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)

5069.04 seconds

Explanation:

The parameter we are looking for is called the Orbital period of the Hubble Space Telescope.

It is given as:

where r = radius of orbit of Hubble Space Telescope

G = gravitational constant =

M = Mass of earth

We are given that:

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

r =

M =

Therefore, T will be:

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

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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