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25 May, 00:10

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?

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  1. 25 May, 02:37
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    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
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