F. If the shuttle's period is synchronized with that of Earth's rotation, what is the height of the shuttle? (1 day = 8.64x104s, ME = 6x1024kg, RE = 6.4x106m)

1.324 × 10⁷ m

Explanation:

The centripetal acceleration, a at that height above the earth equal the acceleration due to gravity, g' at that height, h.

Let R be the radius of the orbit where R = RE + h, RE = radius of earth = 6.4 × 10⁶ m.

We know a = Rω² and g' = GME/R² where ω = angular speed = 2π/T where T = period of rotation = 1 day = 8.64 × 10⁴s (since the shuttle's period is synchronized with that of the Earth's rotation), G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², ME = mass of earth = 6 × 10²⁴ kg. Since a = g', we have

Rω² = GME/R²

R(2π/T)² = GME/R²

R³ = GME(T/2π)²

R = ∛(GME)(T/2π)²

RE + h = ∛(GMET²/4π²)

h = ∛(GMET²/4π²) - RE

substituting the values of the variables, we have

h = ∛(6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg × (8.64 × 10⁴s)²/4π²) - 6.4 × 10⁶ m

h = ∛(2,987,477 × 10²⁰/4π² Nm²s²/kg) - 6.4 × 10⁶ m

h = ∛75.67 × 10²⁰ m³ - 6.4 × 10⁶ m

h = ∛(7567 × 10¹⁸ m³) - 6.4 × 10⁶ m

h = 19.64 × 10⁶ m - 6.4 × 10⁶ m

h = 13.24 × 10⁶ m

h = 1.324 × 10⁷ m

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