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

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this