position vector of the apple and $m_2$ and $\vec{x}_2$ the corresponding quantities for the Earth. Then the equations of motion are
$$m_1 \ddot{\vec{x}}_1 =-\frac{G m_1 m_2}{|\vec{x}_1-\vec{x}_2|^3}(\vec{x}_1-\vec{x}_2),$$
$$m_2 \ddot{\vec{x}}_2 =-\frac{G m_1 m_2}{|\vec{x}_2-\vec{x}_1|^3} (\vec{x}_2-\vec{x}_1),$$
i.e., the force is attractive along the straight line connecting the two masses and its magnitude is proportional to the masses and inversely propoertional to the square of the distance.

As with all central pair-interaction forces also Newton's 3rd Law is fulfilled, i.e., the force on body 1 is the exact opposite as the one on body 2. This is key for the first very simple observation: Just add the two equations of motion. You get
$$m_1 \ddot{\vec{x}}_1 + m_2 \ddot{\vec{x}}_2=0.$$
That's telling you that the center of mass,
$$\vec{R}=\frac{m_1 \vec{x}_1+m_2 \vec{x}_2}{M}, \quad M=m_1 + m_2$$
is moving with constant speed, i.e.,
$$\vec{R}=\vec{V} t + \vec{R}_0, \quad \vec{V}=\text{const}.$$
It's also telling that the total momentum is conserved, i.e.,
$$\vec{P}=m_1 \dot{\vec{x}}_1 + m_2 \dot{\vec{x}}_2=M \dot{\vec{R}}=\text{const}.$$
The interesting question is the relative motion. For that you introduce the relative vector
$$\vec{r}=\vec{x}_1-\vec{x}_2.$$
Then you can express $\vec{x}_1$ and $\vec{x}_2$ in terms of $\vec{R}$ and $\vec{r}$ and plug it into one of the equations of motion. You get
$$\vec{x}_1=\vec{R}+\frac{m_2}{M} \vec{r}.$$
Plugging this into the first equation of motion, you find, because $\ddot{\vec{R}}=0$
$$\frac{m_1 m_2}{M} \ddot{\vec{r}}=-\frac{G m_1 m_2}{r^3} \vec{r}.$$
Now this is the equation of motion of a single particle with mass $\mu=m_1 m_2/M$ in the gravitational field of an infinitely heavy body fixed at the origin. Since $m_1 m_2 =\mu M$ you get
$$\ddot{\vec{r}}=-\frac{G M}{r^3} \vec{r},$$
i.e., the acceleration of the relative motion indeed depends on the sum $M=m_1+m_2$ of the masses.

The further solution of this equation, the "restricted Kepler problem", is a bit more complicated than this pretty simple general considerations.

 

[jeff einstein]

ok, i have got it now this simple answer is way easier to understand compared to the others on the older thread. so the effect I am talking about is actually present but too small to realize

don't overcomplicate everything all I want to know is if this is true. because I have now concluded that however small the mass maybe it still has its own gravity and so must influence its fall toward the earth.

 

[jeff einstein]

two factors affect the fall of the object the distance between the centers and the mass of both objects involved. we can agree that increasing the distance away from the center must influence the rate of "fall" and so likewise increasing or decreasing mass should also affect the rate of fall.

 

[jeff einstein]

A mass by itself doesn't have a "gravity". Gravity is defined as "a fundamental force which causes mutual attraction between all things that have mass." (source)

So for gravity to exist, you need at least two masses. The resulting force $F$ between two masses $m_1$ and $m_2$, separated by a distance $r$ is:
$$F = \frac{Gm_1m_2}{r^2}$$
Where $G$ is the gravitational constant and is equal to 6.674×10⁻¹¹ N⋅m²/kg². Since the force is applied to both ("for every action, there is an equal and opposite reaction"), we get this:

View attachment 337414​

They don't combine, as each mass doesn't have a "gravity" by itself, but the gravity force shared by them is influenced by both masses.You should now understand that they don't.Let's find the acceleration of both masses with what we know.

If we have a 100 g (0.1 kg) apple, with a radius of 5 cm (0.05 m) at a 1m distance from the Earth whose mass is 5.9722 X 10²⁴ kg and radius is 6371 km (6371000 m), the acceleration $a_2$ of the apple $m_2$ can be found with:
$$a_2 = \frac{F_2}{m_2} = \frac{G\frac{m_1m_2}{r^2}}{m_2}= \frac{G}{r^2}m_1$$
Similarly, the acceleration $a_1$ of the Earth $m_1$ will be:
$$a_1 = \frac{F_1}{m_1} = \frac{G\frac{m_1m_2}{r^2}}{m_1}= \frac{G}{r^2}m_2$$
The distance $r$ is the distance center-to-center between the two masses:
$$r = 6371000 + 1 + 0.05 = 6371001.05\text{ m}$$
Thus:
$$a_2 = \frac{6.674 \times 10^{−11}}{6371001.05^2} \times 5.9722 X 10^{24} = 9.8\text{ m/s²}$$
Which is close to the standard acceleration due to gravity on Earth, $g$ (more on this here). And the Earth's acceleration is:
$$a_2 = \frac{6.674 \times 10^{−11}}{6371001.05^2} \times 0.1 = 1.6 × 10^{-25}\text{ m/s²} \approx 0\text{ m/s²}$$
You can try with other objects which are heavier and/or at larger distances and you will see that those numbers will not vary much unless the masses are comparable to the Earth's mass or the distance between them is comparable to the Earth's radius. (For example, other planets.)

Also, if the distance between the two is 0, then the acceleration of both is also 0 since they are just stuck together, pressing against one another (because the gravitational force is still present).

i like this explanation, this is what I am saying, here @jack action has put an approximate sign of 0 and not completely equal. All I am saying is even though the mass of the smaller object may have a very very very small effect on the earth it still is present. I think I am done with this thread after all I think this conclusion is right after all.

 

[Dale]

don't overcomplicate everything all I want to know is if this is true. because I have now concluded that however small the mass maybe it still has its own gravity and so must influence its fall toward the earth.

This was exactly the topic of the previous thread. So the answer to my question:

The previous long thread is this one https://www.physicsforums.com/threads/a-contradiction-of-the-equivalence-principle.1055598/

@jeff einstein what is different from this thread and the previous one? Make sure that you do not wander back to the previous thread which was closed for cause.

Is “nothing”. This thread is closed, for the third time. The questions and answers have not changed from the previous two times you have asked them and we have answered.