Respuesta :
A period of a satellite is the time taken by the satellite to travel round a
body.
The comparison between the periods [tex]T_B[/tex], and [tex]T_A[/tex] is [tex]\underline {T_B = \dfrac{\sqrt{2} }{4 } \cdot T_A}[/tex]
Reason:
The period, T, of a satellite is given as follows;
[tex]T = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \cdot M} }[/tex]
Volume of the planet A = [tex]\dfrac{4}{3} \cdot \pi \cdot r^3[/tex]
Mass of planet A, [tex]m_A[/tex] = [tex]\dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho[/tex]
Volume of the planet B = [tex]\dfrac{4}{3} \cdot \pi \cdot (2 \cdot r)^3 = \dfrac{32}{3} \cdot \pi \cdot r^3[/tex]
Mass of planet B, [tex]m_B[/tex] = [tex]\dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho[/tex]
Period of the satellite on planet A, [tex]T_A[/tex], is given as follows;
[tex]T_A = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{4}{3} \cdot \pi \times \rho} }[/tex]
Period of the satellite on planet B, [tex]T_B[/tex], is given as follows;
[tex]T_B = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{32}{3} \cdot \pi \times \rho} }[/tex]
Therefore, get;
[tex]\dfrac{T_A}{T_B} = \dfrac{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 4 \cdot \pi \times \rho} }}{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 32 \cdot \pi \times \rho} }} = \sqrt{\dfrac{32}{4} } = \sqrt{8} = 2 \cdot \sqrt{2}[/tex]
Therefore, [tex]T_A[/tex] = (2·√2)·[tex]T_B[/tex]
[tex]T_B = \dfrac{T_A}{2 \cdot \sqrt{2} } = \dfrac{\sqrt{2} \cdot T_A}{4 }[/tex]
The comparison between [tex]T_A[/tex] and [tex]T_B[/tex] is therefore;
- [tex]\underline {T_B = \dfrac{\sqrt{2} }{4 } \cdot T_A}[/tex]
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https://brainly.com/question/1448749
