Kepler's third law implies force proportional to mass

[Dazed&Confused]

Homework Statement

Show that Kepler's third law, [itex]\tau = a^{3/2}[/itex], implies that the force on a planet is proportional to its mass.

Homework Equations

3. The Attempt at a Solution [/B]
I haven't really attempted anything. I'm not sure what the question is going for. What can we assume and use?

 

[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?)

 

[Dazed&Confused]

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

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

 

[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.

 

[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.

 

[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.

 

[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 [itex]\ddot{r}[/itex] is near zero then [tex] 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}[/tex]

This is not really what they want I think.

 

[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= a^2/3.

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.

 

[PeroK]

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 [itex]\ddot{r}[/itex] is near zero then [tex] 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}[/tex]

This is not really what they want I think.

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

 

[Dazed&Confused]

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= a^2/3.

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.

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

 

[yihloong]

oh, then i must be misunderstanding the question