Exemplo de Physics, 7.1 Kepler's Laws of Planetary Motion

Kepler’s Third Law

Problema e solução em inglês, como no livro original.

Given that the moon orbits Earth each 27.3 days and that it is an average distance of $3.84 \times 10^{8} m$ from the center of Earth, calculate the period of an artificial satellite orbiting at an average altitude of 1,500 km above Earth’s surface.

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

Estratégia
The period, or time for one orbit, is related to the radius of the orbit by Kepler’s third law, given in mathematical form by $\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$ . Let us use the subscript 1 for the moon and the subscript 2 for the satellite. We are asked to find T₂. The given information tells us that the orbital radius of the moon is $r_{1} = 3.84 \times 10^{8} m$ , and that the period of the moon is $T_{1} = 27.3 days$ . The height of the artificial satellite above Earth’s surface is given, so to get the distance r₂ from the center of Earth we must add the height to the radius of Earth (6380 km). This gives $r_{2} = 1500 km + 6380 km = 7880 km$ . Now all quantities are known, so T₂ can be found.
Solução
To solve for T₂, we cross-multiply and take the square root, yielding

$\begin{array}{l} T_{2}^{2} = T_{1}^{2} {(\frac{r_{2}}{r_{1}})}^{3}; T_{2} = T_{1} {(\frac{r_{2}}{r_{1}})}^{\frac{3}{2}} \\ \begin{matrix} T_{2} = (27.3 d) (\frac{24.0 h}{d}) {(\frac{7880 km}{3.84 \times 10^{5} km})}^{\frac{3}{2}} = 1.93 h . \end{matrix} \end{array}$
7.3

0 de 2 passos abertos

Como foi?

Estratégia

Solução

Estratégia

Solução