Respuesta :
Answer:
A. 5521.86 secs
B. 7680.65 m/s
Explanation:
Parameters given:
Radius of orbit of the space station, R = 6380000 + 370000 = 6750000 m
Mass of earth = 5.97 * 10^24 kg
Gravitational constant, G = 6.67 * 10^(-11) Nm²/kg²
A. Orbital period, T, can be obtained using the formula:
T²/R³ = (4 * pi²) / (G * M)
T²/(6750000³) = (4 * 3.142²) / (6.67 * 10^(-11) * 5.97 * 10^24)
T² = (4 * 3.142² * 6750000³) / (6.67 * 10^(-11) * 5.97 * 10^24)
T² = 30490944.39
T = 5521.86 secs
B. Orbital velocity, v, can be obtained by using the formula:
v² = (G * M) / R
v² = (6.67 * 10^(-11) * 5.97 * 10^(-24)) / 6750000
v² = 58992384
v = 7680.65 m/s
(a) The period of the International Space Station's orbit is 5521.86 s.
(b) Â The speed of the International Space Station in its orbit is 7680.65 m/s.
Given data:
The altitude of Space station is, [tex]h = 370 \;\rm km=370 \times 10^{3} \;\rm m[/tex].
The mass of Earth is, [tex]m = 5.97 \times 10^{24} \;\rm kg[/tex].
The radius of Earth is, [tex]r = 6.38 \times 10^{6} \;\rm m[/tex].
The gravitational constant is, [tex]G = 6.67 \times 10^{-11} \;\rm Nm^{2}/kg^{2}[/tex].
(a)
The standard expression for the period of orbiting is given as,
[tex]T^{2} = \dfrac{4\pi^2 \times (h+r)^{3}}{G \times m}\\\\T^{2} = \dfrac{4\pi^2 \times (370000+6380000)^{3}}{6.67 \times 10^{-11} \times 5.97 \times 10^{24}} \\\\\\T = \sqrt{30490944.39}\\\\T = 5521.86 \;\rm s[/tex]
Thus, the period of the International Space Station's orbit is 5521.86 s.
(B)
Now, the speed of the International Space Station in its orbit is,
[tex]v^{2}=\dfrac{Gm}{r}\\\\v^{2}=\dfrac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6750000}\\\\\\v = \sqrt{58992384}\\\\v = 7680.65 \;\rm m/s[/tex]
Thus, Â the speed of the International Space Station in its orbit is 7680.65 m/s.
Learn more about the orbiting speed here:
https://brainly.com/question/14755913
