Journal of Online Mathematics and Its Applications

Which Planet is Largest?

Phillip Brown and James Braselton

Introduction
Circular Orbits
Elliptical Orbits
Conclusions
References and Resources

1. Introduction

Jupiter, of course! Well, perhaps not--in this article we interpret the question in a new way: which planet occupies the largest space as it revolves about the Sun? As an extreme case, the volume occupied by Mercury in its orbit about the Sun is surely less than the volume occupied by Earth in its orbit about the Sun. After all, Mercury is much smaller than Earth and much closer to the Sun. More thoughtfully, we pondered: "Pluto is much smaller than Earth but much farther away from the Sun. Which planet's orbit about the Sun occupies a greater volume?" More generally, how do the size of a planet and its distance to the Sun affect the orbit volume of the planet? Finally, a planet's distance to the Sun is not constant. What affect does this have?

The graphic below, from NASA, shows the solar system with the planets in their correct relative sizes. It would be nice to show correct relative distances as well, but that's essentially impossible given the enormous variation in the distances and the limits of a computer screen.

Image: SolarSystem.jpg

Kepler's laws describe planetary motion. You can visit the excellent Kepler Mission site for more information about Johannes Kepler and his laws. However, for this article, we just need his first law:

Kepler's first law: Planets move in elliptical orbits with the Sun at one focus.

Table 1 below, from NASA, gives data on the elliptical orbits of the planets.

Table 1. Planetary Data
Planet	Semimajor Axis of Orbit (km)	Mean Radius (km)	Orbit Eccentricity
Mercury	57,909,175	2,439.7	0.20563
Venus	108,208,930	6,051.8	0.0068
Earth	149,597,890	6,378.14	0.016710
Mars	227,936,640	3,397	0.0934
Jupiter	778,412,020	71,492	0.04839
Saturn	1,426,725,400	60,268	0.0541506
Uranus	2,870,972,200	25,559	0.047168
Neptune	4,498,252,900	24,764	0.00859
Pluto	5,906,380,000	1,151	0.2488

2. Circular Orbits

Because the eccentricities of the planets are small, their orbits are often assumed to be circular. The following image shows the orbits of some of the inner planets.

Image: InnerOrbits.jpg

The animation below shows two planets with circular orbits. One is larger but the other is farther from the Sun. Which planet would generate the greater volume? Run the animation and make a guess.

Assuming a circular orbit, a planet's orbit space is the solid obtained by revolving a circular region about the y-axis, which is a torus. Run the animation below to view the process.

So, how do we find the orbit volume? In calculus (see any standard text, such as Stewart), we learn the methods of washers and cylindrical shells to find the volume of a solid obtained by revolving a given region about a given line. For $0 < r < x_{0}$ , we can use either method to find the volume of the solid obtained by revolving the region bounded by the circle ${(x - x_{0})}^{2} + y^{2} = r^{2}$ about the y-axis.

Using the method of cylindrical shells, the volume is:

(1)	$\int_{x_{0} - r}^{x_{0} + r} 4 π x \sqrt{r^{2} - {(x - x_{0})}^{2}} d x = 4 π \int_{- r}^{r} (u + x_{0}) \sqrt{r^{2} - u^{2}} d u = 0 + 4 π x_{0} r^{2} \frac{1}{2} π r^{2} = 2 π^{2} x_{0} r^{2}$

where we have used that $\int_{- r}^{r} u \sqrt{r^{2} - u^{2}} d u = 0$ because $y = u \sqrt{r^{2} - u^{2}}$ is an odd function and $\int_{- r}^{r} \sqrt{r^{2} - u^{2}} d u = \frac{1}{2} π r^{2}$ because the graph of $y = \sqrt{r^{2} - u^{2}}$ is the upper half of the graph of the circle with center at (0, 0) and radius r, $y^{2} + u^{2} = r^{2}$ . (Of course, you could also use the substitutions $w = r \sin (θ)$ or $w = r \cos (θ)$ to evaluate $\int_{- r}^{r} \sqrt{r^{2} - u^{2}} d u$ .)

Let's check our answer using the method of washers:

\int_{- r}^{r} π [{(x_{0} + \sqrt{r^{2} - y^{2}})}^{2} - {(x_{0} - \sqrt{r^{2} - y^{2}})}^{2}] d y = 4 π x_{0} \int_{- r}^{r} \sqrt{r^{2} - y^{2}} d y = 4 π x_{0} \frac{1}{2} π r^{2} = 2 π^{2} x_{0} r^{2}

Good, they agree! Of course, we also recognize our answer as the volume of the circular cylinder of radius r and length $2 π x_{0}$ ; we could obtain this cylinder by cutting the torus and straightening it out:

Image: Cylinder.png

We can now use Equation 1 and the data in Table 1 to approximate the orbit volume of each planet. The x₀ value is obtained from the second column of Table 1 and the r value from the third column. The results (in km³) are given in the second column of Table 2.

Table 2. Orbit volumes of the planets
Planet	Orbit Volume (km³), circular orbit	Orbit Volume (km³), elliptical orbit	Percent Change
Mercury	6.803775 × 10¹⁵	6.731273 × 10¹⁵	1.07709%
Venus	7.822796 × 10¹⁶	7.822705 × 10¹⁶	0.00116%
Earth	1.201277 × 10¹⁷	1.201194 × 10¹⁷	0.00698%
Mars	5.192004 × 10¹⁶	5.180662 × 10¹⁶	0.21189%
Jupiter	7.853336 × 10¹⁹	7.848736 × 10¹⁹	0.05861%
Saturn	1.022925 × 10²⁰	1.022174 × 10²⁰	0.07347%
Uranus	3.702085 × 10¹⁹	3.700025 × 10¹⁹	0.05567%
Neptune	5.445217 × 10¹⁹	5.445117 × 10¹⁹	0.00184%
Pluto	1.544549 × 10¹⁷	1.520362 × 10¹⁷	1.59087%

3. Elliptical Orbits

Of course, the orbits of the planets are not perfect circles. The image below shows the orbits of some of the outer planets.

Image: OuterOrbits.jpg

In this section we will attempt to improve our approximations by finding the volume of the region occupied by a sphere in an elliptical orbit. First, run the animation below. The orbit of the smaller planet is approximately circular while the orbit of the larger planet is elliptical. Which would generate the larger volume?

Run the animation below to see the solid generated by a planet in an elliptical orbit.

Now let's do the math. Consider an elliptical shell with height h and base in the x-z plane given by $\frac{x^{2}}{a^{2}} + \frac{z^{2}}{b^{2}} = 1$ . Using the arc length formula in rectangular coordinates, the circumference of the ellipse is

4 \int_{0}^{a} \sqrt{1 + {[\frac{d}{d x} (\frac{b}{a} \sqrt{a^{2} - x^{2}})]}^{2}} d x = 4 \int_{0}^{a} \sqrt{1 + \frac{b^{2} x^{2}}{a^{2} (a^{2} - x^{2})}} d x

Hence the surface area of the shell can be computed by multiplying the circumference of the ellipse by the height:

(2)	$4 h \int_{0}^{a} \sqrt{1 + \frac{b^{2} x^{2}}{a^{2} (a^{2} - x^{2})}} d x$

Now assume that the major axis is the line segment connecting $(- a, 0)$ and $(a, 0)$ and that the ellipse has eccentricity e, which is defined by

\frac{\sqrt{a^{2} - b^{2}}}{a} = e

Then $b = a \sqrt{1 - e^{2}}$ and rewriting Equation 2 in terms of the length of the major axis and the eccentricity of the ellipse gives us the elliptical shell surface area:

(3)	$4 h \int_{0}^{a} \sqrt{1 + \frac{(e^{2} - 1) x^{2}}{x^{2} - a^{2}}} d x = 4 a h E (e^{2})$

where E(z) is the complete elliptic integral of the second kind:

(4)	$E (z) = \int_{0}^{π / 2} \sqrt{1 - z \sin^{2} (θ)} d θ = \frac{π}{2} - \frac{π}{8} z - \frac{3 π}{128} z^{2} - \frac{5 π}{512} z^{3} - \frac{175 π}{32768} z^{4} - \dots$

You can learn more about elliptic integrals at the MathWorld site. Now, to find the volume of the region occupied by the sphere

{(x - x_{0})}^{2} + y^{2} + {(z - z_{0})}^{2} = r^{2}

as it travels about its elliptical path,

x_{0}^{2} + \frac{z_{0}^{2}}{1 - e^{2}} = r^{2}

in the x-z plane (y = 0), we use the method of elliptical shells and evaluate the volume of the elliptical torus:

(5)

\begin{array}{l} \int_{x_{0} - r}^{x_{0} + r} 4 \cdot 2 \sqrt{r^{2} - {(x - x_{0})}^{2}} (\int_{0}^{x} \sqrt{1 + \frac{(e^{2} - 1) t^{2}}{t^{2} - x^{2}}} d t) d x \\ = 8 \int_{x_{0} - r}^{x_{0} + r} \sqrt{r^{2} - {(x - x_{0})}^{2}} (\int_{0}^{x} \sqrt{1 + \frac{(e^{2} - 1) t^{2}}{t^{2} - x^{2}}} d t) d x = 4 π x_{0} r^{2} E (e^{2}) \end{array}

From the Maclaurin series expansion for E (Equation 4), observe that for $e \approx 0$ , $E (e^{2}) \approx π / 2$ and Equation 5 agrees with Equation 1:

4 π x_{0} E (e^{2}) \approx 4 π x_{0} r^{2} \cdot \frac{π}{2} = 2 π^{2} x_{0} r^{2}

We now use Equation 5, the planetary data in Table 1, and a computer algebra system to find the volume of the elliptical torus for each planet. The x₀ value is obtained from the second column of Table 1, the r value from the third column, and the e value from the fourth column. The approximate orbit volumes (in km³) of the planets are given in the third column of Table 2.

The last column of Table 2 shows the percent by which the volume of the circular orbit exceeds the volume of the elliptical orbit. With the exception of Mercury, Pluto, and (possibly) Mars, we see that the error is quite small, which helps us understand why the assumption of circular orbits is often made.

4. Conclusions

We have used two applications of integration to compute the orbit volumes of the planets. We used the methods of cylindrical shells and washers, assuming circular orbits, and the method of elliptical shells, assuming elliptical orbits. The largest planet, in terms of orbit volume, turns out to be Saturn, not Jupiter! The orbit volumes of Earth and Pluto are very close; Pluto has the larger volume because its enormous distance from the Sun outweighs its much smaller size. The following table gives a complete ranking of the planets (from smallest to largest) in terms of ordinary volume and orbit volume:

Table 3. Planet rankings
Planet	Volume Rank	Orbit Volume Rank
Mercury	2	1
Venus	4	3
Earth	5	4
Mars	3	2
Jupiter	9	8
Saturn	8	9
Uranus	7	6
Neptune	6	7
Pluto	1	5

Our colleague Kevin Phillips pointed out that the volume of the Sun is about 1.41 × 10¹⁸ km³, larger than the orbit volume of five of the nine planets! We should also note that, as we write this, the definition of planet is being reconsidered. There may soon be many additional planets whose orbit volumes can be computed.

5. References and Resources

NASA
Kepler Mission: A search for habitable planets
Stewart, James. Calculus, 5'th edition, Thomson Learning
MathWorld
Young, Hugh D. and Freedman, Roger A. University Physics with Modern Physics, 11'th edition, Addison Wesley
Wolfram Research

Source Files

Mathematica was used to performs many of the calculations illustrated here as well as to generate most of the graphics and all of the animations. You may download the Mathematica code.