This is a f=ma 2018 question and although I know there is a solution online involving pressure i'm pretty sure the more conventional way to approach this problem is with a surface integral

My approach:

$dF= \dfrac{Gm(dM)}{R^2} $

$ \int dF = \int \dfrac{Gm}{r^2} dM$

$ dM=\sigma dS $

$ dS=R^2sin \phi d\theta d\phi $

$F= \int_{0}^{2\pi} \int_{\frac{pi}{2}}^{0} \dfrac{Gm\sigma R^2}{R^2} sin\phi d\phi d\theta $

$F = - 2 \pi G m \sigma \int_{0}^{\frac{pi}{2}} sin \phi d\phi $

$F = - 2 \pi G m \sigma (cos\dfrac{\pi}{2} - cos0 )$

$F = 2 \pi G m \sigma $

My answer is off by a factor of 2 and I don't understand why that is the case.


2 Answers 2


You have summed $dF$ whereas you should have summed $dF \, \cos \theta$ as shown in the diagram below with the summation of $dF \, \sin \theta$ being zero by symmetry.

enter image description here


Looks like you did not take into account that force is a vector quantity, you need some factor in the integral for projection of the force on the axis of symmetry.

  • 1
    $\begingroup$ Im not exactly sure what that means sorry. I do know that force is a vector quantity but I since I was doing the surface integral with respect to Mass I kept my surface integral over a scalar field as taking force as a vector field would give me the flux through the hemisphere instead correct? $\endgroup$
    – hard
    May 14, 2018 at 3:48
  • $\begingroup$ @hard : But you don't need the flux, you need the force. $\endgroup$
    – akhmeteli
    May 14, 2018 at 4:18
  • $\begingroup$ @hard : To get the force, you need to integrate the projection of $dF$ on the symmetry axis. $\endgroup$
    – akhmeteli
    May 14, 2018 at 4:20

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