Question

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?

Answer 1

speed v = 7.47 10³ m /s

Step-by-step 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