Disk Gravity

Classical Mechanics Level 3

Consider a uniform circular disk of mass M M and radius R R . The center of the disk is located 2 2 distance units away from a point particle of mass m m . The particle lies in the same plane as the disk.

What is the magnitude of the gravitational force between them?

Bonus: Compare this to the three-dimensional case (a sphere)

Details and Assumptions:
1) M = m = R = 1 M = m = R = 1
2) Universal gravitational constant G = 1 G = 1


The answer is 0.278.

Steven Chase by Steven Chase , 37, USA

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1 solution

Karan Chatrath
Nov 10, 2020

By parameterizing the disk in polar coordinates, a point lying on the disk is given by:

r ⃗ p = ( r cos ⁡ θ , r sin ⁡ θ , 0 ) \vec{r}_p = \left( r \cos{\theta} , r \sin{\theta} ,0\right)

The location of the point mass:

r ⃗ c = ( 2 , 0 , 0 ) \vec{r}_c = \left( 2 , 0 ,0\right)

The area of the mass element at the general point on the disk is: d S = r d r d θ dS = r \ dr \ d\theta . Using these facts, and applying Newton's law of gravitation leads to:

d F ⃗ = G ( M d S π R 2 ) m ( r ⃗ p − r ⃗ c ∣ r ⃗ p − r ⃗ c ∣ 3 ) d\vec{F} =G \left(\frac{M \ dS}{\pi R^2}\right) m\left( \frac{\vec{r}_p - \vec{r}_c}{\lvert \vec{r}_p - \vec{r}_c \rvert^3}\right)

By visual inspection, one can realise that the resultant Y component of the force will cancel out due to symmetry. Simplifying the above expression for the X component of force leads to:

d F x = 1 π r ( r cos ⁡ θ − 2 ) d r d θ ( ( r cos ⁡ θ − 2 ) 2 + r 2 sin ⁡ 2 θ ) 3 / 2 dF_x = \frac{1}{\pi}\frac{r(r\cos{\theta}-2) \ dr \ d\theta}{\left((r\cos{\theta}-2)^2 + r^2 \sin^2{\theta}\right)^{3/2}}

Therefore:

F x = 1 π ∫ 0 2 π ∫ 0 1 r ( r cos ⁡ θ − 2 ) d r d θ ( ( r cos ⁡ θ − 2 ) 2 + r 2 sin ⁡ 2 θ ) 3 / 2 F_x = \frac{1}{\pi} \int_{0}^{2\pi}\int_{0}^{1}\frac{r(r\cos{\theta}-2) \ dr \ d\theta}{\left((r\cos{\theta}-2)^2 + r^2 \sin^2{\theta}\right)^{3/2}}

The required answer is:

∣ F x ∣ ≈ 0.277933 \boxed{\lvert F_x \rvert \approx 0.277933}

2 Helpful 2 Interesting 2 Brilliant 0 Confused

Thanks for the solution. It's also worth pointing out that if one treats all of the disk mass as being located at the center of mass (as is useful in the 3D case), one gets the wrong answer.

Steven Chase - 7 months ago

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