Calculating Gravity: Mass, Radius & Wall Thickness

[Timothy Schablin]

Consider the 2 objects in the pic. They both have the same mass of 200 kg. They both have a radius of 4 meters. However, the object on the right is hollow, with the walls being 2 meters thick. For the gravity equation, and the object on the right, does one use a radius of 4 squared, or the wall thickness of 2 squared?

 

[phinds]

Consider the 2 objects in the pic. They both have the same mass of 200 kg. They both have a radius of 4 meters. However, the object on the right is hollow, with the walls being 2 meters thick. For the gravity equation, and the object on the right, does one use a radius of 4 squared, or the wall thickness of 2 squared?

View attachment 207047

Outside of either object, the mass acts as a point mass at the center as far as gravitational attraction is concerned.

 

[jack action]

The equation is actually:
$$mg = \frac{GMm}{r^2}$$
Where $r$ is the distance between the 2 centers of mass of the objects of mass $m$ and $M$.

 

[Timothy Schablin]

Just to avoid confusion, we're looking for gravity of each individual object, not the attraction between them.

I was actually thinking of a formula that included the radius of the hollow section (r) and total radius (R).

g = Gm/(R+r)(R-r)

Not so sure tho... Its probably just the normal formula of g = Gm/r^2

 

[phinds]

Just to avoid confusion, we're looking for gravity of each individual object, not the attraction between them.

I was actually thinking of a formula that included the radius of the hollow section (r) and total radius (R).

g = Gm/(R+r)(R-r)

Not so sure tho... Its probably just the normal formula of g = Gm/r^2

So, did you not believe what I told you in post #2 ?

 

[FactChecker]

Just to avoid confusion, we're looking for gravity of each individual object, not the attraction between them.

I was actually thinking of a formula that included the radius of the hollow section (r) and total radius (R).

g = Gm/(R+r)(R-r)

Not so sure tho... Its probably just the normal formula of g = Gm/r^2

It's not obvious, but Newton proved that you can use the center of the object. (see https://en.wikipedia.org/wiki/Shell_theorem )

 

[Timothy Schablin]

Thanks. Shell theorem is good reading.

 

[jack action]

Just to avoid confusion, we're looking for gravity of each individual object, not the attraction between them.

What is the difference? Or am I missing something? The Earth is attracted to a small object as much as the small object is attracted to the Earth.

1st example: Knowing the Earth radius is $r = 6\ 371\ 000\ m$ and its mass $M = 5.972 \times 10^{24}\ kg$ what is the acceleration $g$ felt by a human of mass $m = 75\ kg$ near its surface?
$$mg = \frac{GMm}{r^2}$$
$$g = \frac{GM}{r^2}$$
$$g = \frac{(6.674 \times 10^{-11}) \times (5.972 \times 10^{24})}{(6\ 371\ 000)^2}$$
$$g = 9.82\ m/s^2$$
2nd example: Knowing the Earth radius is $r = 6\ 371\ 000\ m$ and its mass $M = 5.972 \times 10^{24}\ kg$ what is the acceleration $g_h$ felt by the Earth towards a human of mass $m = 75\ kg$ near its surface?
$$Mg_h = \frac{GMm}{r^2}$$
$$g_h = \frac{Gm}{r^2}$$
$$g_h = \frac{(6.674 \times 10^{-11}) \times (75)}{(6\ 371\ 000)^2}$$
$$g_h = 1.23 \times 10^{-22}\ m/s^2$$
The Earth is accelerating towards the human just like the human is accelerating towards the Earth. The Earth will not move because there is probably another human of the same mass on the other side of the planet that balances that acceleration, but the «human gravity» is still there.

 

[Nugatory]

Just to avoid confusion, we're looking for gravity of each individual object, not the attraction between them.

What is the difference? Or am I missing something?

As I read it, "the gravity of each individual object" is intended to mean the acceleration of an ideal test particle under the influence of the object's gravity, that is, the quantity $Gm/r^2$ for these spherically symmetric objects, and the question is what is the approriate value of $r$ to use in the calculation.

 

[jack action]

As I read it, "the gravity of each individual object" is intended to mean the acceleration of an ideal test particle under the influence of the object's gravity, that is, the quantity $Gm/r^2$ for these spherically symmetric objects, and the question is what is the approriate value of $r$ to use in the calculation.

But with the addition of the equation $g=\frac{GM}{r^2}$ found in the OP, that would suggests that $r$ is always the distance from the center of mass of the object to a point on the outside surface of the object (where the ideal test particle is), no matter the shape of that object. That is what I was trying to point out.

My "gravity" - if I have a mass of 75 kg, measure 180 cm high and 40 cm wide - is 6.18 X 10^-9 m/s² ($r$ = 90 cm) for a particle on top of my head and 1.25 X 10^-7 m/s²($r$ = 20 cm) for one on the side of my waist. If I was a sphere, then my ""gravity" would be the same for any point at my surface since the distance $r$ would be the same everywhere.

 

[Khashishi]

Your gravity depends on how far away you are measuring it. It's an inverse square law.

 

[jbriggs444]

My "gravity" - if I have a mass of 75 kg, measure 180 cm high and 40 cm wide - is 6.18 X 10^-9 m/s² ($r$ = 90 cm) for a particle on top of my head and 1.25 X 10^-7 m/s²($r$ = 20 cm) for one on the side of my waist. If I was a sphere, then my ""gravity" would be the same for any point at my surface since the distance $r$ would be the same everywhere.

Since you are not a sphere, Newton's shell theorem does not apply and a simple inverse square calculation will not work. The results above are not correct.

You would have to integrate your body mass density divided by r³ and multiplied by $\vec{r}$ over the volume of your body to get the net gravitational effect on a test particle at a particular point.

 

[Timothy Schablin]

how do i mark this as 'resolved'?

 

[Nugatory]

how do i mark this as 'resolved'?

You don't - that's only for threads in the homework section.