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.

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.