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

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.