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Path of a projectile traveling along a meridian of the earth

  • #1

Homework Statement


I've been working out an example by myself from the book Adv. Engg. Mathematics, E.Kreszig, section 8.7 p441.

The path of a projectile traveling along a meridian of the earth uniformly may be given by
[tex]\vec{r(t)}=Rcos\gamma t\vec{b} + Rsin\gamma t\vec{k}[/tex] where i,j,k are unit vectors along x,y,z axes respectively.

The vector [tex]\vec{b}[/tex] is defined as [tex]\vec{b}=cos\omega t\vec{i} +sin\omega t\vec{j}[/tex]

Differentiating [tex]\vec{r(t)}[/tex] twice to get the expression for acceleration. This comes out to be

[tex]\vec{a(t)}=-R(\gamma ^2+\omega ^2)cos(\gamma t)\vec{b} -2R\gamma sin(\gamma t)\vec{b'} - R\gamma ^2cos(\gamma t)\vec{k}[/tex].

The first term is the centripital acceleration due to the earth and the path of the projectile, the second is the coriolis acceleration. What is the third term?
 

Answers and Replies

  • #2
pam
455
1
It is the z component of the centripetal acceleration.
It should have sin\gamma t.
 
  • #3
Oh. ! thank you. That was incredibly stupid of me.
 

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