Calculating orbital period the hubble space telescope (hst) orbits 569 km above earth’s surface. if hst has a tangential speed of 7,750 m/s, how long is hst’s orbital period? the radius of earth is 6.38 × 106 m.

Orbital period, T = 93.89 minutes

Explanation:

It is given that,

The tangential speed of the Hubble Space Telescope, v = 7750 m/s

The radius of earth,

We need to find the orbital period of the Hubble Space Telescope that orbits 569 km above the Earth's surface i.e. h = 569 km = 569000 m

The orbital speed is given by,

T is the orbital period

R = r + h

R = 6949000 m

T = 5633.78 s

or

T = 93.89 minutes

So, the orbital period of the Hubble Space Telescope is 93.89 minutes. Hence, this is the required solution.

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