Question

1. IF a small planet were discovered whose orbital period was twice that of Earth, how many time farther from the Sun would this planet be?
2. Determine Kepler's constant for any satellite of Earth.

Answer 1

Answer on Question #61300 - Physics - Mechanics | Relativity

Question:

1. IF a small planet were discovered whose orbital period was twice that of Earth, how many time farther from the Sun would this planet be?

2. Determine Kepler's constant for any satellite of Earth.

Solution:

1.

The third Kepler's law is: $\frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3}$

$$\frac {T _ {1} ^ {2}}{T _ {2} ^ {2}} = \frac {a _ {1} ^ {3}}{a _ {2} ^ {3}} \Rightarrow \frac {a _ {1}}{a _ {2}} = \sqrt [ 3 ]{\frac {T _ {1} ^ {2}}{T _ {2} ^ {2}}},$$

where $a_2$ is the radius of Earth's orbit and $a_1$ is the radius of planet's orbit, $T_2$ and $T_1$ are orbital periods of Earth and planet.

$$\frac {a _ {1}}{a _ {2}} = \sqrt [ 3 ]{\frac {T _ {1} ^ {2}}{T _ {2} ^ {2}}} = \sqrt [ 3 ]{\frac {(2 T _ {2}) ^ {2}}{T _ {2} ^ {2}}} = 4 ^ {\frac {1}{3}} \approx 1. 5 8 7 4 \dots$$

2.

Kepler's constant is: $K = \frac{a^3}{T^2} = \frac{GM}{4\pi^2}$

For Earth: $K = \frac{GM}{4\pi^2} = \frac{6.67 \cdot 10^{-11} \cdot 6 \cdot 10^{24}}{4\pi^2} \approx 1.014 \cdot 10^{13}$ ( $\frac{m^3}{s^2}$ )

Answer:

1) $\frac{a_1}{a_2} \approx 1.5874 \ldots$

2) $K \approx 1.014 \cdot 10^{13}$

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