# How to get gravitational force on a gaseous particle?

• vcsharp2003
In summary: There is a standard pair of results you should know.Inside a uniform spherical shell of mass there is no net gravitational force; they all cancel exactly.Outside a uniform spherical shell of mass the net gravitational force is the same as if the shell were collapsed to a single point at its centre.The same pair of results applies in electrostatics, since that is also an inverse square law.
vcsharp2003
Homework Statement
Consider a spherical gaseous cloud of mass density ## \rho (r)## in free space where ##r## is the radial distance from its center. The gaseous cloud is made of particles of equal mass ##m## moving in circular orbits about the common center with the same kinetic energy ##K## . The force acting on the particles is their mutual gravitational force. Mass density is constant in time. If the particle number density is ##n (r ) = \dfrac {\rho (r)} {m}##, then determine it in terms of ## \text {m, K and r}##?
Relevant Equations
##F_g = \dfrac {GMm} {r^2} ##
## F_c = \dfrac {mv^2} {r} ##
## KE = \dfrac {mv^2} {2} ##
This question is very confusing since I don't see two distinct particles that are exerting a gravitational force on each other. Also to complicate matters, a gas is made of many individual particles and I don't know how to determine the gravitational force on a single particle from so many other gaseous particles.

If someone can give me any hint on how to get the gravitational force on a single gas particle from so many other particles then that would help.

I did consider the following analogy, but it's confusing. The spherical gaseous cloud can be considered like a spherical earth. Then, gravitational force between any outside particle and the gaseous cloud would be like the force between Earth and and an external object. But in our case we have no external particle as the particle is part of the spherical gaseous cloud.

The question clearly states that "The force acting on the particles is their mutual gravitational force", but then what should be the big mass M that we should use to mimic a spherical Earth pulling an external object. We know the small mass m is known and its the mass of a particle in circular motion.

Last edited:
Gravitation potential ##\phi(r)## generated by spherical distribution of mass ##\rho(r)## is
$$\phi(r)=-G\int \frac{\rho(r')}{|r-r'|}dV'$$
After some calculation balancing it with centrifugal potential would work.

anuttarasammyak said:
Gravitation potential ##\phi(r)## generated by spherical distribution of mass ##\rho(r)## is
$$\phi(r)=-G\int \frac{\rho(r')}{|r-r'|}dV'$$
After some calculation balancing it with centrifugal potential would work.

What is ## r ## and ##r^\prime## in your above equation?

##r## is position where gravitation potential is considered.
##r'## is position vector of volume element ##dV'## which should be integrated in whole space.

anuttarasammyak said:
##r## is position where gravitation potential is considered.
##r'## is position vector of volume element ##dV'## which should be integrated in whole space.

Is it possible to get gravitational force on a single particle at a distance ## r ## from center of gaseous cloud? I think that would be a good starting point for me since the equation you mentioned is not there in the course material that I am using.

I see. Gravitation potential of a particle whose mass is m
$$\phi(r)=-G\frac{m}{r}$$
Have you got it?

anuttarasammyak said:
I see. Gravitation potential of a particle whose mass is m
$$\phi(r)=-G\frac{m}{r}$$
Have you got it?

I get that part but the question seems to be mentioning circular motion and gravitational force, so I would be inclined to think that the question may have something to do with centripetal force and gravitational force, and not gravitational potential.

vcsharp2003 said:
I get that part but the question seems to be mentioning circular motion and gravitational force, so I would be inclined to think that the question may have something to do with centripetal force and gravitational force, and not gravitational potential.
You are right in thinking that you should treat it much like a particle inside Earth.
Consider the cloud as made of spherical shells. For a given particle at radius r, it will be outside some shells and inside others. What is the net gravitational force on the particle from one such shell?

vcsharp2003
haruspex said:
You are right in thinking that you should treat it much like a particle inside Earth.
Consider the cloud as made of spherical shells. For a given particle at radius r, it will be outside some shells and inside others. What is the net gravitational force on the particle from one such shell?

We have two cases here. First case is when particle is inside the shell and second case is when particle is on or outside the shell.
In first case, the inside particle will experience forces in all directions from each particle in the shell. In second case too forces will act on the particle in all directions. Not too sure how to take it forward using a shell approach? Unless we treat each spherical shell like the Earth with the mass of shell being concentrated in a particle at center of sphere.

vcsharp2003 said:
We have two cases here. First case is when particle is inside the shell and second case is when particle is on or outside the shell.
In first case, the inside particle will experience forces in all directions from each particle in the shell. In second case too forces will act on the particle in all directions. Not too sure how to take it forward using a shell approach?
There is a standard pair of results you should know.
Inside a uniform spherical shell of mass there is no net gravitational force; they all cancel exactly.
Outside a uniform spherical shell of mass the net gravitational force is the same as if the shell were collapsed to a single point at its centre.
The same pair of results applies in electrostatics, since that is also an inverse square law.

vcsharp2003
haruspex said:
Outside a uniform spherical shell of mass the net gravitational force is the same as if the shell were collapsed to a single point at its centre.
The same pair of results applies in electrostatics, since that is also an inverse square law.

So, let's take a shell at a distance of ##r## from the center of sphere that has a thickness ##dr## and a mass ##dm##. Each particle P within this shell is moving in a circular path about the center of sphere.

Then, according to what you mentioned, the shells outer to this shell exert no gravitational force on the particle P since now the particle P is inside all outer shells.

But each inner shell to particle P's shell would exert a force on particle P, since P is now outside each of these inner shells. The mass M would then be the total mass of these inner shells and it would act as a collapsed particle at sphere's center with a mass M.

Now I need to setup some equation(s).

Let M be the total mass of all shells that are inner to particle P's shell. Since the kinetic energy is same for all particles, so for particle P
$$K = \dfrac {mv^2} {2}$$

Also, since the inner shells are collapsed to a particle at center with mass M, the distance between particle P and the collapsed particle is ##r##. So, following equation is the second equation we have where centripetal force is being supplied by the gravitational force for particle P.

$$\dfrac {GMm} {r^2} = \dfrac {mv^2} {r}$$

Last edited:
vcsharp2003 said:
So, let's take a shell at a distance of ##r## from the center of sphere that has a thickness ##dr## and a mass ##dm##. Each particle P within this shell is moving in a circular path about the center of sphere.

Then, according to what you mentioned, the shells outer to this shell exert no gravitational force on the particle P since now the particle P is inside all outer shells.

But each inner shell to particle P's shell would exert a force on particle P, since P is now outside each of these inner shells. The mass M would then be the total mass of these inner shells and it would act as a collapsed particle at sphere's center with a mass M.

Now I need to setup some equation(s).

Let M be the total mass of all shells that are inner to particle P's shell. Since the kinetic energy is same for all particles mo, so for particle P
$$K = \dfrac {mv^2} {2}$$

Also, since the inner shells are collapsed to a particle at center with mass M, the distance between particle P and the collapsed particle is ##r##. So, following equation is the second equation we have where centripetal force is being supplied by the gravitational force.

$$\dfrac {GMm} {r^2} = \dfrac {mv^2} {r}$$
Looks good. And what is the relationship between M and r?

vcsharp2003
haruspex said:
good. And what is the relationship between M and r?

We get ## 2K = mv^2## from the first equation, which we substitute in second equation to get $$\dfrac {GMm} {r^2} = \dfrac {2K} {r}$$
$$M= \dfrac {2Kr} {Gm}$$

The above equation is M in terms of r. Note that K and m are known quantities as given in problem statement.

vcsharp2003 said:
We get ## 2K = mv^2## from the first equation, which we substitute in second equation to get $$\dfrac {GMm} {r^2} = \dfrac {2K} {r}$$
$$M= \dfrac {2Kr} {Gm}$$

The above equation is M in terms of r. Note that K and m are known quantities as given in problem statement.
Yes, but there's also the relationship via ##\rho(r)##.

vcsharp2003
haruspex said:
Yes, but there's also the relationship via ρ(r).

We can look at an inner shell whose thickness is ##dr##. Volume of this infinitesimal shell is ##(4 \pi r^2) dr##. Since mass density for this shell is ##\rho(r)##, so we can say ##\text {mass of this shell = density x volume i.e}## ##dM =\rho(r) \times 4 \pi r^2dr##. Integrating both sides with limits from #r=0 to r= r, we can get the total mass M of all inner shells.

But integrating right hand side gets difficult since we do not know ##\rho(r)## i.e we have no idea of what is ##\rho## in terms of ##r##.

Last edited:
vcsharp2003 said:
integrating right hand side gets difficult since we do not know ρ(r)
So what can you do instead of integrating the RHS?

vcsharp2003
haruspex said:
So what can you do instead of integrating the RHS?
Perhaps, we can integrate and get M = some definite integral. Then compare the above definite integral with expression of M obtained in earlier equation and decide the expression for ##\rho (r)##.

Or we can get ##\dfrac {dM} {dr} = \rho (r) \times 4\pi r^2## instead of taking a definite integral.

vcsharp2003 said:
Or we can get ##\dfrac {dM} {dr} = \rho (r) \times 4\pi r^2## instead of taking a definite integral.
Try that.

vcsharp2003
haruspex said:
Try that.
We have the rate of change of mass M of sphere with respect to its radius r as ##\dfrac {dM} {dr} = \rho (r) \times 4\pi r^2## as derived in post#15.

Also, using the already derived equation for mass of sphere M in post#14, ##M= \dfrac {2Kr} {Gm}## and differentiating both sides of this equation we get ## \dfrac {dM} {dr} = \dfrac {2K} {Gm} ## ( note that we treat gravitational constant ##G##, mass of each gaseous particle ##m## and kinetic energy of each gaseous particle ##K## as constants since these quantities do not depend on the radius of sphere ##r##.

Now, we can equate ## \dfrac {dM} {dr}## obtained in above two different manners, so that ## \rho (r) \times 4\pi r^2 = \dfrac {2K} {Gm} ##, which gives us the function ## \rho (r) = \dfrac {K} {2\pi r^2Gm} ##. Therefore, ## n(r) = \dfrac {\rho(r)} {m} = \dfrac {K} {2\pi r^2Gm^2} ## and this is the answer to the question.

(Kinetic energy of a gaseous particle at all distances from center of spherical cloud is the same according to the problem and its value is ##K##. Normally, I was thinking that kinetic energy of a gaseous particle would change as distance from center changes and probably many others would think likewise.)

Last edited:
vcsharp2003 said:
Kinetic energy of a gaseous particle at all distances from center of spherical cloud is the same according to the problem and its value is K. Normally, I was thinking that kinetic energy of a gaseous particle would change as distance from center changes and probably many others would think likewise.
On Earth, we are familiar with air being cooler at higher altitudes, but this is because of our "greenhouse" gases.
With an atmosphere of diatomic gases, heat and light pass straight through. Correspondingly, it cannot radiate away heat. It can only exchange heat with the ground by conduction and redistribute it by convection.

Lnewqban
haruspex said:
On Earth, we are familiar with air being cooler at higher altitudes, but this is because of our "greenhouse" gases.
With an atmosphere of diatomic gases, heat and light pass straight through. Correspondingly, it cannot radiate away heat. It can only exchange heat with the ground by conduction and redistribute it by convection.
Why would kinetic energy of gaseous particles be same in the gaseous cloud? It may just be an assumption in this problem. I am trying to find some reason why this could be true.

For two different particles at radius ##r_1## and at ##r_2##, the velocity of gaseous particles is the same even though ##r_1 < r_2 ## or ##r_1 >r_2 ##.

vcsharp2003 said:
Why would kinetic energy of gaseous particles be same in the gaseous cloud?
Well, they won't all be exactly the same, of course, but for diatomic and monatomic gas mixtures the temperature can be largely independent of altitude, so the mean KE can be independent of altitude.

vcsharp2003
haruspex said:
Well, they won't all be exactly the same, of course, but for diatomic and monatomic gas mixtures the temperature can be largely independent of altitude, so the mean KE can be independent of altitude.
I was thinking that temperature decides the KE of gaseous particles. Higher temperature means higher KE.

I think in this problem, the gaseous cloud being considered is not that big and it has almost equal temperatures as one goes away from the center of the cloud, which is a good assumption for small gaseous clouds but for large gaseous clouds. Hence, the assumption of equal KE is a good assumption in this problem.

For air above us, if altitudes can become very high then there will be considerable difference between air particle temperatures at ground level and high altitudes. Then, the assumption of equal KE no matter what the altitude would not be a good assumption.

vcsharp2003 said:
Homework Statement:: Consider a spherical gaseous cloud of mass density ## \rho (r)## in free space where ##r## is the radial distance from its center. The gaseous cloud is made of particles of equal mass ##m## moving in circular orbits about the common center with the same kinetic energy ##K## .
Sounds like a trick question. It is not a gaseous cloud. An ideal gas is modeled by a large number of small particles interacting statistically. Here we have a large number of small particles that do not interact other than by their bulk gravitational effect. Otherwise they could not all be in circular orbits.

If you have a bunch of non-interacting small particles in any spherically symmetric distribution whatsoever, a new layer of small non-interacting particles can be layered on the outside. The mass density of that new layer is arbitrary.

A collection of non-interacting particles does not even have a temperature.

vcsharp2003
vcsharp2003 said:
For air above us, if altitudes can become very high then there will be considerable difference between air particle temperatures at ground level and high altitudes.
@jbriggs444 is right that I have muddied the waters by thinking of it as interacting particles. But to answer your point above, as I mentioned in post #20, the decline in temperature with altitude we observe on Earth should not be taken to be what generally happens.
The triatomic etc. gases in our atmosphere can radiate away heat into space. Collisions with other molecules spread this cooling around. This makes the upper layers denser, leading to convection. The rising air that results cools due to expansion, limiting its ability to warm the upper layers. This leads to the lapse rate we observe.
Without the triatomic gases the air would only be able to exchange heat with the ground, and that by conduction. If the ground were all at the same temperature the entire atmosphere could be at that temperature too, and there would be no convection.

Last edited:
jbriggs444 said:
A collection of non-interacting particles does not even have a temperature.

You mean particles have to impact each other for heat to be produced from their KE and only then would the gas have a rise in temperature. Without impact there would be no heat and therefore no temperature. Is that correct?

vcsharp2003 said:
You mean particles have to impact each other for heat to be produced from their KE and only then would the gas have a rise in temperature. Without impact there would be no heat and therefore no temperature. Is that correct?
[I am not expert on thermodynamics, nor an educator. If I've muffed some details, correction is always welcome].

Strictly speaking, temperature is defined only for an ensemble that is at equilibrium with itself. Then we can look at how entropy and energy flow between that collection and another. If there is no interaction then there can be no meaningful equilibrium.

Entropy always increases. Energy is conserved. Thus, heat energy always flows in a direction so that entropy increases. The definition of temperature works out so that heat energy always flows from high temperature to low. [Negative temperatures are possible and are higher than all positive temperatures. It is sometimes convenient to work with inverse temperatures where the ordering of temperatures is consistent].

Temperature

The existence of temperature follows from the zeroth and second laws of thermodynamics: thermal equilibrium is transitive, and entropy is maximum in equilibrium. Temperature is then defined as the thermodynamic quantity that is the shared by systems in equilibrium.

If two systems are in equilibrium then they cannot increase entropy by flowing energy from one to the other. That means that if we flow a tiny bit of energy from one to the other (δU1 = -δU2), the entropy change in the first must be the opposite of the entropy change of the second (δS1 = -δS2), so that the total entropy (S1 + S2) doesn't change. For systems in equilibrium, this leads to (∂S1/∂U1) = (∂S2/∂U2). Define 1/T = (∂S/∂U), and we are done.

Temperature is sometimes taught as, "a measure of the average kinetic energy of the particles," because for an ideal gas U/N = (3/2) kBT. This is wrong as a definition, for the same reason that the ideal gas entropy isn't the definition of entropy.

Last edited:
jbriggs444 said:
Temperature is sometimes taught as, "a measure of the average kinetic energy of the particles," because for an ideal gas U/N = (3/2) kBT. This is wrong as a definition, for the same reason that the ideal gas entropy isn't the definition of entropy.
Photons in photon gas do not interact each other. However https://en.wikipedia.org/wiki/Photon_gas says it has thermodynamic parameters, i.e. temperature or entropy. Even if photons do not interact each other they interact with environment, e.g. in hollow radiation case. I suppose interaction with environment, not interaction between molecules, is a key issue to define its thermodynamic parameters for photons or ideal gas.

jbriggs444
anuttarasammyak said:
Even if photons do not interact each other they interact with environment, e.g. in hollow radiation case. I suppose interaction with environment, not interaction between molecules, is a key issue to define its thermodynamic parameters for photons or ideal gas.
Nice. I'd not considered it, but this seems elegant and even obvious: If it is at equilibrium with a black body then we can say that its temperature is that of the black body.

anuttarasammyak

## 1. How is gravitational force calculated on a gaseous particle?

The gravitational force on a gaseous particle is calculated using Newton's Law of Gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

## 2. What factors affect the gravitational force on a gaseous particle?

The gravitational force on a gaseous particle is affected by the mass of the particle and the mass of the object it is interacting with, as well as the distance between them. The force also depends on the gravitational constant, which is a value that is constant throughout the universe.

## 3. Can the gravitational force on a gaseous particle be negative?

No, the gravitational force on a gaseous particle cannot be negative. Gravitational force is always attractive, meaning it pulls objects towards each other. However, the direction of the force can be negative if it is acting in the opposite direction of a chosen reference point.

## 4. How does the gravitational force on a gaseous particle compare to other forces?

The gravitational force on a gaseous particle is one of the four fundamental forces of nature, along with the strong nuclear force, the weak nuclear force, and the electromagnetic force. It is the weakest of these forces, but it is still a significant force on a large scale, such as in the interactions between planets and stars.

## 5. How does the gravitational force on a gaseous particle change with distance?

The gravitational force on a gaseous particle decreases with distance. This is due to the inverse square law, which states that the force is inversely proportional to the square of the distance between two objects. This means that as the distance between two objects increases, the force between them decreases significantly.

Replies
16
Views
819
Replies
10
Views
991
Replies
12
Views
2K
Replies
15
Views
800
Replies
5
Views
2K
Replies
5
Views
2K
Replies
3
Views
2K
Replies
2
Views
2K
Replies
3
Views
1K
Replies
9
Views
1K