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Movement on a Sphere

  1. Mar 14, 2004 #1


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    Hi all,

    I have a question on the movement of a point on a sphere.

    Consider a point [tex]P(\theta,\phi)[/tex] ([tex]\theta[/tex] and[tex]\phi[/tex] are the latitude and the longitude, respectively)moving from a point [tex]P_1(\theta_1,\phi_1)[/tex] to a point [tex]P_2(\theta_2,\phi_2)[/tex] with a constant speed [tex]v[/tex].

    How to find the latitude and the longitude of [tex]P(\theta,\phi)[/tex] (the expression of [tex]\theta[/tex] and [tex]\phi[/tex]) if we suppose that the trajectory is the shortest distance between [tex]P_1[/tex] and [tex]P_2[/tex] ?

  2. jcsd
  3. Mar 14, 2004 #2
    I'm not quite understanding the question. Do you want the angles as a function of time, or one angle as a function of the other?

    To start working on the problem, you can find the theta and phi components of the velocity from the known P1, P2 and the radius of the sphere. Then it shouldn't be too bad to write down expressions for the two angles as a function of t , since the speed is constant. The phi equation will take a little more trig than the equation for theta. When you get this far, you should be able to answer your question.
  4. Mar 14, 2004 #3


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    thank you for the answer.

    angles as a function of time.

    I see how to do it on a circle but on a sphere I don't arrive to visualize the movement.

    Just to be sure, on a circle of radius R we have that [tex]d \theta=\frac{v}{R} dt[/tex], so by integrating we have [tex]\theta(t)=\frac{v}{R} t + \theta_1[/tex] ?
  5. Mar 14, 2004 #4
    You have the circle approach correct. The same approach will work for the theta component on the sphere. For phi, R will have to be replaced by R cos(theta), but the approach is otherwise the same. You will still have to figure out how to set up the two components of v (theta and phi) to go along a great circle from the initial to the final point.
  6. Mar 17, 2004 #5


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    thanks. I still don't see how things work. Any other help ?

    I have that [tex]x=R \cos(\theta) \cos(\phi), \\ y=R \sin(\theta) \cos(\phi), \\ z=R \sin(\phi)[/tex]

    and [tex]\frac{dr}{dt}=v [/tex]

    what am I missing ?. Is [tex]\tan(\theta)=\frac{r}{R} [/tex] ?
    Last edited: Mar 17, 2004
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