# Can Kepler's 2nd Law be applied to more than one planet?

1. May 27, 2015

### Overt

I saw an explanation for why Jupiter has a slower tangential velocity in its orbit compared to inner planets and it stated:

"Remember that by Kepler’s second law, the planets each sweep-out the same area in the same amount of time. The outer planets’ elliptical orbits are considerably larger than those of the inner planets so, so over any given time period, they only need to complete a much smaller part of their orbit than do the inner planets and thus they have a longer year."

I've always heard and thought of the 2nd law with respect to a given planet, not as an explanation for the velocities of multiple planets in a system. To me it doesn't make sense to do it this way because of the differing initial conditions in each planets' formations.

2. May 27, 2015

### Orodruin

Staff Emeritus
As long as the central body is much heavier than the orbiting one, the second law has the same proportionality constant, which then only depends on Newton's gravitational constant and the mass of the central body.

Edit: I should add that this is true for circular motion. Naturally, you may have a situation with zero angular momentum regardless of the radius.

Last edited: May 27, 2015
3. May 27, 2015

### rolotomassi

The way the planets form isn't important, this is just the mechanics of a 2 body problem. Kepler's Law is a consequence of the much broader principle of conservation of angular momentum, which you most probably have already heard of.

4. May 27, 2015

### Overt

Just to be clear, if we let a stopwatch run for 10 seconds and attach a line connecting the sun to Earth and a line connecting the sun to Jupiter then they will both trace the same area as each other.

5. May 27, 2015

### Orodruin

Staff Emeritus
Sorry, I do not know what I was thinking with ... The proportionality constant does depend on the radius. In fact, it has to depend on the radius just from dimensional analysis.

The argument for why outer planets have a lower velocity is to be found in the relation between the required centripetal force to the gravitational force:
$$m \frac{v^2}r = m \frac{GM}{r^2} \quad \Longrightarrow \quad v \propto \sqrt{1/r}.$$

6. May 27, 2015

### nasu

Indeed, the areal speed is constant for a given planet but varies from planet to planet.
The relation between the distance and velocity follows from Newton's second law, not from Kepler's second law.
Their "explanation" seem to imply that all the planets have the same areal speed which is not true.

7. May 27, 2015

### rolotomassi

You can see a simple little proof as such :
The area swept out of a segment is
$$A = \int \frac{1}{2} r dr \delta \theta \ = \frac{1}{2} r^2 \delta \theta = \frac{1}{2} r^2 \frac{\delta \theta}{\delta t} \delta t \\$$ Where the RHS expression is the fraction swept out in a time dt. Using angular momentum,
$$= \frac{1}{2} r^2 \omega \delta t \ = \frac{L}{2m} \delta t$$
Because their is no torque acting on the planet the angular momentum is constant over time. Hence the area swept out in every time element, dt is constant.

This holds for individual planets. The angular momentum of Jupiter is not equal that of earth. Nor is the area swept out the same.

:::: Let me add. You should google the radius and orbital period and check whether I am indeed correct for yourself !!

Last edited: May 27, 2015