# Kepler's third law implies force proportional to mass

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1. Jul 25, 2016

### Dazed&Confused

1. The problem statement, all variables and given/known data
Show that Kepler's third law, $\tau = a^{3/2}$, implies that the force on a planet is proportional to its mass.

2. Relevant equations
3. The attempt at a solution

I haven't really attempted anything. I'm not sure what the question is going for. What can we assume and use?

2. Jul 25, 2016

### J Hann

Hint: You are given T^2 = k a^3 where T = period, k = constant, a = mean distance
Also, you know F = m v^2 / a
The trick is to eliminate v (how does this relate to T and a?)

3. Jul 26, 2016

### Dazed&Confused

Well in this question a is the semi-major axis. The book tells me (without proof) that $a$ is also the median distance.

I'm not sure why you said $F = mv^2/a$ when $mv^2/r$ is just the centripetal acceleration, and only equal to the force in the case of a circular orbit, right?

4. Jul 26, 2016

### yihloong

in my point of view, the question start by asking you to show, therefore you should start with the derivation/proof for Kepler's third law.
hint: Kepler third law is about the orbits of planets, so think of any equations that applicable to the orbit of a planet.

5. Jul 26, 2016

### Dazed&Confused

Ok well the derivation of Kepler's third law requires knowing the expressions for the semi-major axis and semi-minor axis which also requires Kepler's first law. The derivation also makes use of Kepler's second law.

6. Jul 26, 2016

### yihloong

i think it is safe to assume the mean distance of an elliptical orbit = [ (semi major + semi minor )/2 ] , is approximately equal to the radius of a circular orbit. hence, F=mv2/r or mv2/a is applicable.

7. Jul 26, 2016

### Dazed&Confused

In that case the problem is a bit too easy I think, assuming I've done it correctly. If we are assuming the orbit is nearly circular so that $\ddot{r}$ is near zero then $$F = m \omega^2 r = \frac{m4 \pi^2 r}{\tau^2} = \frac{m 4 \pi^2 r}{k^2r^3} = \frac{m 4 \pi^2}{k^2r^2}$$

This is not really what they want I think.

8. Jul 26, 2016

### yihloong

i don't think you can substitutes t2 = k2r3, because that's what they want to show, so you should have a series of steps which finally leads to t= a2/3.

another hint: derivation of Kepler's third law start with F=mv2/r , or any equivalent formula for centripetal force, due to it's a circular orbit. and F=GMm/r2 due to force of attraction between 2 celestial object. work on the two equation.

9. Jul 26, 2016

### PeroK

Well, if Kepler's laws hold for any ellipical orbit, then they must hold for a circlular orbit; and, as a circle is symmetric, there must be constant speed; and, as the speed is constant, you can derive the force equation. QED

10. Jul 26, 2016

### Dazed&Confused

I think the question is asking to do the opposite, to start from Kepler's third law and end up with the force equation.

11. Jul 26, 2016

### yihloong

oh, then i must be misunderstanding the question