Question

If the Hubble Space Telescope is 598 m above the surface of the Earth and is traveling at 7.56 x 103 m/s, how long does it take the Hubble Space Telescope to orbit the earth? (Radius of the earth RE= 6.38 x 106 m, Mass of the earth ME = 5.98 x 1024 kg)

Answer 1

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:

$T = \sqrt{\frac{4\pi^2r^3 }{GM} }$

where r = radius of orbit of Hubble Space Telescope

G = gravitational constant = $6.67408 * 10^{-11} m^3 kg^{-1} s^{-2}$

M = Mass of earth

We are given that:

r = radius of the earth + distance of HST from earth

r = $6.38 * 10^6 + 598 = 6380598 m$

M = $5.98 * 10^{24} kg$

Therefore, T will be:

$T = \sqrt{\frac{4*\pi^2 * 6380598^3 }{6.67408 * 10^{-11} * 5.98 * 10^{24}}}$

$T = 5069.04 secs$

The orbital period of the Hubble Space Telescope is 5069.04 seconds.

Answer 2

Answer:

88.25822 Minutes

Explanation:

$T=\frac{S}{V}$  Relates T, time period To S, circumference at height of 598m and V velocity.

S = circumference = $2\pi (h+r)=2\pi (598m+3.38*10^6)=40033930.94meters.$

Substituting all this in our equation gives.

$T = \frac{40033930.94m}{7.56*10^3m/s} =5295.49351s=88.25822minutes.$