Question

The German astronomer Johannes Kepler developed three major laws of planetary motion. His third law can be expressed by the equation \(\log T=\frac{3}{2} \log d-3.263\) where \(T\) is the time, in Earth years, for the planet to revolve around the sun and \(d\) is the average distance, in millions of kilometers, from the sun. a) Pluto is on average 5906 million kilometers from the sun. To the nearest Earth year, how long does it take Pluto to revolve around the sun? b) Mars revolves around the sun in 1.88 Earth years. How far is Mars from the sun, to the nearest million kilometers?

Short answer

a) Pluto takes approximately 248 Earth years to revolve around the sun. b) Mars is about 228 million kilometers from the sun.

Step-by-step solution

Identify the given values for part (a)

For part (a), the average distance of Pluto from the sun is given as 5906 million kilometers. This means that d = 5906.

Substitute the value of d into Kepler's third law

Substitute d = 5906 into the equation \(\text{log} T = \frac{3}{2} \text{log} d - 3.263\).

Calculate \(\text{log} d\)

First, find the logarithm of 5906: \( \text{log} d = \text{log} 5906\).

Calculate the expression \(\frac{3}{2} \text{log} d\)

Multiply the logarithm by \( \frac{3}{2} \): \( \frac{3}{2} \text{log} 5906 \).

Compute \(\text{log} T\)

Substitute \( \frac{3}{2} \text{log} 5906 \) into the equation: \( \text{log} T = \frac{3}{2} \text{log} 5906 - 3.263 \).

Find T

To find T in Earth years, solve for T using: \( T = 10^{ \text{log} T} \).

Identify the given values for part (b)

For part (b), the time for Mars to revolve around the sun is given as 1.88 Earth years. This means that T = 1.88.

Substitute the value of T into Kepler's third law

Substitute T = 1.88 into the equation \( \text{log} T = \frac{3}{2} \text{log} d - 3.263 \) to solve for d.

Transform and solve for \(\text{log} d\)

Express the equation in terms of \( \text{log} d \): \( \text{log} d = \frac{ \text{log} T + 3.263}{ \frac{3}{2}} \).

Compute d

Find the value of d by solving the equation: \( d = 10^{ \text{log} d} \).

Key concepts

planetary motion

Planets move in elliptical orbits around the sun. This concept is encapsulated in Kepler's laws of planetary motion.
The third law, which we focus on here, relates the distance of a planet from the sun to its orbital period.
This law shows a mathematical relationship between how far planets are from the sun and how long they take to complete an orbit.
When applying Kepler's third law, it’s essential to recognize that the further away a planet is from the sun, the longer its orbital period.
Understanding this law can help explain phenomena such as the lengthy 248-year orbit of Pluto compared to the 88-day orbit of Mercury.

logarithm

A logarithm is the power to which a number must be raised to get another number.
For example, in Kepler’s third law equation \(\log T = \frac{3}{2} \log d - 3.263\), we use logarithms to manipulate the relationship between the distance from the sun (d) and the orbital period (T).
The base-10 logarithm, commonly seen as \(\log\), simplifies large multiplication and division tasks.
In the context of this exercise, it helps convert multiplicative relationships into additive ones, making it easier to solve for the unknowns.

astronomy calculations

Astronomy calculations involve using mathematical principles to understand celestial phenomena.
Using Kepler's third law, we can predict the behavior of planets in our solar system.
This involves substituting known values into the equation and solving for the unknown.
For instance, if we know the distance of a planet from the sun, we can determine its orbital period, and vice versa.
These calculations help in understanding not just the planets in our solar system but also in analyzing exoplanets orbiting other stars.

distance from the sun

The average distance of a planet from the sun is a crucial factor in determining its orbital period.
In the equation \(\log T = \frac{3}{2} \log d - 3.263\), 'd' represents this average distance, usually measured in millions of kilometers.
For example, Pluto’s average distance from the sun is about 5906 million kilometers.
This distance dictates that it takes Pluto approximately 248 Earth years to orbit the sun.
On the other hand, the closer Mars averages 227.94 million kilometers away, leading to its shorter 1.88-year orbit.

orbital period

The orbital period is the time it takes for a planet to complete one full orbit around the sun.
In Kepler’s third law, 'T' represents this period, measured in Earth years.
Knowing a planet's distance from the sun allows us to calculate its orbital period using the relationship \(\log T = \frac{3}{2} \log d - 3.263\).
Conversely, if the orbital period is known, we can find the average distance from the sun.
These calculations explain why inner planets, which are closer to the sun, have shorter orbital periods compared to distant outer planets.