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Use Kepler's 2nd to show

  1. Sep 12, 2007 #1
    whether a planet moves faster or slower when it is closer to the Sun.

    Now I intuitively know that is [tex]\frac{dA}{dt}[/tex] is constant and the Radius r is changing. Then as r increases it moves slower and as r decreases is moves faster.

    I am having trouble setting up a more formal proof.

    I am thinking I will need to use the relationship [tex]L=\vec r*\vec p[/tex]

    Please send a hint my way.

  2. jcsd
  3. Sep 12, 2007 #2


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    Use right triangles to approximate the area swept...
  4. Sep 12, 2007 #3
    [tex]T = 2\pi \sqrt {\frac{r}{glocal}}[/tex]

    This is a great formula that'll solve a lot of questions for you in times of need. Just remember that it's the same as the pendulum formula, only that g is the strength of Earth's gravitational field at the point of the satellite. This is only valid for orbits with low eccentricities.

    [tex]g[/tex]local[tex] = \frac {9.8} {R^2} [/tex] where R is a number that is in terms of Earth's radius. i.e. 2 radii from Earth's CoM. You can use:

    [tex]g[/tex]local [tex] = \frac{GM} { r^2}[/tex] if you don't like using shortcuts.
  5. Sep 12, 2007 #4
    Opps, that's Kepler's 3rd law.

    The second law is about area, so yeah, use triangles.
  6. Sep 12, 2007 #5
    What you could do is consider two traingles of small but equal areas. One representing the distance to the nearer part and other to the farther part. The areas of these two triangles are same, and one side of each triangle represents the distance covered by the two objects in the same time.

    Areas of triangles are equal,

    1/2*d1*r1=1/2*d2*r2 -- where r1 and r2 are perpendiculars drawn from focus to d1,d2, which thus represent the radius approximately.

    d1*r1 = d2*r2

    If r1<r2, then d1>d2. Thus proved.

    The above proof is a very crude proof I came up with, but still it works quite a bit.

  7. Sep 12, 2007 #6
    The text makes reference to this method and I can sort of follow it, but I have some questions...but I first have to figure out how to word them!

    So if I draw radius r from the Sun focus F to any point on the ellipse, and then a small line extending out from it at 90 degreesand then connect that line to the focus as well...I have a right triangle whose area id approximate to that of the area swept out ....I wish I could draw this in LaTex.

    ...to be continued shortly...
  8. Sep 12, 2007 #7
    Sleek, I think I may be able to work something out of this, however as in my description above I think my right angle is in a different spot...however it may not matter, seeing as there is indeed a right angle and my r is indeed r.
    I will upload a pic.

    Now how can I write this out quantitatively to state that as r increase speed must decrease since dA/dt is constant?


    Attached Files:

    Last edited: Sep 12, 2007
  9. Sep 13, 2007 #8
    In the below image, you can see two traingles, with equal areas and a perpendicular dropped to both of their "bases", which is an approximation of the distance covered by a body. I've exaggerated the size of the triangles, so the real distance and the distance represented by the base of the triangles might look distinct, but we're assuming triangles with very small areas.
    What we do is, from a focus of an ellipse, we draw two lines to a longer and smaller side of the ellipse, which is indeed the radii r1 and r2. Then we construct the bases of both triangles such that its perpendicular to the radii and the areas of both triangles are equal and infinitesimally small. The digram is larger just for clarity.

    By equating areas of both triangles as before, we get

    since r1<r2, d1>d2.

    For a specific time interval, a body travels d1 distance when nearer to focus and d2 when far away such that d1>d2. But it covers both this distance in the same time t. Thus speed nearer to the focus is larger.

    I could not see your attached picture since it is still pending approval.
  10. Sep 13, 2007 #9


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    Hint: rw = v. So dA/dt = (1/2)rv
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