Graph of gravitational quantities against distance

[songoku]

I want to ask the graph of gravitational force against r and gravitational potential energy against r.

First, about gravitational force against r
The shape of the graph is straight line from origin until the surface of the sphere and then curve (decreasing). I understand about the decreasing part because it follows the inverse - square law. But I don't get the part of the straight line.

I saw the derivation:
F = GMm/r² then subs ρ = M / V where V = 4/3 πr³

Final form: F = 4/3 π G ρ m r. From this equation, it is obvious that F is proportional to r so the graph will be straight line.

My questions:
1. Why we substitute M only and not m too?

2. Why can we cross out r³ with r² so it becomes r? What my brain processes is r³ is the cube of radius where r² is the square of distance of a point measured from the center of the sphere. They are not the same so why can be crossed out?Second, gravitational potential energy against r
To find the shape of the graph when r is less than radius of the sphere, I tried to do the same thing (although I don't understand about it):
U = - GMm/r then subs ρ = M / V where V = 4/3 πr³. After crossing out the r, I got U is proportional to r² but my teacher said it was wrong. Where is my mistake?

Thanks

 

[Bandersnatch]

For the first part, try and imagine what is being done in that derivation. You are taking a test mass m, which is going to be held constant (say, an iron ball or something), and drop it down below the surface of the uniform sphere of mass M.

Now, the important bit here is that it can be shown using some geometric considerations* that as you move below the surface, the gravitational contribution from all the mass in the shell above you is zero. In other words, if you had an empty shell and tried to measure the gravitational force inside that shell, it'll come out to zero.
Anything that you ever need to worry about when finding the force of gravity from spherical, unifrom distribution of material, is the mass in a sphere below your feet.

So, what happens is as you go down below the surface, there is less and less mass left to pull you in, which according to Newton's law of gravity is going to reduce the force (M goes down), but at the same time you're getting closer to the centre of mass, which increases the force ($1/R^2$).
The rate of change of the amount of mass that is left is proportional to $R^3$, because the formula for the volume of a sphere has got $R^3$ in it, while the rate of change of the force is proportional to $1/R^2$.
These are both the same radii. You're not thinking here of the full radius of the sphere, but the distance from the centre to the point below the surface where you're measuring how strong the gravity is.

And you're not doing the same thing to the mass m, because it's a) constant throughout the process and b) it cancels out with the $ma$ from Newton's 2nd law, so it'll behave the same regardless of its actual value (as long as it's much less than the mass M).*it's called the shell theorem. Wikipedia has got a good article on it. It's the same theorem that shows why exactly you can treat the whole mass of a spherical object as if it were concentrated in its centre. Intuitively, it's all about the fact that the farther you look from a point in or out of a sphere, the more mass there is to attract you, but at the same time it's farther and thus contributing less. The two effects neatly cancel out. You may need to know integration to follow the maths, although I think I saw a purely geometric explanation once.

As for the second part, I have to admit that I don't know. I looks right to me, but maybe I'm missing something.
edit: No, I'll have to think about it. Maybe somebody else will jump in in the meantime.

 

[Bandersnatch]

O.k., so for the second part, I can't see the reason why the shape of the curve shouldn't be proportional to $r^2$. What you do to get the potential energy, is integrate the force equation, which will net the $r^2$ factor for radii below the surface. The only thing I can think of that your teacher might have objected to, is that the shape of the U curve is not the same as U which is always measured between two points (otherwise it's undefined).

 

[songoku]

For the first part, try and imagine what is being done in that derivation. You are taking a test mass m, which is going to be held constant (say, an iron ball or something), and drop it down below the surface of the uniform sphere of mass M.

Now, the important bit here is that it can be shown using some geometric considerations* that as you move below the surface, the gravitational contribution from all the mass in the shell above you is zero. In other words, if you had an empty shell and tried to measure the gravitational force inside that shell, it'll come out to zero.
Anything that you ever need to worry about when finding the force of gravity from spherical, unifrom distribution of material, is the mass in a sphere below your feet.

So, what happens is as you go down below the surface, there is less and less mass left to pull you in, which according to Newton's law of gravity is going to reduce the force (M goes down), but at the same time you're getting closer to the centre of mass, which increases the force ($1/R^2$).
The rate of change of the amount of mass that is left is proportional to $R^3$, because the formula for the volume of a sphere has got $R^3$ in it, while the rate of change of the force is proportional to $1/R^2$.
These are both the same radii. You're not thinking here of the full radius of the sphere, but the distance from the centre to the point below the surface where you're measuring how strong the gravity is.

And you're not doing the same thing to the mass m, because it's a) constant throughout the process and b) it cancels out with the $ma$ from Newton's 2nd law, so it'll behave the same regardless of its actual value (as long as it's much less than the mass M).*it's called the shell theorem. Wikipedia has got a good article on it. It's the same theorem that shows why exactly you can treat the whole mass of a spherical object as if it were concentrated in its centre. Intuitively, it's all about the fact that the farther you look from a point in or out of a sphere, the more mass there is to attract you, but at the same time it's farther and thus contributing less. The two effects neatly cancel out. You may need to know integration to follow the maths, although I think I saw a purely geometric explanation once.

As for the second part, I have to admit that I don't know. I looks right to me, but maybe I'm missing something.
edit: No, I'll have to think about it. Maybe somebody else will jump in in the meantime.

O.k., so for the second part, I can't see the reason why the shape of the curve shouldn't be proportional to $r^2$. What you do to get the potential energy, is integrate the force equation, which will net the $r^2$ factor for radii below the surface. The only thing I can think of that your teacher might have objected to, is that the shape of the U curve is not the same as U which is always measured between two points (otherwise it's undefined).

I think I understand your explanation about the graph of force against r.

Yeah, I also didn't know why. The method of derivation can be applied for force vs r, gravitational potential vs r, gravitational field vs r. It is weird that it can't be applied to energy vs r.

In my opinion, the graph of energy vs r would be all below x-axis and consists of two parts:
first part: curve (away from x-axis), following the relation U is proportional to (- r²)
second part: curve (towards x-axis), following the relation U is proportional to (-1/r)

Thank you very much for the explanation

 

[pmacias]

for your question and for sharing your thoughts on the graph of gravitational quantities against distance. I can provide some insights and clarifications on your questions.

1. Why we substitute M only and not m too?

In the equation for gravitational force, F = GMm/r^2, both masses M and m are involved. However, when we are looking at the force on a test object (m) due to a larger object (M), we can assume that M is much larger than m. This allows us to treat M as a constant and only substitute in the variable r. This is similar to how we can ignore the mass of Earth when calculating the gravitational force on an object near its surface.

2. Why can we cross out r^3 with r^2 so it becomes r?

In the equation F = 4/3 π G ρ m r, we are looking at the gravitational force on a test object (m) due to a larger object with a known density (ρ) and radius (r). The r^3 term comes from the volume of the larger object, which is not relevant when we are only interested in the force at a specific distance (r). By substituting ρ = M/V and V = 4/3 πr^3, we can simplify the equation to F = 4/3 π G ρ m r and eliminate the r^3 term.

Now, moving on to the graph of gravitational potential energy against r. It is important to note that potential energy is a scalar quantity, meaning it has magnitude but no direction. This is different from force, which is a vector quantity with both magnitude and direction. In the equation for gravitational potential energy, U = - GMm/r, we are looking at the potential energy of a test object (m) in the gravitational field of a larger object (M).

When substituting ρ = M/V and V = 4/3 πr^3, we get U = - 3/5 GMm/r. As you mentioned, this does not match with your teacher's explanation of U being proportional to r^2. This is because potential energy is not directly proportional to distance, but rather it is inversely proportional to distance. In other words, as the distance (r) increases, the potential energy (U) decreases. This can be seen in the shape of the graph, which is a curve that decreases as r increases.

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