The space shuttle releases a satellite into a circular orbit 710 km above the earth. how fast must the shuttle be moving (relative to earth) when the release occurs?

speed  v = 7.47 10³ m /s

Explanation:

The space shuttle has a circular orbit around the earth so it is subject to centripetal acceleration

    a = v² / R

Where the distance R is measured from the center of the earth

   R = y + Re

   R = 710 +6370

   R = 7080 km (1000m / 1km)

   R = 7.080 106 m

We write Newton's second law

    F = m a

    G m1M / R2 = m1 v² / R

    G M / R = v²

    v = √(GM / R)

    v = √ (6.6 10⁻¹¹ 5.98 10²⁴/ 7.08 10⁶

     v = √ (5.575 10⁷)

    v = 0.747 10⁴ m / s

    v = 7.47 10³ m /s