Section   15.1The Solar System: An Overview

1. (Exercises 1 and 2) If a tenth planet were to be orbiting the Sun, it might have a semimajor axis of about 150 AU. If this planet did exist at this distance from the Sun, how long would it take to complete one orbit around the Sun?

Since: T² / R³ = k = 1 year² / AU³

we can rewrite this equation to give:

T² = k R²

For the data given:

T² = 1 year² / AU³ (150 AU)³

T² = 3,375,000 years²

Taking the square root of both sides gives us the actual period in years:

T = 1840 years

2. (Exercises 7 and 8) Find the period of revolution for Earth if the mass of the Sun were 10 times its present value. Assume that the orbital distance of Earth is still 1 AU.

If m equals the mass of the Sun, and G is the universal gravitational constant, Newton showed that Kepler's third law could be written in the following form:

T² / R³ = 4 ² / G m as found in Section   15.1 of the textbook.

Since R remains the same (1 AU) but the mass of the Sun changes, we rearrange this equation to give us a constant that relates the period of revolution to the mass of the Sun:

T₁² m₁ = 4 ² / G R³ = T₂²m₂

Here T₂ (the new period of revolution) can be found where m₂ is the new mass of the Sun.

T₂² m₂ = T₁² m₁

We know that T₁ = 1 year when the mass (m₁) of the Sun is 1 Solar Mass, so we can write:

T₂²(10 Solar Masses) = (1 year)² (1 Solar Mass)

T₂² = 1 year ² (1 Solar Mass)/ (10 Solar Masses)

T₂² = 0.1 year²

Taking the square root of both sides gives us:

T₂ = 0.32 years

The period of revolution of Earth would be only about 1/3 as long as it is now (about 4 months), if the Sun had a mass 10 times larger than it actually does have.

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