# Gravity of a disk acting on a mass on the z axis

• kostoglotov
In summary, the problem involves finding the magnitude of the force of attraction that a lamina with constant density exerts on a body of mass m located at a point (0,0,d) on the positive z axis. Using Newton's Law of Gravitation, the formula for this force is found to be F = 2πGmρd(1/d - 1/√(R^2+d^2)). To find the vertical component of the force exerted by a polar subrectangle, the formula F_v = GmρM_ij/(r^2+d^2)√(1-(r/(r^2+d^2))^2) is used, where M_ij = ρ
kostoglotov

## Homework Statement

A lamina has constant density $\rho$ and takes the shape of a disk with center the origin and radius R. Use Newton's Law of Gravitation to show that the magnitude of the force of attraction that the lamina exerts on a body of mass m located at the point (0,0,d) on the positive z axis is

$$F = 2 \pi Gm \rho d \left(\frac{1}{d}-\frac{1}{\sqrt{R^2+d^2}}\right)$$

[Hint: Divide the disk and first compute the vertical component of the force exerted by the polar subrectangle $R_{ij}$]

## Homework Equations

p = diagonal distance between mass m and a polar subrectangle of the disk.

$$F = G\frac{Mm}{p^2}$$

$F_h$ = horizontal component of the force of attraction

$F_v$ = vertical component of the force of attraction

$$F^2 = F_v^2+F_h^2$$

## The Attempt at a Solution

I made the following diagram to help illustrate my process

First, all the horizontal components of the force F will cancel out due to symmetry. So I just need to find a Riemann Sum for all the vertical components.

$$F_v^2 = F^2 - F_h^2$$

$$F_h = F \sin{\phi} = F \frac{r}{r^2+d^2}$$

$$F_v^2 = F^2 - \left(F \frac{r}{r^2+d^2}\right)^2 = F^2\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)$$

$$F_v = F \sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)}$$

Now, to just consider the polar subrectangle $R_{ij}$

$$F_{v_{ij}} = G\frac{mM_{ij}}{r^2+d^2}\sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)}$$

and I figure that $M_{ij} = \rho \Delta A$

so $dA = rdrd\theta$ when changing to polar form.

I get the double integral

$$F_v = Gm\rho \int_0^{2\pi}\int_0^R \frac{r}{r^2+d^2}\sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)} drd\theta$$

Is this right? Because I can't get Matlab to evaluate it analytically. I've tried trig. sub and u sub on it. It looks like there should be a way through via some sort of fairly standard method, but the change of the infinitesmals always gives me some hideous substitute that just makes the integral more complicated.

Do I need to revisit my model, or do I have the correct double integral and I just need to figure out how to solve it?

kostoglotov said:

## Homework Statement

A lamina has constant density $\rho$ and takes the shape of a disk with center the origin and radius R. Use Newton's Law of Gravitation to show that the magnitude of the force of attraction that the lamina exerts on a body of mass m located at the point (0,0,d) on the positive z axis is

$$F = 2 \pi Gm \rho d \left(\frac{1}{d}-\frac{1}{\sqrt{R^2+d^2}}\right)$$

[Hint: Divide the disk and first compute the vertical component of the force exerted by the polar subrectangle $R_{ij}$]

## Homework Equations

p = diagonal distance between mass m and a polar subrectangle of the disk.

$$F = G\frac{Mm}{p^2}$$

$F_h$ = horizontal component of the force of attraction

$F_v$ = vertical component of the force of attraction

$$F^2 = F_v^2+F_h^2$$

## The Attempt at a Solution

I made the following diagram to help illustrate my process

First, all the horizontal components of the force F will cancel out due to symmetry. So I just need to find a Riemann Sum for all the vertical components.

$$F_v^2 = F^2 - F_h^2$$

$$F_h = F \sin{\phi} = F \frac{r}{r^2+d^2}$$

$$F_v^2 = F^2 - \left(F \frac{r}{r^2+d^2}\right)^2 = F^2\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)$$

$$F_v = F \sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)}$$

Now, to just consider the polar subrectangle $R_{ij}$

$$F_{v_{ij}} = G\frac{mM_{ij}}{r^2+d^2}\sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)}$$

and I figure that $M_{ij} = \rho \Delta A$

so $dA = rdrd\theta$ when changing to polar form.

I get the double integral

$$F_v = Gm\rho \int_0^{2\pi}\int_0^R \frac{r}{r^2+d^2}\sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)} drd\theta$$

Is this right? Because I can't get Matlab to evaluate it analytically. I've tried trig. sub and u sub on it. It looks like there should be a way through via some sort of fairly standard method, but the change of the infinitesmals always gives me some hideous substitute that just makes the integral more complicated.

Do I need to revisit my model, or do I have the correct double integral and I just need to figure out how to solve it?

Check your expression for ##F_h##. You are missing a square root in the value for ##sin(\theta)##. And while you are at it, why don't you just compute ##F_h## using the cosine of that angle?

kostoglotov
kostoglotov said:

## Homework Statement

A lamina has constant density $\rho$ and takes the shape of a disk with center the origin and radius R. Use Newton's Law of Gravitation to show that the magnitude of the force of attraction that the lamina exerts on a body of mass m located at the point (0,0,d) on the positive z axis is

$$F = 2 \pi Gm \rho d \left(\frac{1}{d}-\frac{1}{\sqrt{R^2+d^2}}\right)$$

[Hint: Divide the disk and first compute the vertical component of the force exerted by the polar subrectangle $R_{ij}$]

## Homework Equations

p = diagonal distance between mass m and a polar subrectangle of the disk.

$$F = G\frac{Mm}{p^2}$$

$F_h$ = horizontal component of the force of attraction

$F_v$ = vertical component of the force of attraction

$$F^2 = F_v^2+F_h^2$$

## The Attempt at a Solution

I made the following diagram to help illustrate my process

First, all the horizontal components of the force F will cancel out due to symmetry. So I just need to find a Riemann Sum for all the vertical components.

$$F_v^2 = F^2 - F_h^2$$

$$F_h = F \sin{\phi} = F \frac{r}{r^2+d^2}$$

$$F_v^2 = F^2 - \left(F \frac{r}{r^2+d^2}\right)^2 = F^2\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)$$

$$F_v = F \sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)}$$

Now, to just consider the polar subrectangle $R_{ij}$

$$F_{v_{ij}} = G\frac{mM_{ij}}{r^2+d^2}\sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)}$$

and I figure that $M_{ij} = \rho \Delta A$

so $dA = rdrd\theta$ when changing to polar form.

I get the double integral

$$F_v = Gm\rho \int_0^{2\pi}\int_0^R \frac{r}{r^2+d^2}\sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)} drd\theta$$

Is this right? Because I can't get Matlab to evaluate it analytically. I've tried trig. sub and u sub on it. It looks like there should be a way through via some sort of fairly standard method, but the change of the infinitesmals always gives me some hideous substitute that just makes the integral more complicated.

Do I need to revisit my model, or do I have the correct double integral and I just need to figure out how to solve it?
If you're interested in only the vertical component of the gravitational force, why are you wasting time calculating the horizontal component?

After all, you can use Fv = F cos φ just as easily, instead of going thru the complication of the Pythagorean relation to find cosine.

[EDIT: I think Dick means why don't you just calculate Fv = F cos φ?).

kostoglotov
SteamKing said:
[EDIT: I think Dick means why don't you just calculate Fv = F cos φ?).

Yes, I did mean that. Thanks for the correction!

kostoglotov said:

## Homework Statement

A lamina has constant density $\rho$ and takes the shape of a disk with center the origin and radius R. Use Newton's Law of Gravitation to show that the magnitude of the force of attraction that the lamina exerts on a body of mass m located at the point (0,0,d) on the positive z axis is
$$F = 2 \pi Gm \rho d \left(\frac{1}{d}-\frac{1}{\sqrt{R^2+d^2}}\right)$$
[Hint: Divide the disk and first compute the vertical component of the force exerted by the polar subrectangle $R_{ij}$]

## Homework Equations

p = diagonal distance between mass m and a polar subrectangle of the disk.
$$F = G\frac{Mm}{p^2}$$
$F_h$ = horizontal component of the force of attraction

$F_v$ = vertical component of the force of attraction
$$F^2 = F_v^2+F_h^2$$

## The Attempt at a Solution

I made the following diagram to help illustrate my process

[ IMG]http://i.imgur.com/lP0o2HU.jpg[/PLAIN]

First, all the horizontal components of the force F will cancel out due to symmetry. So I just need to find a Riemann Sum for all the vertical components.
$$F_v^2 = F^2 - F_h^2$$
$$F_h = F \sin{\phi} = F \frac{r}{r^2+d^2}\ \quad\ \ \color{red}{\text{Take note: }\sin\phi=\frac{r}{\sqrt{r^2+d^2\ }}}$$
$$F_v^2 = F^2 - \left(F \frac{r}{r^2+d^2}\right)^2 = F^2\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)$$
$$F_v = F \sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)}$$
Now, to just consider the polar subrectangle $R_{ij}$
$$F_{v_{ij}} = G\frac{mM_{ij}}{r^2+d^2}\sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)}$$
and I figure that $M_{ij} = \rho \Delta A$

so $dA = rdrd\theta$ when changing to polar form.

I get the double integral
$$F_v = Gm\rho \int_0^{2\pi}\int_0^R \frac{r}{r^2+d^2}\sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)} drd\theta$$
Is this right? Because I can't get Matlab to evaluate it analytically. I've tried trig. sub and u sub on it. It looks like there should be a way through via some sort of fairly standard method, but the change of the infinitesimals always gives me some hideous substitute that just makes the integral more complicated.

Do I need to revisit my model, or do I have the correct double integral and I just need to figure out how to solve it?
You made this more complicated than you needed to.

You recognized that the horizontal components cancel, so you only needed to consider the vertical components, and then you proceeded to calculate the individual vertical components by using the individual horizontal components.

This would all have worked if you had used the correct expression for ##\ \sin\phi\ .##

##\displaystyle \ {F_v}^2 = F^2 - \left(F \frac{r}{\sqrt{r^2+d^2\,}}\right)^2 =F^2\left(1 - \left(\frac{r}{\sqrt{r^2+d^2\,}}\right)^2\right)\ ##​

However, you could directly use ##\displaystyle \ F_v = F \cos{\phi}=F\frac{d}{\sqrt{r^2+d^2\ }}\ .##

While I was messing around formatting things, the SteamKing and Dick beat me to this. -- but after all that effort, I'm leaving this up.

Last edited:
kostoglotov
SteamKing said:
If you're interested in only the vertical component of the gravitational force, why are you wasting time calculating the horizontal component?

Because I like to punish myself. :P

## 1. What is the formula for calculating the gravitational force of a disk on a mass on the z-axis?

The formula for calculating the gravitational force of a disk on a mass on the z-axis is F = G * (M * m) / r^2, where F is the force, G is the gravitational constant, M is the mass of the disk, m is the mass of the object, and r is the distance between the center of the disk and the object.

## 2. How does the distance between the disk and the object affect the gravitational force?

The gravitational force is inversely proportional to the square of the distance between the disk and the object. This means that as the distance increases, the force decreases.

## 3. Does the mass of the disk or the mass of the object have a greater impact on the gravitational force?

The mass of the object has a greater impact on the gravitational force. This is because the force is directly proportional to the mass of the object.

## 4. How does the disk's shape affect the gravitational force?

The disk's shape does not affect the gravitational force as long as the mass is evenly distributed. The formula for calculating the force assumes a point mass at the center of the disk, so the shape does not play a role.

## 5. Can the gravitational force of a disk on a mass on the z-axis be negative?

Yes, the gravitational force can be negative if the mass of the disk and the object have opposite signs. This means that the force is attractive if the masses have the same sign and repulsive if they have opposite signs.

Replies
3
Views
983
Replies
3
Views
1K
Replies
2
Views
901
Replies
20
Views
965
Replies
7
Views
2K
Replies
12
Views
3K
Replies
1
Views
1K
Replies
2
Views
783
Replies
5
Views
1K
Replies
17
Views
2K