Answer:
5069.04 seconds
Explanation:
The parameter we are looking for is called the Orbital period of the Hubble Space Telescope.
It is given as:
[tex]T = \sqrt{\frac{4\pi^2r^3 }{GM} }[/tex]
where r = radius of orbit of Hubble Space Telescope
G = gravitational constant = [tex]6.67408 * 10^{-11} m^3 kg^{-1} s^{-2}[/tex]
M = Mass of earth
We are given that:
r = radius of the earth + distance of HST from earth
r = [tex]6.38 * 10^6 + 598 = 6380598 m[/tex]
M = [tex]5.98 * 10^{24} kg[/tex]
Therefore, T will be:
[tex]T = \sqrt{\frac{4*\pi^2 * 6380598^3 }{6.67408 * 10^{-11} * 5.98 * 10^{24}}}[/tex]
[tex]T = 5069.04 secs[/tex]
The orbital period of the Hubble Space Telescope is 5069.04 seconds.
Answer:
88.25822 Minutes
Explanation:
[tex]T=\frac{S}{V}[/tex] Relates T, time period To S, circumference at height of 598m and V velocity.
S = circumference = [tex]2\pi (h+r)=2\pi (598m+3.38*10^6)=40033930.94meters.[/tex]
Substituting all this in our equation gives.
[tex]T = \frac{40033930.94m}{7.56*10^3m/s} =5295.49351s=88.25822minutes.[/tex]
