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.

5,640 s (94 minutes)

Explanation:

the tangential speed of the HST is given by

(1)

where

is the length of the orbit

r is the radius of the orbit

T is the orbital period

In our problem, we know the tangential speed: . The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:

So, we can re-arrange equation (1) to find the orbital period:

Dividing by 60, we get that this time corresponds to 94 minutes.

the answer is b

explanation:

it has to be b, about 97 percent.