Chapter 15, Problem 1GST

a)

To determine

The period (the time it takes to orbit the Sun) of an imaginary planet that orbits the Sun at a distance of 10 AU.

Answer to Problem 1GST

The period of an imaginary planet that orbits the Sun at a distance of 10 AU is 31.6 years.

Explanation of Solution

The square root of the orbital period of a planet is equal to the cube root of its mean solar distance.

Kepler’s third law states that “the planet’s orbital period squared is equal to its mean solar distance cubed”. Thus, the solar distances of the planets can be calculated if their orbital periods are known.

For example, the orbital period of Mars is 1.88 years, and 1.88 squared equals 3.54. The cube root of 3.54 is 1.52, and that is the mean distance from Mars to the Sun, in astronomical units.

This can be expressed in the given formula:

$p^2=d^3$

Here, p denotes orbital period, and d denotes the mean distance from the Sun.

Substitute 1 AU in d.

$\begin{array}{l}{p}^2={\text{ 10}}^3\\ p^2=1000\\ p=\sqrt{1000}\\ p=31.6\text{ years}\end{array}$

Therefore, the orbital period of an imaginary planet is 31.6 years.

b)

To determine

The distance between the Sun and a planet with a period of 5 years.

Answer to Problem 1GST

The distance between the Sun and a planet with a period of 5 years is 2.92 AU.

Explanation of Solution

The square root of the orbital period of a planet is equal to the cube root of its mean solar distance.

Kepler’s third law states that “the planet’s orbital period squared is equal to its mean solar distance cubed”. Thus, the solar distances of the planets can be calculated when their periods of revolution are known.

For example, the orbital period of Mars is 1.88 years, and 1.88 squared equals 3.54. The cube root of 3.54 is 1.52, and that is the mean distance from Mars to the Sun, in astronomical units.

This can be expressed as follows:

$p^2=d^3$

Here, p denotes orbital period, and a denotes the mean distance from the Sun.

The above expression can be written as follows:

$d=p^{\frac{2}{3}}$

Substitute 5 years for p in the above expression.

$\begin{array}{l}d={\left(5\right)}^{\frac{2}{3}}\\ d={\left(5\right)}^{0.667}\\ d=2.92\text{AU}\end{array}$

Therefore, the distance between the Sun and a planet is 2.92 AU.

c)

To determine

Whether one body or both the bodies will complete its orbit around the Sun in less time where one body is twice as large as the other orbiting the Sun at the same distance.

Answer to Problem 1GST

Kepler’s third law does not take mass into consideration. This law states that both bodies will complete their orbits around the sun in the same orbital period. Newton’s law states that a body that contains greater mass will orbit more gradually.

Explanation of Solution

Kepler’s third law states that “the planet’s orbital period squared is equal to its mean solar distance cubed”. Thus, the solar distances of the planets can be calculated when their periods of revolution are known.

For example, the orbital period of Mars is 1.88 years, and 1.88 squared equals 3.54. The cube root of 3.54 is 1.52, and that is the mean distance from Mars to the Sun, in astronomical units.

According to Kepler’s third law, both the bodies will complete their orbits around the Sun in the same period, as the law does not take mass into consideration. However, Newton’s law states that a body with greater mass will orbit more slowly.