Kepler's laws: I can recite them, but can't use them. Is planetary speed constant? Which 'a' in T² ∝ a³?

7 answers & replies

• [+][–]Neel Chatterjee12thSample· 1mo ago

  Law 2 is angular momentum conservation. Gravity is a central force, so the torque about the Sun is zero and $L = mvr\sin\phi$ is constant. The areal velocity is

  $\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$

  So the planet cannot have constant speed. Where it is closer, it must move faster. At perihelion and aphelion the velocity is perpendicular to the radius, so $v_p r_p = v_a r_a$. Closer means faster, in exact inverse proportion to the distance.

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• [+][–]Aman YadavDropperSample· 1mo ago

  (ii) Use Law 3. $T^2 \propto r^3$, so $T \propto r^{3/2}$.

  $T_A/T_B = 4^{3/2} = (\sqrt4)^3 = 8$.

  You squared T on one side and cubed r on the other, that's why you got confused. Never apply 16 or 64 alone. Pick the form $T \propto r^{3/2}$ and do it in one line.

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• [+][–]Rohan Sharma11thSample· 1mo ago

  Isn't Law 2 just conservation of energy? Speed increases because potential energy decreases as it goes near the Sun. i read that somewhere lol

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  • [+][–]Neel Chatterjee12thSample· 1mo ago

    Rohan, energy is also conserved (KE goes up as PE goes down), and it explains the speed change. But the equal-area statement itself comes from angular momentum, not energy. A planet could be in a circular orbit with energy conserved and areas equal, and an elliptical one too; the area law is about L.

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• [+][–]Ishita GoyalDropperSample· 1mo ago

  (iii) It's the semi-major axis, which is half of the longest diameter of the ellipse. It equals the average of the nearest and farthest distances, $a = \dfrac{r_p + r_a}{2}$.

  (iv) A circle is just an ellipse with both foci together at the centre, so $a = r$ and everything works. The laws were not "for ellipses only", they're for any bound orbit of an inverse-square force.

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• [+][–]Rahul DubeyDropperSample· 1mo ago

  Where does Law 3 come from? Do it for a circle, once. Gravity gives the centripetal force:

  $\frac{GMm}{r^2} = m\omega^2r = m\frac{4\pi^2}{T^2}r \Rightarrow T^2 = \frac{4\pi^2}{GM}r^3$

  So the constant is $4\pi^2/GM$, which depends only on the central body (the Sun), not on the planet's mass. Same logic explains why all planets share one constant, and Earth satellites share a different constant. For ellipses r gets replaced by a.

  Third attempt and I still enjoy this derivation. 🙂

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  • [+][–]Aarav Kulkarni11thSample· 1mo agoAuthor

    thank you, ticking all four boxes now. the T ∝ r^(3/2) one-liner is going on my formula card

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