The equation that governs the circular orbit of a satellite is T = 2(pi)√(r3 / Gme) where me is the mass of the earth, which is approximately 6 x 1024kg. Assuming a circular orbit, to achieve an orbit at 20,000km, what would the speed need to be?

Respuesta :

Answer:

[tex]v = 4.47 \times 10^3 m/s[/tex]

Explanation:

As we know that time period is given by the equation

[tex]T = 2\pi \sqrt{\frac{r^3}{GM}}[/tex]

so we know that

[tex]r = 20,000 km = 2\times 10^7 m[/tex]

[tex]M = 6 \times 10^{24} kg[/tex]

here we have

[tex]T = 2\pi\sqrt{\frac{(2 \times 10^7)^3}{(6.67 \times 10^{-11})(6 \times 10^{24})}}[/tex]

[tex]T = 2.81 \times 10^4 s[/tex]

now in order to find the speed we can say

[tex]speed = \frac{distance}{time}[/tex]

[tex]speed = \frac{2\pi r}{T}[/tex]

[tex]v = \frac{2\pi(2 \times 10^7)}{2.81 \times 10^4}[/tex]

[tex]v = 4.47 \times 10^3 m/s[/tex]

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