Kepler's Third Law for Elliptical Orbits: Videos & Practice Problems

Kepler's Third Law applies to elliptical orbits by using the semi-major axis (a) instead of the orbital radius (r). The equation for elliptical orbits is $T^2=\frac{4\pi a^3}{GM}$, where G is the gravitational constant and M is the mass of the star. Understanding eccentricity and distances like aphelion helps calculate the orbital period, demonstrating the relationship between distance and time in celestial mechanics.

Kepler's Third Law for Elliptical Orbits

Kepler's Third Law is a fundamental principle in celestial mechanics that describes the relationship between the orbital period of a planet and its distance from the star it orbits. While it is commonly applied to circular orbits, it also extends to elliptical orbits with a slight modification. In circular orbits, the law is expressed as:

T^2 = \frac{4\pi^2}{G M} r^3

where T is the orbital period, G is the gravitational constant, M is the mass of the star, and r is the orbital radius. For elliptical orbits, the key difference lies in the use of the semi-major axis, denoted as a, instead of the radius r. The modified equation becomes:

T^2 = \frac{4\pi^2}{G M} a^3

The semi-major axis a is defined as half the length of the longest diameter of the ellipse, and it serves as the average distance from the orbiting body to the star. In the case of circular orbits, the semi-major axis and the radius are equal, making this equation a special case of the more general elliptical scenario.

To apply Kepler's Third Law to an elliptical orbit, one must first determine the semi-major axis. This can be calculated using the aphelion distance (the farthest point in the orbit) and the eccentricity e of the orbit. The relationship is given by:

R_a = a(1 + e)

where R_a is the aphelion distance. Rearranging this equation allows us to solve for a when the aphelion distance and eccentricity are known. Once a is determined, it can be substituted back into the modified Kepler's Third Law equation to find the orbital period T.

For example, if a planet orbits a star with a mass of \(4 \times 10^{30}\) kg and has an eccentricity of 0.4, with an aphelion distance of \(1.5 \times 10^{11}\) m, we can calculate the semi-major axis:

1.5 \times 10^{11} = a(1 + 0.4)

Solving for a gives:

a = \frac{1.5 \times 10^{11}}{1.4} \approx 1.07 \times 10^{11} \text{ m}

Substituting this value into the modified Kepler's Third Law equation allows us to calculate the orbital period:

T^2 = \frac{4\pi^2}{6.67 \times 10^{-11} \times 4 \times 10^{30}} (1.07 \times 10^{11})^3

After performing the calculations, we find that T is approximately \(1.35 \times 10^{7}\) seconds, which converts to about 156 days. This result is consistent with the expected orbital period for a planet in such an orbit.

Understanding how to apply Kepler's Third Law to both circular and elliptical orbits is crucial for studying celestial mechanics and the motion of planets.

Comet Halley has a highly elliptical orbit around the Sun, circling once every 75.6 years with its closest point to the Sun being only 0.57AU ("Astronomical Unit", where 1 AU = $1.5\times 10^{11}$ m and represents the average Earth-Sun distance). How far will Comet Halley get from the Sun?

• A. $2.64\times 10^{11}$ m
• B. $2.45\times 10^{13}$ m
• C. $5.43\times 10^{12}$ m
• D. $5.25\times 10^{12}$ m

Answer: (D)

Distant Star orbiting Black Hole

In this problem, we explore the orbital characteristics of a star believed to be orbiting a black hole. The star's orbit is highly elliptical, with a period of 14 years and an eccentricity of 0.9, indicating a very elongated path. The closest point in the orbit to the black hole, known as periapsis, is given as \( r_p = 1.8 \times 10^{13} \) meters. Our goal is to determine the mass of the black hole, denoted as \( M \).

To find the mass, we utilize Kepler's Third Law, which relates the orbital period, the semi-major axis, and the mass of the central body. The law can be expressed as:

\( T^2 = \frac{4\pi^2 a^3}{G M} \)

Rearranging this equation to isolate \( M \) gives us:

\( M = \frac{4\pi^2 a^3}{G T^2} \)

Here, \( T \) is the orbital period, \( G \) is the gravitational constant (\( G = 6.67 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 \)), and \( a \) is the semi-major axis of the orbit. To proceed, we need to calculate the semi-major axis using the periapsis distance and eccentricity.

The relationship between periapsis distance \( r_p \), semi-major axis \( a \), and eccentricity \( e \) is given by:

\( r_p = a(1 - e) \)

Rearranging this equation allows us to solve for \( a \):

\( a = \frac{r_p}{1 - e} \)

Substituting the known values, we find:

\( a = \frac{1.8 \times 10^{13}}{1 - 0.9} = 1.8 \times 10^{14} \, \text{meters} \)

Next, we convert the orbital period from years to seconds for consistency in units:

\( T = 14 \, \text{years} \times 365 \, \text{days/year} \times 86,400 \, \text{seconds/day} = 4.42 \times 10^8 \, \text{seconds} \)

Now, we can substitute \( a \), \( G \), and \( T \) back into the rearranged Kepler's Third Law equation to find the mass \( M \):

\( M = \frac{4\pi^2 (1.8 \times 10^{14})^3}{(6.67 \times 10^{-11}) (4.42 \times 10^8)^2} \)

Calculating this yields:

\( M \approx 1.77 \times 10^{37} \, \text{kg} \)

This mass is approximately 8,800,000 times the mass of our Sun, highlighting the immense scale of the black hole in question. Understanding these calculations not only illustrates the application of gravitational principles but also emphasizes the fascinating dynamics of celestial bodies in our universe.

Here’s what students ask on this topic:

What is Kepler's Third Law for elliptical orbits?

Kepler's Third Law for elliptical orbits states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit. The equation is given by:

$T^2=\frac{4}{\pi }\frac{a^3}{\frac{G}{M}}$

where G is the gravitational constant and M is the mass of the star. This law helps in understanding the relationship between the distance of a planet from its star and its orbital period.

How does Kepler's Third Law differ for circular and elliptical orbits?

For circular orbits, Kepler's Third Law uses the orbital radius (r) in the equation:

$T^2=\frac{4}{\pi }\frac{r^3}{\frac{G}{M}}$

For elliptical orbits, the semi-major axis (a) replaces the orbital radius (r) because the distance to the star varies throughout the orbit. The equation becomes:

$T^2=\frac{4}{\pi }\frac{a^3}{\frac{G}{M}}$

This adjustment accounts for the varying distance in elliptical orbits.

What is the semi-major axis in an elliptical orbit?

The semi-major axis (a) in an elliptical orbit is half the length of the major axis, which is the longest diameter of the ellipse. It represents the average distance between the orbiting object and the star it orbits. The semi-major axis is crucial in applying Kepler's Third Law to elliptical orbits, as it replaces the orbital radius used in circular orbits.

How do you calculate the orbital period using Kepler's Third Law for elliptical orbits?

To calculate the orbital period (T) using Kepler's Third Law for elliptical orbits, you use the equation:

$T^2=\frac{4}{\pi }\frac{a^3}{\frac{G}{M}}$

First, determine the semi-major axis (a) of the orbit. Then, plug in the values for a, the gravitational constant (G), and the mass of the star (M). Solve for T by taking the square root of both sides of the equation. This will give you the orbital period in seconds, which can be converted to other time units if needed.

What role does eccentricity play in Kepler's Third Law for elliptical orbits?

Eccentricity (e) measures the deviation of an orbit from being circular. In Kepler's Third Law for elliptical orbits, eccentricity helps determine the semi-major axis (a) when given distances like aphelion (farthest point) or perihelion (closest point). The relationship is given by:

$r$_a=a(1+e)

and

$r$_p=a(1-e)

where r_a is the aphelion distance and r_p is the perihelion distance. Eccentricity thus influences the shape of the orbit and the calculation of the semi-major axis.