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Otrada [13]
3 years ago
10

Find an exact expression for the orbital period T. Hint: Each planet feels two forces.

Mathematics
1 answer:
Otrada [13]3 years ago
6 0

Answer:

T^2 = (4π^2/GM)a^3

Step-by-step explanation:

The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit” That’s Kepler’s third law. In other words, if you square the ‘year’ of each planet, and divide it by the cube of its distance to the Sun, you get the same number, for all planets.

(The other two are “the orbit of each planet is an ellipse with the Sun at a focus”, and “a line between a planet and the Sun sweeps out equal areas in equal times”.)

Copernicus, Kepler, and Newton dealt a one-two-three knockout blow to the idea – thousands of years old – that the Sun (and planets) moved around the Earth. Copernicus put the Sun at the center, Kepler modified Copernicus’ circular motions (and provided a simple, quantitative description of the actual motion), and Newton explained how it all worked (gravity).

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<u>Given functions</u>:

\begin{cases}p(t)=6t \\ A(p)=\pi p^2 \end{cases}

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\begin{aligned}A(p) & = \pi p^2\\\\ \implies A[p(t)] & = \pi [p(t)]^2\\& = \pi (6t)^2\\& = \pi 6^2 t^2\\& = 36\pi t^2\end{aligned}

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